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19.2: Environmental Effects on NMR Spectra

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Instrumental_Analysis_(LibreTexts)/19%3A_Nuclear_Magnetic_Resonance_Spectroscopy/19.02%3A_Environmental_Effects_on_NMR_Spectra

In the previous section we showed that there is a relationship between the Larmor frequency for a nucleus, \(\nu\), its magnetogyric ratio, \(\gamma\), and the primary applied magnetic field strength, \(B_0\)\[\nu = \frac{\gamma B_0}{2 \pi} \label{env1} \]and we used this equation to show that the Larmor frequency for the 1H nucleus in a magnetic field of \(B_0 = 11.74 \text{ T}\) is 500 MHz. If this is the only thing that determines the frequency where absorption takes place, then all compounds that contain hydrogens will yield a 1H NMR spectrum with a single peak at the same frequency. If all spectra are identical, then NMR provides little in the way of useful information. The NMR spectrum for propane (CH3–CH2—CH3) in shows two clusters of peaks that give us confidence in the utility of NMR. In this case, it seems likely that the cluster of peaks between 250 Hz and 300 Hz, which have a greater total intensity, are for the six hydrogens in the two methyl groups (–CH3) and that the cluster of peaks around 400 Hz are due to the methylene group (–CH2–). In this section we will consider why the location of a nucleus within a molecule—what we call its environment—might affect the frequency at which it absorbs and why a particular absorption line might appear as a cluster of individual peaks instead of as a single peak.Before we consider how a nucleus's environment affects the frequencies at which it absorbs, let's take a moment to become familiar with the scale used to plot a NMR spectrum. The label on the x-axis of the NMR spectrum for propane in raises several questions that we will answer here.From Equation \ref{env1} we see that the frequency at which a nucleus absorbs is a function of the magnet's field strength, \(B_0\). This means the frequency of a peak in an NMR spectrum depends on the value of \(B_0\). One complication is that instruments with identical nominal values for \(B_0\) likely will have slightly different actual values, which leads to small variations in the frequency at which a particular hydrogen absorbs on different instruments. We can overcome this problem by referencing a hydrogen's measured frequency to a reference compound that is set to a frequency of 0. The difference between the two frequencies should be the same on different instruments. For example, the most intense peak in the NMR spectrum for propane, , has a frequency of 269.57 Hz when measured on an NMR with a nominal field strength of 300 MHz, which means that its frequency is 269.57 Hz greater than the reference, which is identified as TMS.The reference compound is tetramethylsilane, TMS, which has the chemical formula of (CH3)4Si in which four methyl groups are in a tetrahedral arrangement about the central silicon. TMS has the advantage of having all of its hydrogens in the same environment, which yields a single peak. Its hydrogen atoms also absorbs at a low frequency that is well removed from the frequency at which most other hydrogen atoms absorb, which makes it easy to identify its peak in the NMR spectrum.An additional complication with the spectrum in is that the frequency at which a particular hydrogen absorbs is different when using a 60 MHz NMR than it is when using a 300 MHz NMR, a consequence of Equation \ref{env1}. To create a single scale that is independent of \(B_0\) we divide the peak's frequency, relative to TMS, by \(B_0\), expressing both in Hz, and then report the result on a part-per-million scale by multiplying by 106. For example, the most intense peak in the NMR spectrum for propane, , has a frequency of 269.57 Hz; the NMR on which the spectrum was recorded had a field strength of 300 MHz. On a parts-per-million scale, which we identify as delta, \(\delta\), the peak appears at\[\delta = \frac{269.57 \text{ Hz}}{300 \times 10^6 \text{ Hz}} \times 10^6 = 0.899 \text{ ppm} \nonumber \]If we record the spectrum of propane on a 60 MHz instrument, then we expect that this peak to appear at 0.899 ppm, or a frequency of\[\nu = \frac{0.899 \text{ ppm} \times (60 \times 10^6 \text{ Hz}}{10^6} = 53.9 \text{ Hz} \nonumber \]relative to TMS.Most hydrogens have values of \(\delta\) between 1 and 13. shows the 1H NMR for propane using a ppm scale. The right side of the ppm scale is described as being upfield, with absorption occurring at a lower frequency, and with a smaller difference in energy, \(\Delta E\), between the ground state and the excited state. The left side of the ppm scale is described as being downfield, with absorption occurring at a higher frequency, and with a greater difference in energy, \(\Delta E\), between the ground state and the excited state.The NMR spectrum for propane in shows two important features: the peaks for the two types of hydrogen in propane are shifted downfield relative to the reference and the methylene hydrogens are shifted further downfield than the methyl hydrogens. Both groups appear as clusters of peaks instead of as single peaks. In this section we consider the source of these two phenomena.In the presence of a magnetic field, the electrons in a molecule circulate, generating a secondary magnetic field, \(B_e\), that usually, but not always, opposes the primary applied magnetic field, \(B_\text{appl}\). The result is that the nucleus is partially shielded by the electrons such that the field it experiences, \(B_0\), usually is smaller than the applied field and\[B_0 = B_\text{appl} - B_e \label{env2} \]The greater the shielding, the smaller the value of \(B_0\) and the further to the right the peak appears in the NMR spectrum. For example, in the NMR spectrum for propane in the cluster of peaks for the –CH3 hydrogens centered at 0.899 ppm shows greater shielding than the cluster of peaks for the –CH2– hydogens that is centered at 1.337 ppm.Chemical shifts are useful for determining structural information for molecules. A few examples are listed in the following table and more extensive tables here. Note that the range of chemical shifts for the methyl and the methylene groups encompass the values for propane in .Chemical shifts are the result of shielding from the magnetic field associated with a molecule's circulating electrons. The splitting of a peak into a multiplet of peaks is the result of the shielding of one nucleus by the nuclei on adjacent atoms, and is called spin-spin coupling. Consider the NMR for propane in , which consists of two clusters of peaks. The six hydrogens in the two methyl groups are sufficiently close to the two hydrogens in the methylene group that the spins of the methylene hydrogens can affect the frequency at which the methyl hydrogens absorb. shows how this works. Each of the two methylene hydrogens has a spin and those spins can both be aligned with the magnetic field, \(B_0\), both be aligned against \(B_0\), or two configurations in which one is aligned with \(B_0\) and one is aligned against \(B_0\), as seen by the arrows. When the two spins are aligned with \(B_0\), the frequency at which the methyl hydrogens absorb is shifted downfield, and when the two spins are aligned against \(B_0\), the frequency at which the methyl hydrogens absorb is shifted upfield; in the remaining two cases, there is no change in the ferquency at with the methyl hydrogens absorb. The result, as seen in is a triplet of peaks in a 1:2:1 intensity ratio.The analysis for the effect of the six methyl hydrogens on the two methylene hydrogens is a bit more complex, but works in the same way. , for example, shows that there are 15 ways to arrange the spins of the six methyl hydrogens such that two are spin down and four are spin up. show the resulting NMR spectrum, which is a set of seven peaks in a 1:6:15:20:15:6:1 intensity ratio. provides the splitting pattern observed for nuclei with \(I = +1/2\), such as 1H. The pattern is defined by the coefficients of a binomial distribution—asking how many different ways you can get X outcomes in Y attempts is at the heart of a binomial distribution—this is easy to represent using Pascal's triangle—which shows us that for six equivalent nuclei we expect to find seven peaks with relative peak areas (or other measure of the signal) of 1:6:15:20:15:6:1. Note that the first and the last entry in any row is 1 and that all other entries in a row, as illustrated for the third entry in the seventh row, are the sum of the two entries in the row immediately above. The pattern also is know as the \(N+1\) rule as the \(N\) equivalent hydrogens will split the peak for an adjacent hydrogen into \(N + 1\) peaks. compares the experimental NMR for propane with its simulatd spectrum based on spin-spin splitting and the 2:6 ratio of methylene hydrogens relative to methyl hydrogens. The overall agreement between the two spectra is pretty good. The splitting of the individual peaks is designated by the coupling constant, J, which is shown in for both the experimental and the calculated spectra. Note that the coupling constant is the same whether we are considering the effect of the methyl hydrogens on the methylene hydrogens, or the effect of the methylene hydrogens on the methyl hydrogens. Values of the coupling constant become smaller the greater the distance between the nuclei.The treatment of spin-spin coupling above works well if the difference in the chemical shifts for the two nuclei is significantly greater than the magnitude of their coupling constant. When this is not true, the splitting patterns can become much more complex and often are difficult to interpret. There are a variety of to simplify spectra, one of which, decoupling, is outlined in . The original spectrum (top) shows two doublets, suggesting that we have two individual nuclei that are coupled to each other. If we irradiate the nucleus on the right at its frequency, we can saturate its ground and excited states such that it ceases to absorb. As a result, the nucleus on the left no longer shows evidence of spin-spin coupling to the nucleus on the right (middle) and appears as a singlet. When we turn off the decoupler (bottom) the spin-spin coupling between the two nuclei returns more quickly than relaxation returns the signal for the nucleus on the right.This page titled 19.2: Environmental Effects on NMR Spectra is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

19.3: NMR Spectrometers

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Earlier in this chapter, we noted that there are two basic experimental designs for recording a NMR spectrum. One is a continuous-wave instrument in which the range of frequencies over which the nucleus of interest absorb is scanned linearly, exciting the different nuclei sequentially. Most instruments, however, use pulses of RF radiation to excite all nuclei at the same time and then use a Fourier transform to recover the signals from the individual nuclei. Our attention in this chapter is limited to instruments for FT-NMR. includes a photograph of 400 MHz NMR and a cut-away illustration of the instrument; together, these show the key components of a FT-NMR: a magnet that provides the applied magnetic field, \(B_0\), a nucleus-dependent probe that provides the radio-frequency signal that yields the magnetic field, \(B_1\), and a way to insert the sample into the instrument. The NMR in the photograph also is equipped with a sample changer that allows the user to load 30 or more samples that are analyzed sequentially.The NMR in is described as having a frequency, \(\nu\), of 400 MHz. The relationship between frequency and the magnet's field strength, \(B_0\), is given by the equation\[\nu = \frac{\gamma B_0}{2 \pi} \label{compent1} \]where \(\gamma\) is the magnetogyric ratio for the nucleus. A NMR's frequency is defined in terms of a nucleus of 1H; thus, a 400 MHz NMR has a magnet with a field strength of\[B_0 = \frac{(2 \pi) \times \nu}{\gamma} = \frac{(2 \pi) \times (400 \times 10^6 \text{ s}^{-1})}{2.86 \times 10^{8} \text{ rad T}^{-1} \text{ s}^{-1}} = 9.4 \text{ T} \nonumber \]Early instruments used a permanent magnet and were limited to field strengths of 0.7, 1.4, and 2.1 T, or 30, 60, and 90 MHz. As higher frequencies provide for greater sensitivity and resolution, modern instruments use a tightly wrapped coil of wire—typically a niobium/tin alloy or a niobium/titanium wire—that becomes superconducting when cooled to the temperature of liquid He (4.2 K). The result is a magnetic field strength of as much as 21 T or 900 MHz for 1H NMR. The magnetic coil is held within a reservoir of liquid He, which, itself, is held within a reservoir of liquid N2.To be useful, the magnetic field must remain stable—that is, it must not drift—and it must be homogeneous throughout the sample. These are accomplished by using a reference to lock the magnetic field in place and by shimming.Samples for NMR are prepared using a solvent in which the protons are replaced with deuterium. For example, instead of using chloroform, CHCl3, as a solvent, we use deuterated chloroform, CDCl3, where D is equivalent to 2H. This has the benefit of providing a solvent that will not contribute to the signals in the NMR spectrum. It also has the benefit that 2H has a spin of \(I = 1\), and a corresponding Larmor frequency. By monitoring the frequency at which 2H absorbs, the instrument can use a feed-back loop to maintain its value by adjusting the magnet's field strength.A magnetic field that is not homogeneous is like a table with four legs, one of which is just a bit shorter than the others. To balance the table, we place a small wedge, or shim, under the shorter leg. When a magnetic field is not homogeneous, small, localized adjustments are made to the magnetic field using a set of shimming coils arranged around the sample. Shimming can be accomplished by the operator by monitoring the quality of the signal for a particular nucleus, however, most instruments use an algorithm that allows the instrument to shim itself.The center of the instrument, which runs from the sample input at the top to the sample probe at the bottom, is open to the laboratory environment and is at room temperature. The sample is placed in a cylindrical tube ), that is made from thin-walled borosilicate glass and is 180 mm long and 5 mm in diameter. The tube is then inserted into a teflon sleeve—called a spinner—as shown in , which is designed to both situate the sample at the proper depth within the sample probe, and to spin the sample about its long axis. This spinning is used to ensure that the sample averages out any inhomogeneities in the magnetic field not resovled by shimming.The sample probe contains the coils needed to excite the sample and to detect the NMR signal as the excited states undergo relaxation. shows two configurations for this; in both configurations, the same coil is used for both excitation and detection. In the design on the left, which uses a permanent magnet, the applied magnetic field, \(B_0\), is oriented horizontally across the sample's diameter and the radio frequency electromagnetic radiation and its field, \(B_1\) is oriented vertically using a spiral coil. In the design on the right, which is used with a superconducting magnet, the applied magnetic field, \(B_0\), is oriented vertically and the pulse of radio frequency electromagnetic radiation and its field, \(B_1\), is oriented horizontally using a saddle coil.In Chapter 19.1 we used the following figure to describe a pulse NMR experiment. Following a pulse that is applied for 1–10 µs, the free-induction decay, FID, is recorded for a period of time that may range from as little as 0.1 seconds to as long as 10 seconds, depending on the nucleus being probed.The FID is an analog signal in the form of a voltage, typically in the µV range. This analog signal must be converted into a digital signal for data processing, which is called an analog-to-digital conversion, ADC. Two important considerations are needed here: how to ensure that the signal—more specifically, the location of the peaks in the NMR spectrum—is not distorted, and how to accomplish the ADC when the frequencies are on the order of hundreds of MHz.An analog-to-digital converter maps the signal onto a limited number of possible values—expressed in binary notation—and are characterized by the number of available bits. A 2-bit ADC convertor, for example, is limited to \(2^2 = 4\) possible binary values of 00, 01, 10, and 11 that correspond to the decimal numbers 1, 2, 3, and 4. Having only four possible values, of course, would distort the FID pattern in from a smoothly varying oscillating signal into a series of steps. Using an ADC convertor with 16 bits allows for 65,536 unique digital values, a significant improvement. Another form of distortion occurs if we do not sample the FID with sufficient frequency. Consider, for example, the simple sine wave in that is shown as a solid line. If we sample this signal only five times over a period of less than four complete cycles, as shown by the five equally-spaced dots in , then the apparent signal is that shown by the dadshed line.According to the Nyquist theorem, to determine accurately the frequency of a periodic signal, we must sample the signal at least twice during each cycle or period. Given a sampling rate of \(\Delta\), the following equation\[\Delta = \frac{1}{2 \nu_\text{max}} \label{adc1} \]defines the highest frequency, \(\nu_\text{max}\), that we can monitor accurately. A sampling rate of six samples per period is more than sufficient to reproduce the real signal in .A peak with a frequency that is greater than \(\nu_\text{max}\) is not absent from the spectrum; instead, it simply appears at a different location. For example, suppose we can monitor accurately any frequency within the window shown in and that we only measure frequencies within this window. A peak with a frequency that is greater than what we can measure accurately by \(\Delta \nu\) appears at an apparent frequency that is \(\Delta \nu\) greater than the frequency window's lower limit. This is called folding.The instrument in is a 400 MHz NMR. This is a range of frequencies that is too large for an analog-to-digital convertor to handle with accuracy. The frequency window of interest to us, however, is typically 10 ppm for 1H NMR (see Chapter 19.2 to review the NMR scale). For a 400 MHz NMR this corresponds to just 4000 Hz, with the useful range running from 400.000 MHz to 400.004 MHz. Subtracting the instrument's frequency of 400 MHz from the signal's frequency limits the latter to the range of 0–4000 Hz, a range that is easy for an ADC to handle.Integrating to determine the area under the peaks provides a way to gain some quantitative information about the sample. shows the integration of the NMR of propane first seen in Chapter 19.2. Integration of the peak for the two methyl groups gives a result of 1766 and integration of the peak for the methylene group gives a result of 710. The ratio of the two is\[\frac{1766}{710} = 2.5 \nonumber \]which is somewhat smaller than the expected 3:1 ratio.This page titled 19.3: NMR Spectrometers is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

19.4: Applications of Proton NMR

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Proton (1H) NMR finds use for both qualitative analyses and quantitative analyses; in this section we briefly consider each of these areas.Proton NMR is an essential tool for the qualitative analysis of organic, inorganic, and biochemical compounds. provides a simple example that shows the relationship between structure and a 1H NMR's peaks. The spectra in this figure are for a set of four simple organic molecules, each of which has a chain of three carbons and an oxygen: 1-propanol, CH3CH2CH2OH, 2-propanol, CH3CH(OH)CH3, propanal, CH3CH2CHO, and propanoic acid, CH3CH2COOH. The first two of these molecules are alcohols, the third is an aldehyde, and the last is an acid. The main spectrum runs from 0–14 ppm, with insets showing the spectra over a narrower range of 0–5 ppm.Each of these molecules has a terminal –CH3 group that is the most upfield peak in its spectrum, appearing between 0.94 – 1.20 ppm. Each of these molecules has a hydrogen that either is bonded to an oxygen or a hydrogen bonded to the same carbon as the oxygen. The hydrogens in the –OH groups of the two alcohols have similar shifts of 2.16 ppm and 2.26 ppm, but the aldehyde hydrogen in the –CHO group and the acid hydrogen in –COOH are shifted further downfield appearing at 9.793 ppm and 11.73 ppm, respectively. The hydrogens in the two –CH2– groups of 1-propanol have very different shifts, with the one adjacent to the –OH group appearing more downfield at 3.582 ppm than the one next to the –CH3 group at 1.57 ppm. Not surprisingly, the –CH– hydrogen in 2-proponal, which is adjacent to the –OH group appears at 4.008 ppm.Comparisons such of this make it possible to build tables of chemical shifts—see Table 19.2.1 in Chapter 19.2 for an example—that can help in determining the identify of the molecule giving rise to a particular NMR spectrum. As this receives extensive coverage in other courses, particularly courses in organic chemistry, we will not provide a more extensive coverage here.A quantitative analysis requires a method of standardization, which for NMR usually makes use of an internal standard. A good internal standard should have high purity and should have a relatively simple NMR spectrum with peaks that do not overlap with the analyte or other species present in the sample. If we are interested in only the relative concentrations of the analyte and the internal standard, then we can use the following formula\[\frac{M_a}{M_{is}} = \frac{I_a}{I_{is}} \times \frac{N_{is}}{N_a} \label{quant1} \]where \(M\) is the molar concentration of the analyte or internal standard, \(I\) is the intensity of the NMR peak for the analyte or internal standard, and \(N\) is the number of nuclei giving rise to the NMR peak for the analyte and the internal standard. Even if we don't know the exact concentration of the internal standard, if we know that its concentration is the same in all samples, then we can determine the relative concentration of analyte in a collection of samples.If we are interested in determining the absolute concentration of analyte in a sample, then we must know the absolute concentration of the internal standared; when true, then Equation \ref{quant1} becomes\[M_a = \frac{I_a}{I_{is}} \times \frac{N_{is}}{N_a} \times M_{is} \label{quant2} \]Determining the purity of an analyte, \(P_a\), in a sample, we can use the equation\[P_a = \frac{I_a}{I_{is}} \times \frac{N_{is}}{N_a} \times \frac{M_a}{M_{is}} \times \frac{W_{is}}{W_a} \times P_{is} \label{quant3} \]where \(W\) is the weight of the internal standard or the sample that contains our analyte.This page titled 19.4: Applications of Proton NMR is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

19.5: Carbon-13 NMR

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The relatively slow development of instrumentation for 13C NMR spectra is the result of its limited sensitivity compared to 1H NMR. This difference in sensitivity is due to two key differences between the nuclei 1H and 13C: their relative abundances and their relative magnetogyric ratios. While 1H comprises 99% of all hydrogen, 13C accounts for just 1% of all carbon. The strength of an NMR signal also depends on the difference in energy, \(\Delta E\), between the ground state and the excited state, which is a function of the magnetogyric ratio, \(\gamma\)\[\Delta E = h \nu = \frac{\gamma B_0}{2 \pi} \label{carbon1} \]The greater the difference in energy, the greater the difference in the population between the ground and the excited states, and the greater the signal. The magnetogyric ratio, \(\gamma\), for 1H is \(4 \times\) greater than that for 13C. As a result of these two factors, 1H NMR is approximately \(6400 \times\) more senstitive than 13C. The development of magnets with higher field strengths and the capabilities of signal averaging (see Chapter 5 on signals and noise) when using Fourier transforms to gather and analyze data, make 13C feasible. shows 13C NMR spectrum for three related molecules: p-nitrophenol, o-nitrophenol, and m-nitrophenol. There are three things to make note of from this figure. First, each spectrum consists of a set of peaks, each of which is a singlet, suggesting that no spin-spin coupling is taking place. Second, the number of peaks in each spectrum is the same as the number of unique types of carbon—four unique carbons for p-nitrophenol and six each for m-nitrophenol and o-nitrophenol—which suggests that chemical shifts in 13C provide useful information about the environment of the carbon atoms and, therefore, the molecule's structure. And, third, unlike 1H, there is no relationship between the intensity of a 13C peak and the number of carbon atoms. This is particularly evident when comparing the intensity of the peaks for the carbon bonded to the –NO2 group and the carbon bonded to the –OH group, which are significantly less intense than the peaks for other carbons. We will consider each of these observations in the remainder of this section.In 13C NMR there is no coupling between adjacent carbon atoms because it is unlikely that both are 13C, the only isotope of carbon that is NMR active (the odds that two adjacent carbons are both 13C is \(0.01 \times 0.01\), or \(0.0001\) or \(0.010\%\)). Coupling does take place between 13C and 1H when the hydrogen atoms are attached to the carbon atom. Such coupling follows the same N+1 rule as in 1H NMR; thus, a quartenary carbon (R4C) appears as a singlet, a methine carbon (R3CH) appears as a doublet, a methylene carbon (R2CH2) appears as a triplet, and a methyl carbon (RCH3) appears as a quartet. Even with the extensive range of ppm values over which 13C peaks appear—chemical shifts for 13C spectra run from 250 – 0 ppm instead of 14 – 0 ppm for 1H spectra—a compound with many different types of carbon atoms, each with 1 – 3 hydrogen atoms results in a complex spectrum. For this reason, 13C NMR spectra are acquired in a way that prevents coupling between 13C and 1H. This is called proton decoupling.The most common method of proton decoupling is to use a second RF generator to irradiate the sample with a broad-band of RF signals that spans the range of frequencies for the protons. As described earlier in Section 19.3, the effect is to saturate the proton's ground and excited states, which prevents the protons from absorbing energy and from coupling with each other and with the carbons atoms. The 13C spectra in are examples of decoupled spectra.Just as with 1H NMR spectra, tables of chemical shifts for 13C peaks aids in determining a molecules structure. Table \(\PageIndex{1}\) provides ranges of chemical shifts for different types of carbon atom. A set of tables is available here.The most intense peak in the 13C NMR spectrum for m-nitrophenol (see above) is the carbon in the benzene ring labeled as position 4, with an intensity of 1000 (the intensity scale is normalized here to a maximum value of 1000, but for this section, let's take it as an absolute value). Because the spectrum was acquired with proton decoupling turned on, the peak for this carbon appears as a singlet. If we turn proton decoupling off, then we expect the peak to appear as a doublet as this carbon has one hydrogen attached to it. We might reasonably expect to find that each peak has an intensity of 500, giving a total intensity of 1000. The actual intensities of the peaks for this carbon, however, are smaller than expected. Put another way, when we turn proton decoupling on, the intensity of a 13C line increases more than expected and the more hydrogens, the greater the effect. This is called nuclear overhauser enhancement (NOE).NOE is the result of the relative populations of the ground and excited states. The technical details are more than we will consider here, but the extent of the total enhancement of the peak intensities is proportional to the ratio of the magnetogyric ratios of the irradiated nucleus (1H) and the observed nucleus (13C), which for a 1H decoupled 13C NMR results in a total enhancement of the intensity of approximately 200%. As magnetogyric ratios can be negative, as is the case for 15N, a decoupled spectrum can result in less intense peaks. One important consequence of NOE, is that integrated peak areas are not proportional to the number of identical carbon atoms, which is a loss of information.Although our focus in this chapter is on 1H and 13C NMR, other nuclei, such as 31P, 19F, and 15N are useful for the study of chemically and biochemically important molecules.This page titled 19.5: Carbon-13 NMR is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

19.6: Two-Dimensional Fourier Transform NMR

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The 1H and 13C spectra up to this point are shown in one dimension (1D), which is the frequency absorbed by the analyte's nuclei expressed in ppm. These spectra were acquired by applying a brief RF pulse to the sample, recording the resulting FID, and then using a Fourier transform to obtain the NMR spectrum. In addition to 1D experiments, there are a host of 2D experiments in which we apply a sequence of two or more pulses, recording the resulting FID after applying the last pulse. In this section we will consider one example of a 2D NMR experiment in some detail: 1H – 1H correlation spectroscopy, or 1H – 1H COSY. shows the basic experimental details for the COSY experiment. The pulse train is shown in (a) and consists of a first pulse that is followed by a delay time, \(t_1\), in which the nuclear spins are allowed to precess and relax. This is followed by the application of a second pulse and the measurement of the resulting FID during the acquisition period, \(t_2\), that consists of \(n_{t_2}\) individual data points. The COSY experiment consists of a sequence of \(n_{t_1}\) individual pulse trains, each with a different \(t_1\). The result (b) is a matrix with \(n_{t_1}\) rows and \(n_{t_2}\) columns. To process the data, each row in this matrix is Fourier transformed, the resulting \(n_{t_1} \times n_{t_2}\) matrix is transposed to a \(n_{t_2} \times n_{t_1}\) matrix and then Fourier transformed again to give (c) a \(n_{t_2} \times n_{t_2}\) matrix that shows the intensity of the signal for all possible combinations of applied frequencies. shows the 1H – 1H COSY spectrum for ethyl acetate. Instead of just annotating the two axes with numerical values of the frequencies, they are displayed by superimposing the 1D 1H NMR spectrum for ethyl acetate in ppm. The points that fall on the diagonal line are just the three frequencies where ethyl acetate has 1H peaks. Of more interest are the points that fall on either side of the diagonal line—these are called cross peaks—as these show pairs of hydrogens that are coupled (or correlated) to each other. The cross peaks in the COSY spectrum are symmetrical about the diagonal line and show the correlation between the hydrogens on the methyl carbon and and the methylene carbon that are adjacent to each other (the ethyl part of ethyl acetate). The remaining methyl group is not coupled to the other hydrogen's in the ethyl acetate, so there is no cross peak at the intersection of \(\delta = 2.038 \text{ ppm}\) and \(\delta = 1.260 \text{ ppm}\). The information about coupling from cross peaks assists in interpreting complex 1H NMR spectra.COSY is one example of a homonuclear 2D NMR experiment because it examines coupling between identical nuclei, such as 1H in . There are many other 2D NMR experiments, each of which uses a sequence of pulses—some that use two pulses, as in COSY, and others that use three or more pulses—and a data analysis algorithm. Table \(\PageIndex{1}\) provides details on some of these methods.This page titled 19.6: Two-Dimensional Fourier Transform NMR is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

2.1: Basic Terminology and Laws of Electricity

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Current, \(I\), is a movement of charge over time and is expressed in amperes, \(A\), where 1 ampere is equivalent to 1 coulomb/sec. In this section we review the convention used to describe currents in electrical circuits and review four laws of electricity.If we connect one end of a wire to the positive terminal of a battery and connect the other end to the negative terminal of the same battery, then electrons will move through the and a current will flow through the wire. The electrons move from the battery's negative terminal through the wire to the battery's positive terminal. The direction of the current, however, runs from the battery's positive terminal to the battery's negative terminal; that is, current is treated as if it is the movement of positive charge. This probably strikes you as odd, but it simply reflects the original understanding of current from a time before the electron was identified. shows the difference between these two ways of thinking about current.There are four basic laws of electricity that are important to us in this chapter: Ohm's law, Kirchhoff's laws, and the power law. Let's take a brief look at each.Ohm's law explains the relationship between current, \(I\), measured in amps (\(A\)), resistance, \(R\), measured in ohms (\(\Omega\)), and potential, \(V\), measured in volts (\(V\)), and is written as\[V = I \times R \label{ohm} \]The voltage is measured between any two points in a circuit using a voltmeter.The first of Kirchoff's two laws states that the sum of the currents at any point in a circuit must equal zero.\[ \sum{I} = 0 \label{kirch1} \]The second law states that the sum of the voltages in a closed loop must equal zero.\[ \sum{V} = 0 \label{kirch2} \]When a current passes through a resistor, the temperature of the resistor increases and power (energy per unit time) is lost. The amount of power lost, \(P\), is the product of current and voltage, with units of joules/sec\[P = I \times V \label{power1} \]or, substituting in Ohm's law (Equation \ref{ohm}), we can express power as\[P = I^2 \times R = \frac{V^2}{R} \label{power} \]An excellent resource for this section and other sections in this chapter is Principles of Electronic Instrumentation by A. James Diefenderfer and published by W. B. Saunders Company, 1972.This page titled 2.1: Basic Terminology and Laws of Electricity is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

2.2: Direct Current (DC) Circuits

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A direct current, which is the focus of this section, is one that flows in one direction. An alternating current, which is the focus of the next section, is one that periodically switches direction.There are two basic direct current circuits of importance to us: those with two or more resistors connected in a series, and those with two or more resistors arranged parallel to each other. Other direct current circuits can be understood in terms of these two basic circuits. shows an example of a simple DC circuit in which three resistors, with resistances of R1, R2, and R3, are connected in series to the two ends of battery that has a potential of V. A switch is included in the circuit that is used to close the loop and allow a current to flow from the battery's positive terminal to its negative terminal.Kirchoff's first law requires that the sum of the currents at any point in the circuit is zero. Consider point b. If the current that arrives from point a is \(I\), then the current that leaves b is \(-I\), where the sign tells us about the direction of the current with respect to the point. This requires that \(I_a - I_c = 0\), which means that the current has the same absolute value at all points in the circuit.Application of Kirchhoff's second law requires that the sum of the voltages in this circuit equal 0, which is true if the sum of the voltage across each of the three resistors is equal to the voltage of the battery. The voltage across each resistor is given by Ohm's law\[V = IR_1 + IR_2 + IR_3 \label{series1} \]If we divide both sides of Equation \ref{series1} by the current, then we have\[V = I \times (R_1 + R_2 + R_3) = IR_s \label{series2} \]where \(R_s\) is the circuit's effective resistance, which is the sum of the resistances of the individual resistors.One of the useful properties of this circuit is that the voltage drop across an individual resistor is proportional to that resistor's contribution to \(R_s\). Consider the points in labeled \(a\) and \(b\) that are on opposite sides of the first resistor in this series. The drop in voltage across this resister, \(V_{ab}\), is\[V_{ab} = I R_1 \label{series3} \]Dividing Equation \ref{series3} by Equation \ref{series1}\[\frac{V_{ab}}{V} = \frac{IR_1}{IR_s} = \frac{R_1}{R_s} \label{series4} \]and\[V_{ab} = V \times \frac{R_1}{R_s} \label{series5} \]The circuit in is an example of a simple voltage divider in that it divides the battery's voltage into parts and allows us to use a single battery to select one of several possible voltages. For example, the voltage at between points a and b is\[V_{ab} = V \times \frac{R_1}{R_s} \label{series6} \]the voltage at between points a and c is\[V_{ac} = V \times \frac{R_1 + R_2}{R_s} \label{series7} \]and the voltage at between points a and d is\[V_{ad} = V \times \frac{R_1 + R_2 + R_3}{R_s} = V \times \frac{R_s}{R_s} = V \label{series8} \] shows an example of a simple DC circuit in which three resistors, with resistances of R1, R2, and R3, are connected parallel to each other. A switch is included in the circuit that is used to close the loop and allow a current to flow from the battery's positive terminal to its negative terminal.If we apply Kirchoff's first law to the current at the point identified as a, then the sum of the currents must equal zero and\[\sum{I} = 0 = I - I_1 - I_2 - I_3 \label{parallel1} \]where I is the current entering point a and \(I_1\), \(I_2\), and \(I_3\) are the currents passing through the three resistors. Rearranging Equation \ref{parallel1} and substituting into Ohm's law gives\[\frac{V}{R_p} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3} \label{parallel2} \]where \(R_p\) is the circuit's effective resistance, which is equivalent to\[\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \label{parallel3} \]or to\[G_p = G_1 + G_2 + G_3 \label{parallel4} \]where \(G\) is a resistor's conductance, which is the inverse of its resistance.One of the useful properties of this circuit is that it serves as a current divider. The current passing through the resistor \(R_1\) is\[\frac{I_1}{I} = \frac{V/R_1}{V/R_p} = \frac{1/R_1}{1/R_p} = \frac{G_1}{G_p} \label{parallel5} \]Multiplying through by the total current gives\[I_1 = I \times \frac{G_1}{G_p} \label{parallel6} \]The treatment above considers circuits that contain only resistors in series or resistors in parallel. A circuit that contains both resistors in series and resistors in parallel can often be simplified to an equivalent circuit that has only resistors in series or in parallel, or that consists of single resistor. provides an example. The circuit at the far left shows two parallel resistors, \(R_2\) and \(R_3\), that, together, are in series with a third resistor, \(R_1\).Using Equation \ref{parallel3} we can replace the two resistors in parallel with a single resistor, \(R_4\), where\[\frac{1}{R_4} = \frac{1}{R_2} + \frac{1}{R_3} \label{complex1} \]giving the equivalent circuit shown in the middle. Finally, we can use Equation \ref{series2} to replace the two resistors in series with the single resistor, \(R_5\), as shown on the far right, where\[R_5 = R_1 + R_4 \label{complex2} \] shows a digital multimeter that is used to measure voltage or current (amongst other possible measurements that we will not consider here). The measurement of voltages and currents always contains some error, the magnitude of which we consider here.To measure an unknown voltage of \(V_x\) with an internal resistance of \(R_x\), we include the meter with its internal resistance of \(R_m\) as part of a voltage divider circuit. We read the voltage displayed on the meter, \(V_m\), and use Equation \ref{series5} to determine \(V_x\)\[V_m = V_x \times \frac{R_m}{R_m + R_x} \label{meter1} \]If we do not know the value of \(R_x\), which is often the case, then we can still report an accurate value for \(V_x\) if \(R_m >> R_x\), as we can then write\[V_m = V_x \times \frac{R_m}{R_m + R_x} \approx V_x \times \frac{R_m}{R_m} \approx V_x \label{meter2} \]The percent error, \(E_x\), in \(V_x\)\[E_x = \frac{V_m - V_x}{V_x} \times 100 = - \frac{R_m}{R_m + R_x} \times 100 \label{meter3} \]For example, suppose that \(R_m = 10^3 \times R_x\), then the measurement error is\[E = - \frac{R_x}{(10^3 \times R_x) + R_x} \times 100 = - \frac{1}{10^3 + 1} \times 100 = -0.0999\% \label{meter4} \]or approximately –0.1%.To measure an unknown current, \(I_x\), we include the meter in a current divider circuit in which some of \(I_x\) is drawn through a load resistor, \(R_l\), of known value, and the remaining current is drawn through a known standard resistance set by the meter, \(R_m\). Using Equation \ref{parallel5} for a current divider, the fraction of \(I_x\) that passes through the meter is\[\frac{I_m}{I_x} = \frac{R_m + R_l}{R_m} \label{meter5} \]Solving for \(I_x\) gives\[I_x = I_m \times \left( \frac{R_m}{R_m + R_l} \right) = I_m \times \left(1 + \frac{R_m}{R_l}\right) \label{meter6} \]If the resistors are selected such that \(\frac{R_m}{R_l} << 1\), then the current displayed on the meter, \(I_m\), is an accurate measure of \(I_x\). The percent error in the reported current is\[E_x = - \frac{R_m}{R_m + R_l} \times 100 \label{meter7} \]For example, suppose that \(R_m = 10^{-3} \times R_l\), then the measurement error is\[E = - \frac{10^{-3} \times R_l}{(10^{-3} \times R_l) + R_l} \times 100 = -\frac{10^{-3}}{10^{-3} + 1} \times 100 = -0.0999\%\label{meter8} \]or approximately \(-0.1\%\).This page titled 2.2: Direct Current (DC) Circuits is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

2.3: Alternating Current Circuits

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A direct current has a fixed value that is independent of time. An alternating current, on the other hand, has a value that changes with time. This change in current follows a pattern that we can characterize by it period—the time, \(t_p\), for one complete cycle—or by its frequency, \(f\), which is the reciprocal of its period\[f = \frac{1}{t_p} \label{sine1} \]Frequency is reported in hertz (Hz), which is equivalent to one cycle per second.Although we can draw many periodic signals—and will do so in later chapters—the simplest periodic signal is a sine wave: as shown on the right side of , the sine is a propagating wave whose amplitude, \(A\), is a function of time, \(t\), which we write as \(a(t)\).The left side of provides a rotating vector representation of the sine wave (a representation we will encounter again in Chapter 19 on NMR spectroscopy). The vector is the arrow that extends from the center of the circle to the circle's edge. It is rotating to the left with an angular velocity given by \(\omega\) and that is expressed in radians per the sine wave's period, \(t_p\); thus\[\omega = \frac{2 \pi}{t_p} = 2 \pi f \label{sine2} \]where \(f\) is the frequency. The amplitude of the sine wave as a function of time, \(a(t)\), is equivalent to the projection of the rotating vector onto the x-axis; thus\[a(t) = A\sin{\omega t} = A\sin{2 \pi f t} \label{sine3} \]In the context of this chapter, the amplitude is either a current, \(i\), or a voltage, \(v\).\[i(t) = I\sin{\omega t} = I\sin{2 \pi f t} \label{sine4} \]\[v(t) = V\sin{\omega t} = V\sin{2 \pi f t} \label{sine5} \]where \(I\) is the maximum, or peak current, and \(V\) is the maximum, or peak voltage.Equations \ref{sine4} and \ref{sine5} require that a sine wave's time-dependent amplitude, \(a(t)\), has a value of zero when \(t = n \pi\), where \(n\) is an integer. There is no reason to insist on this and two sine waves can be separated from each other in time, as shown in , by a phase angle, \(\Phi\). The equation for the sine wave when \(\Phi \ne 0\) becomes\[a(t) = A\sin{(\omega t + \Phi)} = A\sin{(2 \pi f t + \Phi)} \label{sine6} \]One complication of an alternating current is that the net current over the course of a single cycle is zero. This is a problem for us because the equation for power in a resistor is\[P = \frac{I^2}{R} \ne 0 \label{sine7} \] shows several ways to report current in AC circuits.The root-mean-square current, \(I_{rms}\), is defined as\[I_{rms} = \sqrt{\frac{I_p^2}{2}} = \sqrt{2} \times \frac{I_p}{2} = 0.707 \times I_p \label{sine8} \]and yields the same power in an AC circuit as a direct current of equal value in a DC circuit. The average current, \(I_{avg}\), is\[I_{avg} = \frac{1}{\pi} \int_{0}^{\pi}I_p \sin{\omega t}\, dt = \frac{2 I_p}{\pi} = 0.6371 \times I_p \label{sine9} \]A capacitor is a component of circuits that is capable of storing charge. shows the design of a typical capacitor and its symbol when constructing an electrical circuit. The capacitor consists of two conducting plates separated by a thin layer of an insulating, or dielectric material. The plates have areas of \(A\) and are separated by a distance, \(d\). The dielectric material has a dielectric constant, \(\epsilon\). A simple capacitor might consist of two pieces of a metal foil separated by air, which serves as as the dielectric material. A capacitor's ability to store charge, \(Q\), is given by\[Q = C \times V \label{cap1} \]where \(V\) is the voltage applied across the two plates and where \(C\) is the capacitor's capacitance, which, in turn, is defined as\[C = \frac{\epsilon A}{d} \label{cap2} \]Capacitance is measured in units of farads, where one farad is equal to one coulomb per volt. shows a resistor, with a resistance of \(R\), and a capacitor, with a capacitance of \(C\), in series with a voltage source, with a voltage of \(V\).When the switch it closed, current flows as the capacitor builds up a charge. From Kirchoff's laws, we know that\[V = v_R + v_C = iR + \frac{Q}{C} \label{cap3} \]where \(v_R\) and \(v_C\) are, respectively, the time-dependent voltages across the resistor and the capacitor. Because \(V\) has a fixed value, any increase in \(v_C\) as the capacitor is charged is offset by a decrease in \(V_r\). Given that the values of \(v_C\) and \(v_R\)—and the associated currents—are time-dependent, we can differentiate Equation \ref{cap3} with respect to time\[\frac{dV}{dt} = 0 = \left( R \times \frac{di}{dt} \right) + \left( \frac{1}{C} \times \frac{dq}{dt} \right) = \left(R \times \frac{di}{dt}\right) + \frac{i}{C} \label{cap4} \]Rearranging Equation \ref{cap4} gives\[\frac{di}{i} = - \frac{1}{RC}dt \label{cap5} \]Integrating both sides of this equation\[ \int_{I_{0}}^{i} \frac{1}{i} di = -\frac{1}{RC} \int_{0}^{t} dt \label{cap6} \]leads to the following relationship between the current at time \(t\) and the initial current, \(I_0\)\[i_t = I_0 \times e^{-t/RC} \label{cap7} \]which tells us that the current decreases exponentially as the capacitor becomes fully charged. Replacing the current in equation \ref{cap7} with \(\frac{V}{R}\) and substituting back into Equation \ref{cap3}\[v_C = V_0 \left( 1 - e^{-t/RC} \right) \label{cap8} \]shows us that during the time the capacitor is being charged, the current flowing through the capacitor is decreasing exponentially to its limit of zero, and the voltage across the capacitor is increasing exponentially to its limit of the applied voltage.The value \(RC\) in Equation \ref{cap7} and in Equation \ref{cap8} is the circuit's time constant. It takes approximately five time constants for the capacitor to fully charge or fully discharge. shows the voltage across the capacitor, \(v_C\), as it is allowed to charge and to discharge. Time is shown in terms of the number of elapsed time constants, and voltage is expressed as a fraction of the maximum voltage. The dashed line shows that the time constant, \(RC\), is equivalent to \(0.63 \times\) the maximum voltage.If we replace the DC voltage source in with an AC source, then the capacitor will undergo a continuous fluctuation in its voltage and current as a function of time. We know, form Equation \ref{cap1} that charge, \(Q\), is the product of capacitance, \(C\), and voltage,\(V\), which we can write as a derivative with respect to time.\[\frac{dq}{dt} = C \times \frac{dv}{dt} \label{ac1} \]In an AC circuit, as we learned earlier in Equation \ref{sine4}, the current, which is equivalent to \(dq/dt\) is\[i = I_p \sin{2 \pi f t} \label{ac2} \]where \(I_p\) is the peak current. Substituting into Equation \ref{ac1} gives\[i = I_p \sin{2 \pi f t} = C \times \frac{dv}{dt} \label{ac3} \]Rearranging this equation and integrating over time gives the time-dependent voltage across the capacitor, \(v_C\), as\[v_C = \frac{I_p}{2 \pi f C} \left( -\cos{2 \pi f t} \right) \label{ac4} \]We can rewrite this equation in terms of a sine function instead of a cosine function by recognizing that the two are 90° out of phase with each other; thus\[v_C = \frac{I_p}{2 \pi f C} \left( \sin{2 \pi f t -90} \right) = V_p \left( \sin{2 \pi f t -90} \right) \label{ac5} \]Comparing Equation \ref{ac2} and Equation \ref{ac5}, we see that the current and the voltage are 90° out-of-phase with each other; shows this visually.From Equation \ref{ac5} we see that\[V_p = \frac{I_p}{2 \pi f t} \label{ac6} \]Dividing both sides by \(I_p\) gives\[\frac{V_p}{I_p} = X_C = \frac{1}{2 \pi f t} \label{ac7} \]where \(X_C\) is the capacitor's reactance, which, like a resistor's resistance, has units of ohms. Unlike a resistor, however, a capacitor's reactance is frequency dependent and, given the reciprocal relationship between \(X_C\) and \(f\), it becomes smaller at higher frequencies.In a RC circuit, both the resistor and the capacitor contribute to the circuit's impedence of the alternating current. Because the contribution of the capacitor is 90° out-of-phase to the contribution from the resistor, the net impedence, \(Z\), is\[Z = \sqrt{R^2 + X_C^2} \label{ac8} \]as shown in where the vector that represents \(Z\) is the hypotonus of a right triangle defined by the resistor's resistance and the capacitor's reactance.Substituting in Equation \ref{ac7} shows the effect of frequency on impedence.\[Z = \sqrt{R^2 + \left( \frac{1}{2 \pi f t} \right)^2} \label{ac9} \]Writing Ohm's law in terms of impedence, \(V_p = I_p \times Z\), and substituting it into Equation \ref{ac9}, defines \(I_p\) and \(V_p\) in terms of impedence.\[V_p = I_p \times \sqrt{R^2 + \left( \frac{1}{2 \pi f t} \right)^2} \label{ac10} \]\[I_p = \frac{V_p}{\sqrt{R^2 + \left( \frac{1}{2 \pi f t} \right)^2}} \label{ac11} \]The frequency dependence of an RC circuit provides us with the ability to attenuate some frequencies and to pass other frequencies. This allows for the selective filtering of an input signal. Here we consider the design of a low-pass filter that removes higher frequency signals, and the design of a high-pass filter that removes lower frequency signals. shows that (a) a low-pass filter places the resistor before the capacitor and measures the output voltage, \(V_{out}\), across the capacitor, and that (b) a high-pass filter places the capacitor before the resistor and measures the output voltage, \(V_{out}\), across the resistor.For the low-pass filter in , the ratio of the voltage across the capacitor, \((V_p)_{out}\), to the peak input voltage, \((V_p)_{in}\), is equal to the fraction of the circuit's impedence, \(Z\), attributed to the capacitor's reactance, \(X_C\), as expected for a voltage divider that consist of elements in series.\[\frac{(V_p)_{out}}{(V_p)_{in}} = \frac{X_C}{Z} = \frac{(2 \pi f C)^{-1}}{\sqrt{R^2 + \left( \frac{1}{2 \pi f C}\right)^2}} \label{lowpass1} \] shows the frequency response for a low-pass filter with a \(1 \times 10^6 \text{ Hz}\) resistor and a \(1 \times 10^{-6} \text{ F}\) capacitor, removing all frequencies greater than approximately \(10^1\) Hz.For the high-pass filter in , the ratio of the voltage across the resistor, \((V_p)_{out}\), to the peak input voltage, \((V_p)_{in}\), is equal to the fraction of the circuit's impedence, \(Z\), attributed to the resistor's resistance, \(R\), as expected for a voltage divider that consist of elements in series.\[\frac{(V_p)_{out}}{(V_p)_{in}} = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + \left( \frac{1}{2 \pi f C}\right)^2}} \label{lowpass2} \] shows the frequency response for a low-pass filter with a \(1 \times 10^5 \text{ Hz}\) resistor and a \(1 \times 10^{-7} \text{ F}\) capacitor, removing all frequencies less than approximately \(10^{-1}\).This page titled 2.3: Alternating Current Circuits is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

2.4: Semiconductors

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A semiconductor is a material whose resistivity to the movement of charge falls somewhere between that of a conductor, through which we can move a charge easily, and an insulator, which resists the movement of charge. Some semiconductors are elemental, such as silicon and germanium (both of which we examine more closely in this section) and some are multielemental, such as silicon carbide.A conductor, such as aluminum or copper, has a resistivity on the order of \(10^{-8} - 10^{-6} \ \Omega \cdot m\), which means that its resistance to the movement of electrons is sufficiently small that it carries a current without much effort. An insulator, such as the mineral quartz, \(\ce{SiO2}\), has a resistivity on the order of \(10^{15} - 10^{19} \ \Omega \cdot m\). Silicon, on the other hand, has a resistivity of approximately \(640 \ \Omega \cdot m\) and germainum has a resistivity of about \(0.46 \ \Omega \cdot m\).The inverse of resistivity is conductivity.Silicon and germanium are in the same group as carbon. If we use a simplified view of atoms, we can treat silicon and germanium as having four valence electrons and an effective nuclear charge, \(Z_{eff}\), of\[\begin{align*} (Z_{eff})_\ce{Si} &= Z - \text{ number of core electrons} \\[4pt] &= 14 - 10 \\[4pt] &= +4 \\[4pt] (Z_{eff})_\ce{Ge} &= Z - \text{ number of core electrons} \\[4pt] &= 32 - 28 \\[4pt] &= +4 \end{align*} \]We can increase the conductivity of silicon and germanium by adding to them—this is called doping—a small amount of an impurity. Adding a small amount of In or Ga, which have three valence electrons instead of four valence electrons, leaves a small number of vacancies, or holes, in which an electron is missing. Adding a small amount of As or Sb, which have five valence electrons instead of four valence electrons, leaves a small number of extra electrons. shows all three possibilities.If we apply a potential across the semiconductor doped with As or Sb, the extra electron moves toward the positive pole, creating a small current, and leaving behind a vacancy, or hole. If we apply a potential across the semiconductor doped with In or Ga, electrons enter the semiconductor from the negative pole, occupying the vacancies, or holes, and creating a small current. In both cases, electrons move toward the positive pole and holes move toward the negative pole. We call an As or Sb doped semiconductor an n-type semiconductor because the primary carrier of charge is an electron; we call an In or Ga doped semiconductor an p-type semiconductor because the primary carrier of charge is the hole.A diode is an electrical device that is more conductive in one direction than in the opposite direction. A diode takes advantage of the properties of the junction between a p-type and an n-type semiconductor.Let's use to make sense of how a semiconductor diode works. The figure is divided into two parts: the left side of the figure, parts (a), (b), and (c), show the behavior of the semiconductor diode when a foward bias is applied, and the right side of the figure, parts (d), (e), and (f), show its behavior when a reverse bias is applied. For both, the semiconductor diode consists of a junction between a n-type semiconductor, which has an excess of electrons and carries a negative charge, and a p-type semiconductor, which has an excess of holes and, thus, a positive charge; this is shown in (a) and (d). How the semiconductor is manufactured is not of important to us.To effect a forward bias, we apply a positive potential to the p-type region and apply a negative potential to the n-type region. As we see in (b), the holes in the p-region move toward the junction and the electrons in the n-region move toward the junction as well. When holes and electrons meet they combine and are eliminated, which is why we see fewer holes and electrons in (c). Additional electrons flow into the n-region and electrons are pulled away from the p-region, as seen in (c), resulting in a current. To effect a reverse bias, we switch the applied potentials so that the p-region has the negative potential and the n-region has the positive potential. The result, as seen in (e) is a brief current as the holes and electrons move away from each other. leaving behind, in (f), a depletion zone that has essentially no electrons or holes. shows a plot of current as a function of voltage for a semiconductor diode. In forward bias mode the current increases exponentially with an increase in applied voltage, but remains at essentially zero when operated under a reverse bias. The use of a sufficiently large negative potential, however, does result in an sudden and dramatic increase in current; the potential at which this happens is called the breakdown voltage.This page titled 2.4: Semiconductors is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

20.1: Molecular Mass Spectra

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shows a mass spectrum for p-nitrophenol, C6H5NO3, which has a nominal (integer) mass of 139 daltons. If we send a beam of energetic electrons through a gas phase sample of p-nitrophenol, it loses an electron, which we write as the reaction\[\ce{C6H5NO3} + e^{-} \rightarrow \ce{C6H5NO3^{+•}} + 2e^{-} \label{pnp1} \]where the product is a radical cation that has a charge of \(+1\) and that retains the nominal mass of 139 daltons. We call this the molecular ion—highlighted here in green—and it has a mass-to-charge ratio (m/z) of 139.Some of the terminology in this chapter was covered earlier in Chapter 11 on atomic mass spectrometry. See the first section of that chapter for a discussion of atomic mass units (amu) and daltons (Da), and of mass-to-charge ratios.If reaction \ref{pnp1} is all that happens when p-nitrophenol interacts with an energetic electron, then it would not provide much in the way of useful information. The radical cation \(\ce{C6H5NO3^{+•}}\), however, retains sufficient excess energy from the initial electron-molecule collision that it is in an excited state. In returning to its ground state, the molecular ion undergoes a series of fragmentations that result in the formation of ions—called daughter ions—with different mass-to-charge ratios. A plot that shows the relative intensity of these ions as a function of their mass-to-charge ratios is called a mass spectrum. The most abundant fragment in the spectrum—shown here in red and which is called the base peak—is assigned a relative intensity of 100; the intensity of all other ions is reported relative to the base perk.A molecule's fragmentation patterns provides rich information about its structure. compares the mass spectra for o-nitrophenol, m-nitrophenol, and p-nitrophenol. All three molecules have clusters of fragment ions at similar mass-to-charge ratios, but the relative abundance of the ions in these clusters varies quite a bit from molecule-to-molecule. For example, the pink rectangle in each spectrum highlights peaks with mass-to-charge ratios from approximately 104 m/z to 115 m/z. All three molecules share the property of producing fragment ions with these mass-to-charge ratios; the relative abundance of the fragment ions, however, varies substantially between the three molecules with o-nitrophenol and p-nitrophenol having a major peak at 109 m/z, but m-nitrophenol showing no more than a trace peak at 109 m/z. We will return to the use of mass spectrometry for determining structure information later in this chapter.This page titled 20.1: Molecular Mass Spectra is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

20.2: Ion Sources

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Since a mass spectrum shows the relative abundance of ions with different mass-to-charge ratios, a mass spectrometer must include a way to generate ions. More specifically, it needs a method that generates the initial ion as it, once formed, will undergo fragmentation without additional help from the analyst (which does not mean the analyst cannot assist in that fragmentation; see discussion of tandem mass spectrometry in Section 20.4). In this section we consider several common ion sources. We can describe these sources using two characteristic properties: (a) the physical state of the species that is initially ionized (gas, liquid, or solid phase), and (b) whether ionization favors the formation of fragment ions or the formation of molecular ions (hard sources or soft sources).The electron ionization (EI) source, also known as an electron impact source, uses a beam of energetic electrons to ionize the analyte. As shown in \(\PageIndex{1}\), the sample is volatilized prior to entering the ion source as gas phase molecules, M(g). A heated tungsten filament is used to generate electrons, which are pulled toward a positively charged anode. This electron beam intersects with the gas phase molecules at 90° where ionization occurs\[ \ce{M}(g) + e^- \rightarrow \ce{M^{+•}}(g) + 2e^- \label{ei1} \]The molecular ions, \(\ce{M^{+•}}(g)\), are then swept into the mass analyzer using a set of accelerating plates (not shown here).The electron beam in has a lot of energy due to the significant potential difference between the cathode and the anode, which may be as much as 70 V. The kinetic energy of these electrons is equivalent to the product of the electron's charge in Coulombs, the applied potential in volts, and Avogadro's number\[e \times V \times N_A = 1.6 \times 10^{-19} \times 6.022 \times 10^{-23} = 6.7 \times 10^{6} \text{ J/mol} \nonumber \]or 6,700 kJ/mol. This energy is much greater than typical bond energies, which range from approximately 150–600 kJ/mol for single bonds, from approximately 500–750 kJ/mol for double bonds, and from approximately 800–1100 kJ/mol for triple bonds. The significant difference between the energy of the electrons and bond energies explains why electron ionization spectra are rich in fragment ions, as we saw earlier in Section 20.1 for o-nitrophenol, m-nitrophenol, and p-nitrophenol. This extensive fragmentation is useful in determining an analyte's structure—which is an advantage of a hard ionization method—but at the possible cost of the loss of the molecular ion peak for some analytes. For example, shows the electron ionization mass spectrum from 1-decanol, C10H22O, which has a nominal mass of 158 daltons. The small peak at m/z = 157 is for the fragment ion C10H21O+; the molecular ion is not observed in this spectrum.Electron ionization is a hard source because the electron beam's energy results in easy fragmentation. In chemical ionization, we introduced a reagent molecule, such as methane, into the electron ionization (CI) source so that it is present at level that is \(1000 \times\) to \(10,000 \times\) greater than the analyte. At this higher concentration, it is the reagent molecule that is ionized; for example, when using CH4 as the reagent gas, ions such as \(\ce{CH4+}\) and \(\ce{CH3+}\) form. These ions then react with additional methane molecules\[\ce{CH4+}(g) + \ce{CH4}(g) \rightarrow \ce{CH5+}(g) + \ce{CH3}(g) \label{ci1} \]\[\ce{CH3+}(g) + \ce{CH4}(g) \rightarrow \ce{C2H5+}(g) + \ce{H2}(g) \label{ci2} \]to form \(\ce{CH5+}\) and \(\ce{C2H5+}\), species that are sufficiently reactive that they easily transfer a hydrogen to a molecule of the analyte, MH\[\ce{CH5+}(g) + \ce{MH}(g) \rightarrow \ce{MH2+}(g) + \ce{CH4}(g) \label{ci3} \]to give a molecular ion that we identify as [M + H]+ and that has a mass that is one amu unit greater than that for M. Alternatively, they can easily remove a hydrogen from a molecule of the analyte, MH\[\ce{C2H5+}(g) + \ce{MH}(g) \rightarrow \ce{M+}(g) + \ce{C2H6}(g) \label{ci4} \]to give a molecular ion that we identify as [M – H]– and that has a mass that is one amu less than that for M. Because formation of the molecular ion occurs indirectly and less energetically, fragmentation is suppressed, leading to a mass spectrum with a molecular ion peak and with only a small number of other ions. shows the mass spectrum for 1-decanol when using chemical ionization with \(\ce{CH4}\) as the reagent gas.Electron impact and chemical ionization are gas phase sources because the sample is volatilized before it enters the mass spectrometer's inlet. In electrospray ionization (ESI), the sample is a liquid and ions desorb from that matrix in the mass spectrometer's inlet system. The liquid sample is pulled into the spectrometer's inlet through a capillary needle, forming a mist of droplets. The application of a large potential across this inlet assures that the droplets carry positive charges. These charged droplets then enter a chamber where they undergo desolvation, which decreases the size of the droplet and increases their charge density (see ). As this charge density increases, the droplets eventually become unstable, for reasons that are not fully understood, and ionized gas-phase ions desorb from the droplets and enter into the mass analyzer.A typical electrospray ionization mass spectrum for a small molecule is shown in for the compound (4-aminophenyl)arsonic acid. As we saw in for chemical ionization, a soft ionization source results in a limited amount of fragmentation and a strong peak for the molecular ion, which here includes a proton transfer to give the [M + H]+ peak at a m/z of 218 amu.Electrospray ionization is particularly useful for biological molecules, such as peptides and proteins, because the soft ionization ensures that molecular weight information is retained. Because these molecules are large, they readily pick up multiple protons, forming multiply charged ions of the general form [M + zH]z+ where z is the number of protons added. shows a hypothetical spectrum for the molecule M and Table \(\PageIndex{1}\) provides the corresponding m/z values for the mass spectrum's peaks.If we take the mass-to-charge ratios for any two adjacent peaks, \(M_i\) and \(M_j\), where the peak \(M_i\) has the greater value for m/z, and if we assume that \(M_j\) has one additional hydrogen atom, giving it a charge that is one unit higher, then\[Z_i = \frac{M_j - 1}{M_i - M_j} \label{findz} \]where \(Z_i\) is the charge on the ion \(M_i\). Table \(\PageIndex{2}\) shows the calculated charges for the ions \(M_1\) to \(M_7\).Here is a derivation for Equation \ref{findz}. Suppose the molecule of interest has a molecular weight of \(m\). If the charge on the ion responsible for peak \(M_i\) is \(Z\), then it must be the case that the peak's mass is equal to m + Z as it has Z extra hydrogens and it must be the case that its mass-to-charge ratio is\[M_i = \frac{m + Z}{Z} \nonumber \]and its molecular weight, \(m\), is\[m = (M_i \times Z) - Z \nonumber \]In the same way, the peak \(M_j\) has a charge of \(Z + 1\) and\[M_j = \frac{m + Z + 1}{Z + 1} \nonumber \]\[m = (M_j \times Z )+ M_j - Z - 1 \nonumber \]Setting the two equations for \(m\) equal to each other and solving for \(Z\) gives\[(M_i \times Z) - Z = (M_j \times Z) + M_j - Z - 1 \nonumber \]\[(M_i \times Z) = (M_j \times Z) + M_j - 1 \nonumber \]\[(M_i \times Z) - (M_j \times Z) = M_j - 1 \nonumber \]\[Z = \frac{M_j - 1}{M_i - M_j} \nonumber \]The molecular weight, \(m\), is given by the equation\[m = (M_i \times Z) - Z \label{findmw} \]Table \(\PageIndex{3}\) shows the molecular weights for the ions \(M_1\) to \(M_7\) and their average value. The simulated mass spectrum was created by setting the molecular weight to 10,000 amu and with charges ranging from +9 to +16.Matrix-assisted laser desorption ionization (MALDI) is a soft ionization source for obtaining the mass spectrum for biologically important molecules, such as proteins and peptides. illustrates the basic steps in obtaining a MALDI spectrum. The sample is first mixed with a small molecule—which is called the matrix—to form a solution; the matrix usually is present in a 10:1 ratio. A drop of this mixture is placed on a sample probe and allowed to dry, leaving the sample in a solid form. A pulsed laser beam (\(\lambda = 237\) nm is typical) is focused on the solid sample–matrix mixture. The matrix absorbs the laser pulse and the absorption of the laser's energy volatilizes both the matrix and the sample. Ionization of the sample forms molecular ions, usually [M + H]+ ions, which are then swept into the mass analyzer.When the sample is a digestion of a protein, then it is a mixture of peptides, each of which appeara as a [M + H]+ peak in the resulting mass spectrum. For example, a peptide with the sequence AWSVAR (alanine–tryptophan–serine–valine–alanine–arginine) will appear as a peak with a mass of 689.8 daltons. To find this value we add together the molecular weights of the amino acids, account for the loss of a molecule of water for each peptide bond that forms, and then account for the hydrogen that gives the [M + H]+ ion. In this case we have\[ \ce{[M + H]^+} = 89.1 + 204.2 + 105.1 + 117.1 + 89.1 + 174.2 - (5 \times 18.0) + 1 = 689.8 \text{ amu} \nonumber \]where the term \(5 \times 18.0\) accounts for the loss of five molecules of \(\ce{H2O}\) when forming the five peptide bonds.Fast atom bombardment (FAB) bears some similarity to MALDI: the sample is mixed with a liquid matrix (often glycerol) and bombarded with a beam of xenon or argon atoms (instead of a laser). Desorption of the sample from its matrix forms gas phase ions that are swept into the mass analyzer. Spectra usually contain both a molecular ion and fragmentation patterns.This page titled 20.2: Ion Sources is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

20.3: Mass Spectrometers

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A mass spectrometer has four essential elements: a means for introducing the sample to the instrument, a means for generating a mixture of ions, a means for separating the ions, and a means for counting the ions. In Chapter 20.2 we introduced some of the most important ways to generate ions. In this section we turn our attention to sample inlet systems and to separating and counting ions. You may wish to review Chapter 11 where we considered these topics in the context of atomic mass spectrometry. As we noted in Chapter 11, a mass spectrometer must operate under a vacuum to ensure that ions can travel long distances without undergoing undesired collisions that affect their ion energy.When the sample is a gas or a volatile liquid, it is easy to transfer a portion of the sample into a reservoir as a gas maintained at a relatively small pressure. The sample is then allowed to enter into the mass spectrometer's ion source through a diaphragm that contains a pin-hole, drawn in by holding the ion source at lower pressure.Solid and non-volatile liquids are sampled by inserting them directly into the ion source through a vacuum lock that allows the mass spectrometer to remain under vacuum except for the ion source where the sample is inserted. The sample is placed in a capillary tube or a small cup at the end of a sample probe, and then moved into the ion source. The sample probe includes a heating coil that is used, along with the instrument's vacuum, to help volatilize the sample.Of particular importance are inlet systems that couple a chromatographic or an electrophoretic instrument to a mass spectrometer, providing a way to separate a complex mixture into its individual components and then using the mass spectrometer to determine the composition of those components. The interface between a gas chromatograph and a mass spectrometer (GC-MS) must account for the significant drop in pressure from atmospheric pressure to a pressure of 10–8 torr; the interface for LC-MS and for EC-MS must provide a way to remove the liquid eluent, to volatilize the samples, and to account for the drop in pressure. See Chapters 27, 28, and 30 for more details.The purpose of the mass analyzer is to separate the ions by their mass-to-charge ratios. Ideally we want the mass analyzer to allow us to distinguish between small differences in mass and to do so with a strong signal-to-noise ratio. As we learned in Chapter 7 when introducing optical spectroscopy, these two desires usually are in tension with each other, with improvements in resolution often coming with an increase in noise.The resolution between two peaks, \(R\), in mass spectrometry is defined as the ratio of their average mass to the difference in their masses\[R = \frac{\overline{m}}{\Delta m} \label{resolution} \]The following table shows how resolution varies as a function of the average mass and the difference in mass. A resolution of 1,000, for example is sufficient to resolve two ions with an average mass of 100 amu that differ by 0.1 amu, or two ions that have an average mass of 1,000 amu that differ by 1 amu.\[\overline{m} \rightarrow \nonumber \]\[ \Delta m \downarrow \nonumber \]When a beam of ions passes through a magnetic field, its path is altered, as we see in . The ions experience an acceleration as they exit the ion source and enter the mass analyzer with a kinetic energy that is given by the equations\[\ce{KE} = z e V \label{msa1} \]\[\ce{KE} = \frac{1}{2} mv^2 \label{msa2} \]where \(z\) is the ion's charge (usually +1), \(e\) is the electronic charge in Coulombs, \(V\) is the applied voltage responsible for the acceleration, \(m\) is the ion's mass, and \(v\) is the ion's velocity after acceleration. Equation \ref{msa1} shows us that all ions with the same charge have the same kinetic energy. Equation \ref{msa2}, then, tells us that ions with a greater mass will move more slowly.An ion's path through the magnetic field is determined by two forces. The first of these forces is the magnetic force, \(F_M\), that acts on the ion, which is\[F_M = B z e v \label{msa3} \]where \(B\) is the magnetic field strength. The second of these forces is the centripetal force, \(F_C\), that acts on the ion as it moves along its curved path, which is\[F_C = \frac{mv^2}{r} \label{msa4} \]where \(r\) is the magnet's radius of curvature. An ion can only navigate these opposing forces if \(F_M\) and \(F_C\) are equal to each other. This requires that\[B z e v = \frac{mv^2}{r} \label{msa5} \]Solving for \(v\) gives\[v = \frac{B z e r}{m} \label{msa6} \]Substituting back into Equation \ref{msa2} and solving for the mass-to-charge ratio gives\[\frac{m}{z} = \frac{B^2 r^2 e}{2V} \label{msa7} \]Equation \ref{msa7} tells us that for any combinaton of magnetic field strength, \(B\), and accelerating voltage, \(V\), only one mass-to-charge ratio has the correct value of \(r\) to reach the director. Ions that are too heavy or ions that are too light, will collide with the sides of the mass analyzer before they reach the detector. The mass spectrum is recorded by holding \(V\) and \(r\) constant and varying the magnetic field strength, \(B\). The resolution of a magnetic sector instrument is usually less than 2000.The resolution of a magnetic sector instrument suffers from limitations that affect its ability to narrow the range of kinetic energies—and, thus, velocities—possessed by the ions when they exit the ion source and enter the mass analyzer. The double-focusing mass analyzer in compensates for this by placing an electrostatic analyzer before the magnetic analyzer, separating the two by a slit. The electrostatic analyzer consists of two curved metal plates, one of which is held at a positive potential and one at a negative potential. As ions pass betweeen the plates, those ions that have too much energy and those that have too little energy fail to pass through the slit that separates the electrostatic analyzer from the magnetic analyzer. In this way, the distribution of energies—and, thus, velocities—is tightened, improving the resolution achieved by the magnetic sector analyzer. Depending on its design, a double-focusing analyzer can achieve a resolution as large as 100,000.The quadrupole mass analyzer was introduced in Chapter 11 and the treatment here is largely the same. A quadupole mass analyzer is compact in size, low in cost, easy to use, and easy to maintain. As shown in , it consists of four cylindrical rods, two of which are connected to the positive terminal of a variable direct current (dc) power supply and two of which are connected to the power supply's negative terminal; the two positive rods are positioned opposite of each other and the two negative rods are positioned opposite of each other. Each pair of rods is also connected to a variable alternating current (ac) source operated such that the alternating currents are 180° out-of-phase with each other. An ion beam from the source is drawn into the channel between the quadrupoles and, depending on the applied dc and ac voltages, ions with only one mass-to-charge ratio successfully travel the length of the mass analyzer and reach the transducer; all other ions collide with one of the four rods and are destroyed.To understand how a quadrupole mass analyzer achieves this separation of ions, it helps to consider the movement of an ion relative to just two of the four rods, as shown in for the poles that carry a positive dc voltage. When the ion beam enters the channel between the rods, the ac voltage causes the ion to begin to oscillate. If, as in the top diagram, the ion is able to maintain a stable oscillation, it will pass through the mass analyzer and reach the transducer. If, as in the middle diagram, the ion is unable to maintain a stable oscillation, then the ion eventually collides with one of the rods and is destroyed. When the rods have a positive dc voltage, as they do here, ions with larger mass-to-charge ratios will be slow to respond to the alternating ac voltage and will pass through the transducer. The result is shown in the figure at the bottom (and repeated in ) where we see that ions with a sufficiently large mass-to-charge ratios successfully pass through the transducer; ions with smaller mass-to-charge ratios do not. In this case, the quadrupole mass analyzer acts as a high-pass filter.We can extend this to the behavior of the ions when they interact with rods that carry a negative dc voltage. In this case, the ions are attracted to the rods, but those ions that have a sufficiently small mass-to-charge ratio are able to respond to the alternating current's voltage and remain in the channel between the rods. The ions with larger mass-to-charge ratios move more sluggishly and eventually collide with one of the rods. As shown in , in this case, the quadrupole mass analyzer acts as a low-pass filter. Together, as we see in , a quadrupole mass analyzer operates as both a high-pass and a low-pass filter, allowing a narrow band of mass-to-charge ratios to pass through the transducer. By varying the applied dc voltage and the applied ac voltage, we can obtain a full mass spectrum.Quadrupole mass analyzers provide a modest mass-to-charge resolution of about 1 amu and extend to \(m/z\) ratios of approximately 2000.In a time-of-flight mass analyzers, , ions are created in small clusters by applying a periodic pulse of energy to the sample using a laser beam or a beam of energetic particles to ionize the sample. The small cluster of ions are then drawn into a tube by applying an electric field and then allowed to drift through the tube in the absence of any additional applied field; the tube, for obvious reasons, is called a drift tube. All of the ions in the cluster enter the drift tube with the same kinetic energy, KE, which we define as\[\text{KE} = \frac{1}{2} m v^2 =z e V \label{tof1} \]The time, \(T\), that it takes the ion to travel the distance, \(L\), to the detector is\[T = \frac{L}{v} \label{tof2} \]Substituting Equation \ref{tof2} into Equation \ref{tof1}\[T = \sqrt{\frac{m}{z}} \times \sqrt{\frac{1}{2eV}} \label{tof3} \]shows us that the time it takes an ion to travel through the drift tube is proportional to the square rate of its mass-to-charge ratio. As a result, lighter ions move more quickly than heavier ions. Flight times are typically less than 30 µs. A time-of-flight mass analyzer provide better resolution than a quadrupole mass analyzer, but is limited to sources that can be pulsed. A linear time-of-flight analyzer, such as that in , provide a resolution of approximately 4,000; other configurations can achieve resolutions of 10,000 or better. The time-of-flight analyzer is well-suited for MALDI ionization as the time between pulses of the laser provides the time needed for detection to occur. provides an illustration of an ion trap mass analyzer, which consists of three electrodes—a central ring electrode and two conical end cap electrodes—that create a cavity into which ions are drawn. The ions in the cavity experience stabilizing and destabilizing forces that affect their movement within the cavity. Ions that adopt stable orbits remain in the cavity. By varying the potentials applied to the electrodes, ions with different mass-to-charge ratios enter into destabilizing orbits and exit through a small hole at the bottom of the trap. An ion trap typcially provides a resolution of 1,000.The ion cyclotron resonance (ICR) analyzer is a form of an ion trap but operates in a way that retains all ions within the trap. When a gas phase ion is placed within an applied magnetic field, the ions move in a circular orbit that is perpendicular to the applied field ). In discussing the magnetric sector analyzer, we showed that the velocity, \(v\), of an ion in an applied magnetic field with a strength of \(B\) is a function of the radius of the ion's motion, \(r\), and its charge\[v = \frac{B z e r}{m} \label{icr1} \]Solving for the ratio \(v / r\) gives the ion's cyclotron frequency, \(w_c\), as\[w_c = \frac{v}{r} = \frac{z e B}{m} \label{icr2} \]When an ion moving in a circular orbit, as shown by the smaller of the two circular orbits in , absorbs energy equal to its cyclotron frequency, \(w_c\), its velocity, \(v\), and the radius of its orbit, \(r\) both increase to maintain a constant value for \(w_c\); the result is an ion that moves in a circular orbit of greater radius. As \(w_c\) depends on the mass-to-charge ratio, all ions of equal \(m/z\) experience the same change in their orbit, while ions with other mass-to-charge ratios are unaffected. Ions in the larger orbits eventually return to their original circular orbit as a result of collisions in which they lose energy.The trap itself, as seen in , is defined by two pairs of plates (four in all). The transmitter plates are used to apply the potential that alters the orbits of the ions. Movement of the ions generates a current in the receiver plates that serves as the signal, as seen in , that is positive when the ion is closer to one receiver plate and negative when it is closer to the other receiver plate. The initial magnitude of the current is proportional to the number of ions with the mass-to-charge ratio.The ion cyclotron resonance analyzer is usually operated by applying a short pulse of energy that varies linearly in its frequency. This sets all ions into motion, with each mass-to-charge ratio yielding a current response similar to that in . Collectively, these individual current-time curves gives a time domain spectrum that we can covert into a frequency domain spectrum by taking the Fourier transform. The frequency domain spectrum yields the mass spectrum through Equation \ref{icr2}. FT-ICR instruments are capable of achieving resolutions of 1,000,000.This page titled 20.3: Mass Spectrometers is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

20.4: Applications of Molecular Mass Spectrometry

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In a qualitative analysis our interest is in determining the identity of a substance of interest to us. By itself, mass spectrometry is a powerful tool for determining the identity of pure compounds. The analysis of mixtures, however, is possible if we use a mass spectrometer as the detector for a separation technique, such as gas chromatography, or if we string together two or more mass analyzers in sequence.There are several ways to use a mass spectrum to identify a compound, including identifying its molecular weight, using isotopic ratios, examining fragmentation patterns, and by searching through data bases.Using Molecular Weight Information. A molecular ion peak, M+•, when it is present, a [M + H]+ peak or a [M – H]+ peak, provides information about the compound's molecular weight. When using a low resolution mass analyzer, this may be sufficient to distinguish between molecular ions with, for example, a nominal mass of 95 amu and a nominal mass of 96 amu, but insufficient to distinguish between molecular ions with a more precise mass of 96.0399 amu and 96.0575 amu. When using a high resolution mass analyzer, the difference between the last pair of molecular ions may be feasible.Using Isotopic Ratios. The molecule cycloheptene has the formula C7H12 and a nominal mass of 96 amu, and the molecule cyclohexenone has the formula C6H8O and a nominal mass of 96 amu. Although both molecules will produce a molecular ion with the same nominal mass-to-charge ratio, each will also have a peak with a nominal mass of M + 1 due to the presence of isotopes of carbon, hydrogen, and oxygen. Because cycloheptene and cyclohexenone have different chemical formulas, the relative heights of their M + 1 peaks are different. Here is how we can work this out.For every 100 atoms of 12C there are 1.08 atoms of 13C (that is, 1.08% of the carbon atoms are 13C), for every 100 atoms of 1H there are 0.015 atoms of 2H, and for every 100 atoms of 16O there are 0.04 atoms of 17O. For cycloheptene, this means that the relative height of its M + 1 peak to its M peak is\[(7 \times 1.08) + (12 \times 0.015) = 7.74 \nonumber \]and for cyclohexenone we have\[(6 \times 1.08) + (8 \times 0.015) + (1 \times 0.04) = 6.64 \nonumber \]Here we see that a careful examination of the relative height of the M + 1 peak provides a way to distinguish between C7H12 and C6H8O even though they have the same nominal masses. On-line calculators are available—this link provides one example—that you can use to calculate the full isotopic abundance patterns, including M + 2, M + 3, and other peaks. Isotopic patterns are particularly useful for identifying the presence of chlorine and bromine in a molecule because each has one isotope with a significant abundance: for chlorine, 37Cl has an abundance of 32.5% relative to 35Cl, and for bromine, 81Br has an abundance of 98.0% of 79Br.Using Fragmentation Patterns. shows the mass spectrum of p-nitrophenol, which we first considered in Section 20.1. A molecule's mass spectrum is unique and contains information that we can use to deduce its structure. Interpretation of a mass spectrum relies on identifying possible sources for the loss of mass, such as the a \(\Delta m\) of 30 amu corresponding to the loss of NO, or a \(\Delta m\) of 46 amu corresponding to the loss of NO2. Some mass-to-charge ratios are recognized as evidence for a particular ion, such as C5H5+ at a mass-to-charge ratio of 65. The interpretation of fragmentation patterns is covered elsewhere in the curriculum, particularly in organic chemistry, and is not given more consideration here.Using Computer Searching. Large databases of mass spectra are available (see here for a source from NIST). A peak table of mass-to-charge ratios and peak intensities for a sample is entered into an algorithm that searches the database and identifies the most likely matches.Mass spectrometry is a powerful analytical technique when the sample we are analyzing is pure (or if impurities are of sufficiently low concentration that they have little effect on the mass spectrum). For a mixture of two or more analytes, the interpretation of the mass spectrum is difficult, if not impossible. To analyze such a mixture, we need a means of separating the analytes from each other. One approach is to interface a mass spectrometer to a gas chromatograph or a liquid chromatograph. The GC, or LC separates the mixture into its component parts with the mass spectrometer serving as the detector. See Chapter 27 and Chapter 28 for further details about GC-MS and LC-MS.Another approach to working with a complex sample is to use two or more mass analyzers in what is called tandem mass spectrometry. For example, if we place three quadrupole mass analyzers in a sequence, we can use a soft ionization source to generate mostly molecular ions of the form [M + H]+ for each of the sample's analytes, and then let the first quadrupole separate these molecular ions by the differences in their mass-to-charge ratio. The [M + H]+ molecular ions for one of the analytes is then selectively passed into the second quadrupole where it is allowed to undergo fragmentation by collision with a gas, such as He. Finally, these fragment ions are passed along to the third quadrupole where the mass spectrum is obtained. By sequentially passing each of the molecular ions from the first quadrupole through the second and third quadrupoles, we are able to obtain mass spectra for each molecule in the mixture.As a detector for other instrumental methods, such as gas chromatography and liquid chromatography, mass spectrometry provides for a quantitative analysis by monitoring either the total ion count or by monitoring ions of a single mass-to-charge ratio, which is known as selective ion monitoring. As an Independent method for determining an analyte's concentration, mass spectrometry is less attractive due to the difficulty of controlling the amount of sample or standard introduced into the instrument and the affect of the sample's matrix on fragmentation. The use of an internal standard improves precision and accuracy.This page titled 20.4: Applications of Molecular Mass Spectrometry is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

21.1: Introduction to the Study of Surfaces

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Thus far we have considered methods for analyzing the bulk properties of samples, such as determining the identity or concentration of an ion in a solution, of a molecule in a gas, or of several elements in a solid. In doing so, we did not concern ourselves with the sample's homogeneity or heterogeneity. In this chapter we give consideration to how we can gather information about the composition of a sample's surface and how it differs from the sample's bulk composition. But first, let's consider several important questions.A surface is a boundary, or interface, between two phases, such as a solid and a gas (the type of interface of particular interest to us in this chapter). This is a helpful, but not a sufficient description. Also of interest is the question of depth. Is a surface just the outermost layer of atoms, ions, or molecules, or does it extend several layers into the sample? In what ways might the composition of a sample at the surface differ from its composition in the sample's bulk interior? And, what about variations in composition across a surface? Is the surface itself homogeneous or heterogeneous in its composition? Different analytical methods will sample the surface to different depths and with different surface areas, which means the volume of sample analyzed will vary from method-to-method. For this reason, we usually define a sample's surface as what is analyzed by the analytical method we are using. shows the crystal structure of AgCl(s), which consists of a repeating pattern of Ag+ ions and Cl– ions. If you look at the ions in interior of the structure, you will see that each Ag+ ion is surrounded by six Cl– ions, and each Cl– ion is surrounded by six Ag+ ions. On the surface, however, we see that Cl– ions and Ag+ ions no longer are surrounded by six ions of opposite charge. As a result, the Ag+ ions and Cl– ions on the surface are more chemically reactive than those in the interior and can serve as sites for interesting chemistry. The chemical and physical properties of a sample's surface are likely to be very different than the sample's bulk properties.Suppose we are interested in studying the surface of a piece of zinc metal using a probe that samples just the outermost layer of atoms and that samples a circular surface area that is 1 µm2. How many atoms of Zn might we expect our probe to encounter? Here is some useful information about zinc: it has a molar mass of 65.38 g/mol, it has a density of 7.14 g/cm3, and it has an atomic radius of approximately 0.13 nm. From this we calculate the atoms per unit volume as\[\frac{7.14 \text{ g}}{\text{cm}^{3}} \times \frac{100 \text{ cm}}{\text{m}} \times \frac{1 \text{m}}{10^9 \text{ nm}} \times \frac{1 \text{ mol}}{65.38 \text{ g}} \times \frac{6.022 \times 10^{23} \text{ atoms}}{\text{mol}} = \frac{6.6 \times 10^{15} \text{ atoms}}{\text{cm}^2 \text{ nm}} \nonumber \]The units in the denominator may look odd to you, but writing them this way emphasizes that we are interested both in the depth from which information is received (given here in nanometers, nm) and in the surface area from which information is received (given here in square centimeters, cm2). Multiplying this value by the thickness of an atomic layer of zinc, which is twice its atomic radius, suggests we are analyzing approximately\[\frac{6.6 \times 10^{15} \text{ atoms}}{\text{cm}^2 \text{ nm}} \times 0.26 \text{ nm} = 1.7 \times 10^{15} \frac{\text{atoms}}{\text{cm}^2} \nonumber \]If we multiply this value by the surface area that we are sampling from, then we are interacting with approximately\[ 1.7 \times 10^{15} \frac{\text{atoms}}{\text{cm}^2} \times \left(\frac{100 \text{ cm}^2}{\text{m}}\right)^2 \times \left(\frac{1 \text{m}}{10^6 \text{µm}} \right)^2 = 1.7 \times 10^7 \text{ atoms of Zn} \nonumber \]Although 17 million may seem like a large number, it is not a particularly large number of atoms on which to carry out an analysis. Now, suppose the surface has a 10 ppm impurity of copper atoms; that is, there are 10 copper atoms for every 106 zinc atoms. In this case, our probe of the sample involves just\[1.7 \times 10^7 \text{ atoms of Zn} \times \frac{10 \text{ atoms of Cu}}{10^6 \text{ atoms of Zn}} = 170 \text{ atoms of Cu} \nonumber \]As a comparison, if we analyze a sample in which the analyte is present at a concentration that is \(1 \times 10^{-6} \text{ mol/L}\) using an analytical method that gathers information from a volume that is just 1 mm3, then we are sampling\[\frac{1 \times 10^{-6} \text{ mol}}{\text{L}} \times \frac{1 \text{L}}{1000 \text{ cm}^3} \times \left( \frac{1 \text{cm}}{10 \text{ mm}}\right)^3 \times 1 \text{ mm}^3 \times 6.022 \times 10^{23} \text{ mol}^{-1} = 6.0 \times 10^{11} \text{ particles of analyte} \nonumber \]An additional challenge when we attempt to analyze a surface is that a freshly exposed surface becomes contaminated with an absorbed layer of gas molecules almost instantly when sitting on a laboratory bench, and in a few seconds to a few minutes at pressures in the range of 10–6 torr to 10–8 torr. Analysis of a surface requires careful attention to how the surface is prepared.Compared to many of the methods in Chapters 6–20 and in Chapters 22–34, the use of a probe that samples from a small area allows for moving the probe across the surface—this is called rastering—developing a two-dimensional image of the surface. When using an energetic beam that can etch a hole in the sample, we can obtain information at depth—a process called depth profilling—that provides information in a third dimension. These are particularly important strengths of surface analytical methods.To study a surface, we put energy into it in the form of a beam of photons, electrons, or ions and then we measure the energy that exits the surface in the form of a beam of photons, electrons, or ions. Table \(\PageIndex{1}\) shows some of the possibilities. Also included in this table are methods in which an applied field generates a response from the surface. Entries in bold receive attention in this chapter. Surface enhanced Raman spectroscopy received a brief mention in Chapter 18. Note that Auger electron spectroscopy appears twice as the emission of electrons can follow the input of X-ray photons or electrons.energy out \(\rightarrow\)energy in \(\downarrow\)surface enhanced Raman spectroscopy (SERS)extended X-ray absorption fine structure (EXAFS)X-ray photoelectron spectroscopy (XPS)Auger electron spectroscopy (AES)UV-photoelectron spectroscopy (UPS)energy dispersive X-ray spectroscopy (EDS)electron microprobe (EM)Auger electron spectroscopy (AES)scanning electron microscopy (SEM)low energy electron diffraction (LEED)Rutherford back scattering (RBS)secondary ion mass spectrometry (SIMS)There are other ways to probe a surface by putting energy into it, including the application of thermal energy and the use of neutral species. See the text Methods of Surface Analysis, Czanderna, A. Editor, Elsevier: Amsterdam and the article "Analytical Chemistry of Surfaces" by D. M. Hercules and S. H. Hercules, J. Chem. Educ. 1984, 61, 402–409 for detailed reviews. Although neither is a recent publication, both provide an excellent introduction to surface analysis.This page titled 21.1: Introduction to the Study of Surfaces is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

21.2: Spectroscopic Surface Methods

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In this section we consider three representative surface analytical methods: X-ray photoelectron spectroscopy, in which the input is a beam of X-ray photons and the output is electrons; Auger electron spectroscopy, in which the input is either a beam of electrons or of X-ray photons and the output is electrons; and secondary-ion mass spectrometry, in which the input is a beam of ions and the output is ions.In X-ray photoelectron spectroscopy, which is also known as electron spectroscopy for chemical analysis (ESCA), we measure the kinetic energy of electrons ejected from a sample following the absorption of X-ray photons. The resulting spectrum is a count of these emitted electrons as a function of their energy.We can explain the origin of X-ray photoelectron spectroscopy using the photoelectric effect. shows the energy level diagram for an element's 1s, 2s, and 2p core-level electrons along with their KLM designations (see Chapter 12.1 for a previous discussion of this way of designating electrons). A nearly monoenergetic X-ray beam of known energy is focused on the sample, which results in the ejection of a core-level photoelectron, as shown in . The kinetic energy of this emitted electron, \(E_{KE}\), is related to its binding energy to the nucleus, \(E_{BE}\), by the equation\[E_{KE} = h \nu - E_{EB} - \Phi_w \label{xps1} \]where \(h \nu\) is the energy of the X-ray photon and \(\Phi_w\) is the spectrometer's work function (the energy needed to remove the electron from the surface and into the vacuum). The most common sources of X-rays are the Mg \(K_{\alpha}\) line with an energy of 1253.6 eV or the Al \(K{\alpha}\) line with an energy of 1486.6 eV.The core-level vacancy created by the photoelectron leaves the atom with an unstable electron configuration. Relaxation to the ground state occurs when this vacancy is filled by an electron from a higher energy shell, with the excess energy released as either the emission of a second electron or the fluorescent emission of a characteristic X-ray, as seen in . The secondary electron in is called an Auger electron. provides an example of an XPS survey spectrum for aluminum oxide, \(\ce{Al2O3}\), using the \(K_{\alpha}\) line for aluminum as the source of X-rays. The peak table gives the binding energies of the peaks for aluminum and for oxygen using the \(K_{\alpha}\) for Al and, for comparison, the \(K_{\alpha}\) line for Mg. Note the difference in how the major peaks are labeled. The photoelectrons ejected in the process shown in are designated by the element and the ns notation that specifies the orbital from which the electron was ejected. Auger electrons are designated using the KLM notation, specifying the initial vacancy created by the absorbed photon, the source of the electron that fills that vacancy, and the source of the ejected Auger electron. The OKLL peak in this spectrum is consistent with the scheme shown in . When the source of the second and third electrons is from the valance shell, then notation is sometimes written as KVV; the OKLL peak here could be designated as OKVV.There are a few additional interesting features to note in the survey spectrum for Al2O3. One is the presence of a peak for carbon even though the sample, Al2O3, does not contain carbon. The prevalence of carbon in the atmosphere means that trace levels of carbon appear in almost all XPS spectra.A second feature is the increase in the signal on the high binding energy (low kinetic energy) side of peaks, which is particularly visible here for the O1s peak the C1s peak, and the Al2s peak. The source of this background is electrons that fail to escape the sample without undergoing inelastic collisions that result in a loss of kinetic energy and, given Equation \ref{xps1}, are recorded as if they have a larger than expected binding energy. Because X-rays penetrate more deeply into the sample than the depth from which electrons can travel without undergoing an inelastic collision, this background is unavoidable. Note that the background is more significant at higher binding energies (smaller kinetic energies).A third feature is that the binding energy of an XPS peak is independent of the X-ray source, but the binding energy for an Auger peak varies with the X-ray source's energy (see table in ). The kinetic energy of the O1s, Al2s, and Al2p photoelectrons is the difference between the energy of the X-ray photon, \(h \nu\), and each electron's binding energy, BE; if we change the X-ray source, then \(h \nu\) and KE increase in value, but the BE remains fixed. For the OKLL Auger electron, the kinetic energy depends on the difference in the binding energies of the three electrons involved\[\text{KE} \approx \text{BE}_K - \text{BE}_L - \text{BE}_L \label{xps2} \]and is, therefore, independent of the energy of the X-ray source. Given Equation \ref{xps1}, if the KE remains constant, then an increase in the energy of the X-ray photon, \(h \nu\), means that the apparent BE must increase. This shift in binding energy when using a different source is one way to identify a peak as resulting from Auger electrons.The basic instrumentation for XPS is shown in . The most common X-ray sources, as noted above, are Mg (1253.6 eV) and Al (1486.6 ev), which have the advantage of relatively narrow line-widths (0.7 and 0.9 eV, respectively) and, therefore, a relatively narrow range of energies. Higher energy sources are available, such as Ag (2984.4 ev), but at the cost of a wider line-width (2.6 eV). A system of electron lenses collects and focuses the ejected electrons onto the entrance slit of a hemispherical analyzer. The path of an electron through the analyzer depends upon its kinetic energy. By varying the potentials applied to the hemispherical analyzer's inner and outer plates, electrons of different kinetic energies reach the detector. A sputtering gun is an optional feature that can be used to clean the surface of the sample or to remove successive layers of the sample, allowing for the gathering of spectra at various depths within the sample. Calibration of the spectrometer's binding energy scale, which accounts for the spectrometer's work function, is made using specific lines for one or more conductive metals; examples include Au 4f7/2 at 83.95 eV and Cu 2p3/2 at 932.63 ev. The peak for carbon that appears in almost all XPS spectra provides an additional way to check the calibration of the binding energy scale.X-ray photoelectron spectroscopy is a particularly useful tool for determining the composition and structure of a sample. XPS also can provide information about how the composition of a sample varies with depth and quantitative information about a sample's components.Qualitative Analysis. One of the strengths of X-ray photoelectron spectroscopy is the ability to determine the elements that make up a sample's surface. shows a survey scan from 0 eV to 1100 eV of a ceramic material. In addition to the O1s peak, we see strong peaks for Si and Al—probably an aluminosilicate ceramic—and small peaks for a variety of elements: La, Ba, Mn, Sn, Ca, Cl, P, and Mg. NIST maintains an extensive, and searchable, database of XPS peaks that help in identifying the elements in a sample.Chemical Shifts. An element's binding energy is sensitive to its chemical environment, particularly with respect to oxidation states and structure. For example, Table \(\PageIndex{1}\) provides the binding energy for chlorine's 2p line in three potassium salts drawn from the NIST database with all values from the same literature source. Using KCl, in which chlorine has an oxidation state of –1, as a baseline, the Cl 2p peak for KClO3 is shifted by +8.4 eV and the Cl 2p peak for KClO4 is shifted by +10.3 eV. The direction of the shift makes sense, as we expect that it will require more energy to remove an electron from an element that has a more positive oxidation state.Chemical shifts also reflect changes in chemical structure. , for example, shows a high resolution scan for the oxygen 2p peak for same sample of aluminum oxide, Al2O3, in . The surface of a metal oxide often has three distinct sources of oxygen: oxides, which make up the bulk of the sample, hydroxides that form at the surface following the chemisorption of water, and water that is physically absorbed to the surface. Curve-fitting of the raw data shows the contribution of each type of oxygen to the raw data and, through the peak areas, their relative abundance.Wagner Plots. Both the binding energy of an X-ray photoelectron and the kinetic energy of an Auger electron convey information about the element from which the electrons were emitted. A Wagner plot shows both the binding energy for a photoelectron that leaves a particular core-level vacancy and the kinetic energy of the Auger electron whose origin arises from the filling of this core-level vacancy. shows an example of a Wagner plot for copper based on its 2p3/2 X-ray photoelectron and its LMM Auger electron. The diagonal lines are called the modified Auger parameter, which is defined as the sum of the XPS binding energy and the AES kinetic energy. Values for 20 compounds are included in this plot. Of interest here is the clustering of the individual compounds. For example, all of the samples for which copper has an oxidation state of +1 (shown as magenta squares) have similar binding energies between 932 eV and 933 eV, but with more variable kinetic energies, which range from 914 eV to 917 eV. Most of the compounds in which copper has an oxidation state of +2 (shown as blue diamonds) have modified Auger parameters between approximately 1850 eV and 1851 eV, although there is signifiant variation in their individual binding energies and kinetic energies. The two metals (shown as green circles) and the commpounds CuS and CuSe occupy a similar space within the Wagner plot. Interestingly, both CuS and CuSe are transition metal chalcogenides and have semiconducting properties.Information at Depth. X-rays penetrate into a sample to a depth that is greater than the distance the photoelectron can travel without losing energy to inelastic collisions. We can take advantage of this to vary the depth from which we gather information. shows how this is accomplished by changing the angle from which we collect and analyze photoelectrons. The length of the solid black line is the distance an electron can travel without losing energy in an inelastic collision. When the detector is placed at 90° to the sample's surface, the sampling depth is at its greatest. Adjusting the detector so that it is at 30° to the surface, results in its detecting electrons from a depth that is just half of that when the detector is at 90°.Quantitative Analysis. The intensity of an XPS peak—either is peak height or its peak area—is proportional to the number of atoms of the element responsible for the peak. This allows for determining the relative concentration, \(C_x\), of an element in a sample; thus\[C_x = \frac {I_x / S_x} {\sum{(I_i / S_i)} } \label{xps3} \]where \(I_x\) is the peak intensity for the element, \(S_x\) is the sensitivity factor for the element, and \(I_i\) and \(S_i\) are the peak intensities and sensitivity factors for all other elements in the sample. Sensitivity factors account for differences in the ease with which photoelectrons are produced and escape from the sample. Published tables of sensitivity factors are available, although they may vary some from instrument-to-instrument. Sensitivity factors are referenced to a standard line, typically C1s, which is assigned a sensitivity factor of 1.00.In many cases we are interested only in the relative concentration of just two elements. In this case, we write Equation \ref{xps3} as\[\frac{C_x}{C_y} = \frac{I_x / S_x}{I_y / S_y} \label{xps4} \]where \(x\) and \(y\) are the two elements. For example, using the data in , the Si2p peak has a height of 30 mm and the Al2p peak has a peak height of 20 mm. Their sensitivity factors are, respectively 0.817 and 0.737. Using these values, we find that\[\frac{C_\text{Si}}{C_\text{Al}} = \frac{30/0.817}{20/0.737} = 1.4 \nonumber \]there are approximately \(1.4 \times\) as many atoms of silicon as there are atoms of aluminum.In we learned that following the ejection of a photoelectron from an atom's core, the now unstable atom releases energy by either emitting a secondary electron from a higher energy orbital or by releasing a photon. In XPS we measure the intensity of the ejected photoejected electrons as a function of their binding energy; in Auger spectroscopy we measure the intensity of these secondary electrons as a function of their kinetic energies.The instrumentation for AES can be coupled with an XPS spectrometer or be a stand-alone instrument. In either case, the basic instrumentation is similar to that shown in , although AES spectra are usually initiated using an electron gun as a source instead of an X-ray source. One advantage to using an electron gun is that it can be focused into a smaller beam and can then easily rastered across a surface, allowing for imaging of the sample's surface. Depth profiling, using an ion beam to remove layers of the sample, is another common use of AES. provides an example of a typical AES spectrum, in this case for the mineral calcite, CaCO3. The raw data, on the left, shows the intensity of the signal as a count of electrons with a particular kinetic energy. The large background signal is from electrons that lose kinetic energy as the result of inelastic collisions. The two broad features between approximately 100 eV and 600 eV are the Auger peaks for calcium and for oxygen. The peaks in this case are easy to see because the two analytes are present in bulk. For a trace-level analyte, the Auger peaks in a normal plot may be difficult to see. For this reason, Auger spectra are usually presented by plotting the derivative of the raw data giving the spectrum on the right. provides an example of how AES is used to study changes in the composition of a single crystal of CaCO3 that was allowed to equilibrate with a solution containing Mg2+ ions. The sample was mounted on a sample probe and the AES spectrum recorded. An Ar+ ion beam was used to remove layers of the sample while the spectrometer was used to record spectra of the sample. As expected, the surface is enriched in Mg2+ ions that diffused into the crystal with the relative concentration of Mg2+ decreasing. The relative abundance of Ca2+ increases with depth; the relative concentration of oxygen remains more or less constant with depth.In SIMS we bombard the surface of a sample with an energetic primary beam—typically 5-20 keV—of an ion, such as Ar+, O2+, or Cs+. The primary ion penetrates the sample's surface, ejecting a variety of particles, including neutral atoms, electrons, and, more importantly, secondary ions (singly-charged or multiply charged cations and anions, and clusters of ions). These secondary ions are characterized by their mass-to-charge ratios and by their kinetic energies. The use of a primary ion beam of Cs+ favors the formation of secondary ions that are anions, and a primary beam of O2+ favors the formation of secondary ions that are cations.The instrumentation for SIMS includes an ion gun for generating the primary beam and a mass spectrometer for analyzing the secondary ions. In static mode, the primary ion beam is run using a low current density that minimizes the extent to which the sample's outermost layers are removed. In dynamic mode, the primary beam is run at a higher current density that removes more of the sample's surface. Dynamic SIMS is well suited to depth profiling. High mass resolution is obtained using a time-of-flight mass analyzer or a double-focusing mass analyzer; see Chapter 20 for a discussion of different types of mass analyzers.SIMS is well suited for imaging as the positively charged primary ion can be rastered across the sample's surface. provides an example of imaging a surface using SIMS by measuring the yield of 14N12C ions while rastering the primary ion beam across a 10 µm \(\times \) 10 µm section of the sample.This page titled 21.2: Spectroscopic Surface Methods is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

21.3: Scanning Electron Microscopy

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In optical microscopy we use photons to provide images of a sample. Although extraordinarily useful and powerful, the ability to resolve features in optical microscopy is limited by the source of light; in general, we can distinguish between two objects if they are separated by a distance that is greater than the wavelength of the photons being used. The maximum resolution for an optical microscope is about 0.2 µm (200 nm), which means we can use an optical microscope to view a human hair (20-200 µm), an eukaryotic cell (10-100 µm), a chloroplast (5-8 µm), and a mitochondrion (1-3 µm), but not a ribosome (0.01 µm-0.02 µm). In this section we will consider the electron microscope, which has a resolution limit of approximately 0.2 nm, or approximately \(1000 \times\) more than an optical microscope. In Section 21.4, we will examine two additional types of non-optical microscopy.In scanning electron microscopy we raster a beam of high-energy electrons over a surface using a two-dimensional grid. shows the basic instrumental needs. The electron gun usually is just a simple tungsten wire that releases electrons when it is heated resistively. Other sources include solid-state crystals of lanthanum hexaboride (LaB6) or cerium hexaboride (CeB6) and the field emission gun, which uses a tungsten wire with a tip that has a radius of about 100 µm. Regardless of their source, these electrons are accelerated to an energy of 1-40 keV and passed through a series of lens that narrow and focus them into a beam with a diameter that falls within a range of 1 nm to 1000 nm (0.001 µm to 1 µm). A set of coiled scan controls deflects the electron beam in a raster pattern across the sample's surface (see inset at the bottom left of ). An electron detector monitors the electrons that scatter back from the sample; the type of detector used varies with the type of emission from the sample that we choose to monitor—see the next sub-heading for types of emission—but typically are scintillation devices when monitoring electrons and energy-dispersive detectors when monitoring X-rays. suggests that the only type of signal is the measurement of electrons that are scattered back toward the detector. The interaction between the electron beam and the sample, however creates a variety of signals, including both electrons and X-rays. illustrates the types of emission that follows from the interaction of the electron beam with the sample. The electron beam penetrates approximately 1-2 µm into the sample. As you might expect from the previous section on electron spectroscopy, the interaction of an electron beam with a sample results in the emission of some Auger electrons; these electrons come from a volume near the vacuum-sample interface. Of more importance are secondary electrons and backscattered electrons.As the electron beam penetrates into the sample, the electrons undergo collisions with the sample's atoms. Some of these collisions are elastic in which the electron changes its direction, but retains its kinetic energy. With sufficient time, these electrons eventually undergo a collision in which they cross the sample-vacuum interface and exit the solid. These backscattered electrons are collected and passed along to the detector. Other electrons undergo inelastic collisions, losing kinetic energy and, eventually, become embedded in the sample. Backscattered electrons come from a depth as great as 50% of the depth to which the electron beam penetrates. Another source of electrons comes from a process in which the electron beam induces the ejection of electrons from the sample's conduction band. These secondary electrons are less numerous than backscattered electrons and they also come from a much shallower depth, typically 5-50 nm.The electron beam also stimulates the release of X-rays, including the characteristic X-rays of the sample's elements, a broad continuum, and fluorescent X-ray emission. See Chapter 12 for more details about atomic X-ray emission.As the electron beam is rastered across the sample, the intensity of the backscattered electrons from a specific position on the sample that reaches the detector is stored in the corresponding pixel on the instrument's monitor. The image created in this way is not an optical picture, but a digitized electronic reproduction of the sample's surface. The extent of magnification depends on the length of the detector's monitor relative to the length of a single scan across the sample; scanning a shorter distance results in a greater magnification. An optical microscope usually provides a maximum magnification of \(1000 \times\); an SEM can achieve a magnification of \(1,000,000 \times\). shows four examples of applications of scanning electron microscopy for the measurement of particle size (upper left), for the evaluation of nanowires (upper right), for characterizing the channels in a microfluidic device (lower left), and for examining the tip of a cantilever and tip used for atomic force microscopy. Other applications include biological samples, films and coatings, fibers, and powders, to name a few.This page titled 21.3: Scanning Electron Microscopy is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

21.4: Scanning Probe Microscopes

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in the last section we considered how we can image a surface using an electron beam. In this section we consider a very different approach to developing an image of a surface, one in which we bring a probe close to the surface and examine how the probe interacts with the surface. One advantage of this approach is that the interaction between the probe and the surface can include attraction and repulsion, which opens up a third dimension to the image.In the scanning tunneling microscope we take advantage of the ability of a current to pass through the gap between the tip of a conducting probe and a conducting sample when the probe and the sample are held at different potentials. shows the basic arrangement in which the probe has, ideally, a single atom at its tip. The tunneling current, \(I_t\) is given by\[I_t = V e^{-Cd} \label{stm1} \]where \(V\) is the applied voltage, \(d\) is the distance between the probe's tip and the sample, and \(C\) is a constant whose value depends upon the composition of the probe and the sample. The exponential decrease in the tunneling current with distance means that a small change in the position of probe's tip relative to the sample results in a significant change in the signal, providing vertical resolution on the order of 0.1 nm. Probes are fashioned using tungsten wires or platinum-iridium wires.Scanning tunneling microscopy images are created by moving the probe back-and-forth across the sample while measuring the current. The signal is acquired in one of two modes: constant current or constant height. In constant current mode, the probe's tip is brought near the surface and the current measured, which establishes a setpoint. As the probe moves across the sample, it is raised or lowered to maintain the setpoint current. The result is measure of the distance, \(d\), between the probe's tip and the sample along the z-axis as a function of the xy position of the probe's tip. In constant height mode, the distance, \(d\), between the probe's tip and the sample is held constant, and the current, \(I_t\), is measured as a function of the xy position of the probe's tip. Constant height mode allows for faster data acquisition, but is limited to samples that have flat surfaces.Positioning of the sample and the probe's tip relative to each other is accomplished by either moving the probe or moving the sample. In either case, the control of movement in accomplished using a piezoelectric scanner. A piezoelectric material, as shown in experiences a change in its length when a dc potential is applied across its sides, either extending its length or contracting its length. shows a configuration of cylindrical piezoelectric scanner in which the cylinder's upper half controls movement along the z-axis, and the cylinder's lower half is used to control movement along the x-axis, the y-axis, or both.One limitation of STM is that the sample must be conductive. It is possible to image a non-conducting sample if it first coated with a conductive material, such as gold, although such coatings can mask surface features.Unlike the scanning tunneling microscope, the atomic force microscope does not require a conducting sample and imaging is achieved without a current flowing between the sample and the tip of the probe. Instead, as shown in , the probe is attached to the end of a flexible cantilever. The tip of the probe (see photograph in ) is pyramidal in shape and extends about 10 µm from its base on the cantilever. The tip of the probe has a diameter on the order of 10 nm and is made of silicon, Si, or silicon nitride, Si3N4. The cantilever typically is 100-500 µm in length. The probe is scanned across the sample's surface and the position of the probe relative to the surface is determined by reflecting the beam from a diode laser off the probe-end of the cantilever to a detector.The force in atomic force is the interaction between the probe's tip and the sample, which may be a force of attraction or a force of repulsion. When the probe's tip is in contact with the sample—known as contact mode—there is a force of repulsion between them. Because the cantilever has a smaller force constant than the atoms in the probe's tip, the cantilever bends. Moving the sample stage to maintain a constant deflection of the laser off of the cantilever provides an image of the sample's surface. Contact mode allows for rapid scanning and work well for samples with rough surfaces, although it may damage samples with softer surfaces.In non-contact mode, the probe's tip is brought close to the sample's surface, but not allowed to come into contact with it. The cantilever is place into an oscillatory motion. The amplitude of this oscillation is proportional to the force of attraction between the probe's tip and the sample, which varies with the distance between the probe's tip and the sample. Moving the sample stage to maintain a constant oscillation provides an image of the sample's surface. Non-contact mode AFM generally provides lower resolution images, but is less damaging to the sample.A third mode for collecting data is called intermittent or tapping mode. In this mode the cantilever is set to oscillate at its resonant frequency with the probe's tip coming into contact with the sample's surface when it reaches the bottom of the cantilever's oscillation. The frequency of the oscillation is sensitive to the distance between the probe's tip and the sample. Moving the sample stage to maintain the resonant frequency provides an image of the sample's surface.You can view a gallery of scanning tunneling microscopy images here, and a gallery of atomic force microscopy images here.This page titled 21.4: Scanning Probe Microscopes is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

22.1: Electrochemical Cells

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A schematic diagram of a typical electrochemical cell is shown in . The electrochemical cell consists of two half-cells, each of which contains an electrode immersed in a solution of ions whose activities determine the electrode’s potential. A salt bridge that contains an inert electrolyte, such as KCl, connects the two half-cells. The ends of the salt bridge are fixed with porous frits, which allow the electrolyte’s ions to move freely between the half-cells and the salt bridge. This movement of ions in the salt bridge completes the electrical circuit, allowing us to measure the potential using a potentiometer.The reason for separating the electrodes is to prevent the oxidation reaction and the reduction reaction from occurring at the same electrode. For example, if we place a strip of Zn metal in a solution of AgNO3, the reduction of Ag+ to Ag occurs on the surface of the Zn at the same time as a portion of the Zn metal oxidizes to Zn2+. Because the transfer of electrons from Zn to Ag+ occurs at the electrode’s surface, we can not pass them through the potentiometer.Current moves through the cell in as a result of the movement of two types of charged particles: electrons and ions. First, when zinc, Zn(s) underoges an oxidation reaction\[\mathrm{Zn}(s) \rightleftharpoons \text{ Zn}^{2+}(a q)+2 e^{-} \label{ox_rxn} \]it releases two electrons. These electrons move through the circuit that connects the metallic Zn electrode in the left half-cell to the metallic Ag electrode in the right half-cell, where it effects the reduction of Ag+(aq).\[\mathrm{Ag}^{+}(a q)+e^{-} \rightleftharpoons \mathrm{Ag}(s) \label{red_rxn} \]If this is all that happens, then the half-cell on the left will develop an excess of positive charge as Zn2+(aq) ions accumulate and the half-cell on the right will develop an excess of negative charge due to the loss of Ag+(aq). The salt bridge provides a way to continue the movement of charge, and thus the current, with the K+ ions moving toward the right half-cell and Cl– ions moving toward the left half-cell.The net reaction for the electrochemical cell in is\[\mathrm{Zn}(s)+2 \mathrm{Ag}^{+}(a q) \rightleftharpoons 2 \mathrm{Ag}(s)+\mathrm{Zn}^{2+}(\mathrm{aq}) \label{net_rxn} \]which simply is the result of adding together the reactions in the two half-cells after adjusting for the difference in electrons. As shown by the arrows in the figure, when we connect the electrodes to the potentiometer, current spontaneously flows from the left half-cell to the right half-cell. We call this a galvanic cell. If we apply a potential sufficient to reverse the direction of the current flow, resulting in a net reaction of\[2 \mathrm{Ag}(s)+\mathrm{Zn}^{2+}(\mathrm{aq}) \rightleftharpoons \mathrm{Zn}(s)+2 \mathrm{Ag}^{+}(a q) \nonumber \]then we call the system an electrolytic cell. A galvanic cell produces electrical energy and an electrolytic cell consumes electrical energy.The half-cell where oxidation takes place is called the anode and, by convention, it is shown on the left for a galvanic cell. The half-cell where reduction takes place is called the cathode and, by convention, it is shown on the right for a galvanic cell.When we oxidize or reduce an analyte at the electrode in one half-cell, the electrons pass through the potentiometer to the electrode in the other half-cell where a corresponding reduction or oxidation reaction takes place. In either case, the current from the redox reactions at the two electrodes is called a faradaic current. A faradaic current due to the reduction of an analyte is called a cathodic current and carries a positive sign. An anodic current results from the analyte’s oxidation and carries a negative sign.In addition to the faradaic current from a redox reaction, the current in an electrochemical cell includes non-faradaic sources. Suppose the charge on an electrode is zero and we suddenly change its potential so that the electrode’s surface acquires a positive charge. Cations near the electrode’s surface will respond to this positive charge by migrating away from the electrode; anions, on the other hand, will migrate toward the electrode. This migration of ions occurs until the electrode’s positive surface charge and the excess negative charge of the solution near the electrode's surface are equal. Because the movement of ions and the movement of electrons are indistinguishable, the result is a small, short-lived non-faradaic current that we call the charging current. Every time we change the electrode’s potential, a short-lived charging current flows.Even in the absence of analyte, a small, measurable current flows through an electrochemical cell. This residual current has two components: a faradaic current due to the oxidation or reduction of trace impurities and a non-faradaic charging current. Methods for discriminating between the analyte’s faradaic current and the residual current are discussed later in this chapter.As noted in the previous section, when we apply a potential to an electrode it develops a positive or negative surface charge, the magnitude of which is a function of the metal and the applied potential. Because the surface carries a charge, the composition of the layer of solution immediately adjacent to the electrode changes with, for example, the concentration of cations increasing and the concentration of anions decreasing if the electrode's surface carries a negative charge. As we move away from the electrode's surface, the net potential first decreases in a linear manner, due to the imbalance of the cations and anions, and then in an exponential manner until it reaches zero. This structured surface is called the electrical double layer and consists of an inner layer and a diffuse layer. Anytime we change the potential applied to the electrode, the structure of the electrical double layer changes and a small charging current flows.The magnitude of a faradaic current is determined by the rate at which the analyte is oxidized at the anode or reduced at the cathode. Two factors contribute to the rate of an electrochemical reaction: the rate at which the reactants and products are transported to and from the electrode—what we call mass transport—and the rate at which electrons pass between the electrode and the reactants and products in solution.There are three modes of mass transport that affect the rate at which reactants and products move toward or away from the electrode surface: diffusion, migration, and convection. Diffusion occurs whenever the concentration of an ion or a molecule at the surface of the electrode is different from that in bulk solution. For example, if we apply a potential sufficient to completely reduce \(\text{Ag}^+\) at the electrode surface, the result is a concentration gradient similar to that shown in . The region of solution over which diffusion occurs is the diffusion layer. In the absence of other modes of mass transport, the width of the diffusion layer, \(\delta\), increases with time as the \(\text{Ag}^+\) must diffuse from an increasingly greater distance.Convection occurs when we mix the solution, which carries reactants toward the electrode and removes products from the electrode. The most common form of convection is stirring the solution with a stir bar; other methods include rotating the electrode and incorporating the electrode into a flow-cell.The final mode of mass transport is migration, which occurs when a charged particle in solution is attracted to or repelled from an electrode that carries a surface charge. If the electrode carries a positive charge, for example, an anion will move toward the electrode and a cation will move toward the bulk solution. Unlike diffusion and convection, migration affects only the mass transport of charged particles.Although provides a useful picture of an electrochemical cell, it is not a convenient way to represent it. Imagine having to draw a picture of each electrochemical cell you are using! A more useful way to describe an electrochemical cell is a shorthand notation that uses symbols to identify different phases and that lists the composition of each phase. We use a vertical slash (|) to identify a boundary between two phases where a potential develops, and a comma (,) to separate species in the same phase or to identify a boundary between two phases where no potential develops. Shorthand cell notations begin with the anode and continue to the cathode. For example, we describe the electrochemical cell in using the following shorthand notation.\[\text{Zn}(s) | \text{ZnCl}_2(aq, a_{\text{Zn}^{2+}} = 0.0167) || \text{AgNO}_3(aq, a_{\text{Ag}^+} = 0.100) | \text{Ag} (s) \nonumber \]The double vertical slash (||) represents the salt bridge, the contents of which we usually do not list. Note that a double vertical slash implies that there is a potential difference between the salt bridge and each half-cell.What are the anodic, the cathodic, and the overall reactions responsible for the potential of the electrochemical cell in ? Write the shorthand notation for the electrochemical cell.The oxidation of Ag to Ag+ occurs at the anode, which is the left half-cell. Because the solution contains a source of Cl–, the anodic reaction is\[\mathrm{Ag}(s)+\mathrm{Cl}^{-}(aq) \rightleftharpoons\text{ AgCl}(s)+e^{-} \nonumber \]The cathodic reaction, which is the right half-cell, is the reduction of Fe3+ to Fe2+.\[\mathrm{Fe}^{3+}(a q)+e^{-}\rightleftharpoons \text{ Fe}^{2+}(a q) \nonumber \]The overall cell reaction, therefore, is\[\mathrm{Ag}(s)+\text{ Fe}^{3+}(a q)+\text{ Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}(s)+\text{ Fe}^{2+}(a q) \nonumber \]The electrochemical cell’s shorthand notation is\[\text{Ag}(s) | \text{HCl} (aq, a_{\text{Cl}^{-}} = 0.100), \text{AgCl} (\text{sat’d}) || \text{FeCl}_2(aq, a_{\text{Fe}^{2+}} = 0.0100), \text{ Fe}^{3+}(aq,a_{\text{Fe}^{3+}} = 0.0500) | \text{Pt} (s) \nonumber \]Note that the Pt cathode is an inert electrode that carries electrons to the reduction half-reaction. The electrode itself does not undergo reduction. This page titled 22.1: Electrochemical Cells is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

22.2: Potentials in Electroanalytical Cells

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If a Zn(s) electrode in a solution of Zn2+(aq) in an electrochemical cell is at equilibrium, the current is zero and the potential is fixed in value. If we change the potential from its equilibrium value, current will flow as the system moves to its new equilibrium position. Although the initial current is quite large, it decreases over time, reaching zero when the reaction reaches equilibrium. The current, therefore, changes in response to the applied potential. Alternatively, we can pass a fixed current through the electrochemical cell, forcing the oxidation of Zn(s) to Zn2+(aq), effecting a change in the potential. In short, if we choose to control the potential, then we must accept the resulting current, and we must accept the resulting potential if we choose to control the current.Because a redox reaction involves a transfer of electrons from a reducing agent to an oxidizing agent, it is convenient to consider the reaction’s thermodynamics in terms of the electron. For a reaction in which one mole of a reactant undergoes oxidation or reduction, the net transfer of charge, q, in coulombs is\[q=n F \label{q} \]where n is the moles of electrons per mole of reactant, and F is Faraday’s constant (96485 C/mol). The free energy, ∆G, to move this charge given an applied potential, E, is\[\Delta G=E q \label{deltaG1} \]The change in free energy (in kJ/mole) for a redox reaction, therefore, is\[\Delta G=-n F E \label{deltaG2} \]where ∆G has units of kJ/mol. The minus sign in Equation \ref{deltaG2} is the result of a different convention for assigning a reaction’s favorable direction. In thermodynamics, a reaction is favored when ∆G is negative, but a redox reaction is favored when E is positive. Substituting Equation \ref{deltaG2} into the thermodynamic equation that relates the free energy to its standard state value\[\Delta G = \Delta G^{\circ} + RT\ln Q_r \label{deltaG3} \]gives\[-n F E = -n F E^{\circ}+R T \ln Q_r \label{nfe} \]Dividing by –nF leads to the Nernst equation\[E=E^{\circ}-\frac{R T}{n F} \ln Q_r \label{nerst1} \]where Eo is the potential under standard‐state conditions (more on this in Section 22.3). Substituting appropriate values for R and F, assuming a temperature of 25 oC (298 K), and switching from the natural logarithm (ln) to the base 10 logarithm (log) gives the potential in volts as\[E=E^{\mathrm{o}}-\frac{0.05916}{n} \log Q_r \label{nernst2} \]The term \(Q_r\) in the previous equations is the reaction quotient, which has the same mathematical form as the reaction's equilibrium constant expression, but uses the instantaneous amounts of reactants and products in place of their equilibrium values. For the cell in +2 \mathrm{Ag}^{+}(a q) \rightleftharpoons 2 \mathrm{Ag}(s)+\mathrm{Zn}^{2+}(\mathrm{aq}) \label{net_rxn} \]and Equation \ref{nernst2} becomes\[E = E^{\circ} - \frac{0.05916}{2} \log \frac {\left[ \ce{Zn^{2+}} \right]} {\left[ \ce{Ag+} \right]^2} \label{zn_ag} \]Equation \ref{zn_ag} shows us how the potential changes as the concentrations of Zn2+ and Ag+ change.As we will see in 22.3, \ref{zn_ag} actually is expressed in terms of activities instead of concentrations. The appendix in Chapter 35.7 explains what activity is, why it is important to make a distinction between activity and concentration, and when it is reasonable to use concentrations in place of activities.A junction potential develops at the interface between two ionic solutions if there is a difference in the concentration and the mobility of the ions. Consider, for example, a porous membrane that separates a solution of 0.1 M HCl from a solution of 0.01 M HCl a). Because the concentration of HCl on the membrane’s left side is greater than that on the right side of the membrane, H+ and Cl– will diffuse in the direction of the arrows. The mobility of H+, however, is greater than that for Cl–, as shown by the difference in the lengths of their respective arrows. Because of this difference in mobility, the solution on the right side of the membrane develops an excess concentration of H+ and a positive charge b). Simultaneously, the solution on the membrane’s left side develops a negative charge because there is an excess concentration of Cl–. We call this difference in potential across the membrane a junction potential and represent it as Ej.The magnitude of a junction potential depends upon the difference in the concentration of ions on the two sides of the interface, and may be as large as 30–40 mV. For example, a junction potential of 33.09 mV has been measured at the interface between solutions of 0.1 M HCl and 0.1 M NaCl [Sawyer, D. T.; Roberts, J. L., Jr. Experimental Electrochemistry for Chemists, Wiley-Interscience: New York, 1974, p. 22]. A salt bridge’s junction potential is minimized by using a salt, such as KCl, for which the mobilities of the cation and anion are approximately equal. We also can minimize the junction potential by incorporating a high concentration of the salt in the salt bridge. For this reason salt bridges frequently are constructed using solutions that are saturated with KCl. Nevertheless, a small junction potential, generally of unknown magnitude, is always present.This page titled 22.2: Potentials in Electroanalytical Cells is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

22.3: Electrode Potentials

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We began this chapter by examining the electrochemical cell in }\) is oxidized to \(\ce{Zn^{2+}(aq)}\) and \(\ce{Ag^{+}(aq)}\) is reduced to \(\ce{Ag(s)}\), as shown by the following reaction.\[2 \mathrm{Ag}(s)+\mathrm{Zn}^{2+}(\mathrm{aq}) \rightleftharpoons \mathrm{Zn}(s)+2 \mathrm{Ag}^{+}(a q) \nonumber \]The reaction proceeds as written because the reduction of Ag+(aq) to Ag(s)\[\mathrm{Ag}^{+}(a q)+e^{-} \rightleftharpoons \mathrm{Ag}(s) \label{red_ag} \]is more thermodynamically favorable than the reduction of \(\ce{Zn^{2+}(aq)}\) to \(\ce{Zn(s)}\)\[\text{ Zn}^{2+}(aq)+2 e^{-} \rightleftharpoons \mathrm{Zn}(s) \label{red_zn} \]But, how do we know this is true? In this section we answer this question by taking a close look at electrode potentials.The potential of an electrochemical cell is the difference between the potential at the cathode, \(E_\text{cathode}\), and the potential at the anode, \(E_\text{anode}\), where both potentials are defined in terms of a reduction reaction (and are called reduction potentials); thus\[E_\text{cell} = E_\text{cathode} - E_\text{anode} \label{cell_pot} \]\[\mathrm{H}^{+}(a q)+e^{-}=\frac{1}{2} \mathrm{H}_{2}(g) \label{she} \]which is the reaction that defines the standard hydrogen electrode, or SHE.The SHE consists of a Pt electrode immersed in a solution in which the activity of hydrogen ion is 1.00 and in which the partial pressure of H2(g) is 1.00 atm ). A conventional salt bridge connects the SHE to the indicator half-cell. The short hand notation for the standard hydrogen electrode is\[\text{Pt}(s), \text{ H}_{2}\left(g, f_{\mathrm{H}_{2}}=1.00\right) | \text{ H}^{+}\left(a q, a_{\mathrm{H}^{+}}=1.00\right) \| \label{she_cell} \]and the standard-state potential for the reaction \ref{she} is, by definition, 0.000 V at all temperatures.Although the standard hydrogen electrode is the standard against which all other potentials are referenced, it is not practical for routine use as it is difficult to prepare and maintain. Instead, we use one of several other reference electrodes. The two most common of these alternative reference electrodes are the calomel, or Hg/Hg2Cl2 electrode, which is based on the following redox couple between Hg2Cl2 and Hg (calomel is the common name for Hg2Cl2)\[\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)+2 e^{-}\rightleftharpoons2 \mathrm{Hg}(l)+2 \mathrm{Cl}^{-}(a q) \nonumber \]and the Ag/AgCl reference electrode, which is based on the reduction of AgCl to Ag\[\operatorname{AgCl}(s)+e^{-} \rightleftharpoons \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q) \nonumber \]A more detailed examination of these two reference electrodes is found in Chapter 23.1.To determine the potential for the reduction of Zn2+(aq) to Zn(s) we make it the cathode in the following electrochemical cell\[\text{Pt}(s), \text{ H}_{2}\left(g, P_{\mathrm{H}_{2}}=1.00\right) | \text{ H}^{+}\left(a q, a_{\mathrm{H}^{+}}=1.00\right) \| \ce{Zn^{2+}}\left(a q, a_{\mathrm{Zn}^{2+}}=x\right) | \ce{Zn}(s) \label{she_zn} \]where x is the activity of Zn2+ in its half-cell. For example, when \(a_{\mathrm{Zn}^{2+}} = 1.00\), the potential of the electrochemical cell is \(-0.763 \text{V}\). If we find that the potential for the electrochemical cell\[\ce{Zn}(s) | \ce{Zn^{2+}} (aq, a_{\mathrm{Zn}^{2+}} = 1.00) \| \ce{Ag+} (aq, a_{\mathrm{Ag}^{+}} = 1.00) | \ce{Ag}(s) \nonumber \]is +1.562 V, then knowing that\[E_{cell} = E_{\ce{Ag+} / \ce{Ag}} - E_{\ce{Zn^{2+}} / \ce{Zn}} = E_{\ce{Ag+} / \ce{Ag}} - (-0.763 \text{V}) \nonumber \]gives \(E_{\ce{Ag+} / \ce{Ag}} = +0.799\). In this way, we can build tables of potentials for individual half-reactions.In Section 22.2 we noted the following relationship between an electrochemical potential, \(E\), and the Gibbs free energy, \(\Delta G\)\[\Delta G = - n F E \label{dg} \]which tells us that a positive potential corresponds to a thermodynamically favorable reaction. Knowing that the potential for the electrochemical cell in Equation \ref{she_zn} is \(-0.763 \text{V}\) tells us that the reduction of Zn2+(aq) to Zn(s) is not thermodynamically favorable relative to the reduction of H+(aq) to H2(g); that is, we do not expect the reaction\[\ce{Zn^{2+}}(aq) + \ce{H2}(g) \rightleftharpoons 2 \ce{H+}(aq) + \ce{Zn}(s) \label{zn_h} \]to occur; however, with a potential of +0.799 V, we do expect the reaction\[2 \ce{Ag+}(aq) + \ce{H2}(g) \rightleftharpoons 2 \ce{H+}(aq) + 2 \ce{Ag}(s) \label{ag_h} \]to occur. Or, looking at this another way, we expect that Zn(s), but not Ag(s), will dissolve in acid.In Chapter 22.2 we wrote the Nernst equation for the reaction\[\mathrm{Zn}(s)+2 \mathrm{Ag}^{+}(a q) \rightleftharpoons 2 \mathrm{Ag}(s)+\mathrm{Zn}^{2+}(\mathrm{aq}) \label{net_rxn} \]in terms of the concentrations of Zn2+(aq) and Ag+(aq)\[E = E^{\circ} - \frac{0.05916}{2} \log \frac {\left[ \ce{Zn^{2+}} \right]} {\left[ \ce{Ag+} \right]^2} \label{zn_ag_conc} \]Although there are times when we will write the Nernst equation in terms of concentrations, thermodynamic functions are more correctly written in terms of the activities of ions. Under ideal conditions, individual ions and molecules of gases behave as independent particles. When this is true, then an ion's activity and concentration are equal and we can write the Nernst equation using concentrations; under other conditions, then the Nernst equation is more correctly written in terms of activities\[E = E^{\circ} - \frac{0.05916}{2} \log \frac {a_{\ce{Zn^{2+}}}} {\left( a_{\ce{Ag+}} \right)^2} \label{zn_ag_activity} \]where \(a_{\ce{Zn^{2+}}}\) and \(a_{\ce{Ag+}}\) are the activities of Zn2+ and Ag+. Equation \ref{zn_ag_activity} shows us how the potential changes as the activities of Zn2+ and Ag+ change.If you are not familiar with activity, or need a reminder on the relationship between activity and concentration, then see the appendix in Chapter 35.7, which explains what activity is, why it is important to make a distinction between activity and concentration, and when it is reasonable to use concentrations in place of activities.The standard electrode potential, \(E^{\circ}\), for a half-reaction is the potential when all species are present at unit activity or, for gases, unit fugacity. Its value is independent of how we choose to write the half-reaction; that is, the standard state potential for the reduction of Ag+(aq) to Ag(s), which is the cathode in the electrochemical cell in + e^{-} \rightleftharpoons \ce{Ag}(s) \label{ag1} \]or as\[2 \ce{Ag+}(aq) + e^{-} \rightleftharpoons 2 \ce{Ag}(s) \label{ag2} \]At first glance, this seems counterintuitive; however, if we calculate the potential when the activity of Ag+ is 0.50 we get\[E = E^{\circ} - \frac{0.05916}{1} \log \frac{1}{a_{\ce{Ag+}}} = 0.799 - \frac{0.05916}{1} \log \frac{1}{0.50} = 0.781 \text{V} \nonumber \]when using reaction \ref{ag1}, and\[E = E^{\circ} - \frac{0.05916}{2} \log \frac{1}{(a_{\ce{Ag+}})^2} = 0.799 - \frac{0.05916}{2} \log \frac{1}{0.50^2} = 0.781 \text{V} \nonumber \]The appendix in Chapter 35.8 provides a table of standard state reduction potentials for a wide variety of half-reactions at 298 K.Although standard electrode potentials are valuable, they are several important limitations to their use, which we outline here.One important limitation is that that the Nernst equation is defined in terms of the activity of ions instead of their concentrations. Although it is easy to prepare a solution for which the concentration of Na+ is 0.100 M using NaCl—just weigh out 5.844 g of NaCl and dissolve in 1.00 L of water—it is much more challenging to prepare a solution for which the activity of Na+ is 0.100. For this reason, in calculations we usually substitute concentrations for activities when using the Nernst equation. This simplification generally is okay for dilute solutions where the difference between activities and concentrations are small.A standard state potential tells us about the equilibrium position of a redox half-reaction reaction under standard state conditions. If one or more of the species in the half-reaction are involved in other equilibrium reactions, then these reactions will affect the value of the standard potential. For example, Fe2+ and Fe3+ form a variety of metal-ligand complexes with Cl– which explains why \(E_{\ce{Fe^{3+}}/\ce{Fe^{2+}}}^{\circ}\) is 0.771 in the absence of chloride ion, but is 0.70 in 1 M HCl.One way to compensate for using concentrations and partial pressures in place of activities and fugacities, and to compensate for other equilibrium reactions, is to replace the standard state potentials, \(E^{\circ}\) with a formal potential, \(E^{\circ \prime}\), that is measured using concentrations of 1.00 for ions, partial pressures of 1.00 for gases, and for a specific concentration of other reagents. The table below, which is adapted from the appendix in Chapter 35.8, provides formal potentials for Fe3+/Fe2+ half-reaction in five different solvents.0.70 in 1 M \(\ce{HCl}\)0.767 in 1 M \(\ce{HClO4}\)0.746 in 1 M \(\ce{HNO3}\)0.68 in 1 M \(\ce{H2SO4}\)0.44 in 0.3 M \(\ce{H3PO4}\)The reduction of Fe3+ to Fe2+ consumes an electron, which is drawn from the electrode. The oxidation of another species, perhaps the solvent, at a second electrode is the source of this electron. Because the reduction of Fe3+ to Fe2+ consumes one electron, the flow of electrons between the electrodes—in other words, the current—is a measure of the rate at which Fe3+ is reduced. One important consequence of this observation is that the current is zero when the reaction \(\text{Fe}^{3+}(aq) \rightleftharpoons \text{ Fe}^{2+}(aq) + e^-\) is at equilibrium. If redox half-reaction cannot maintain an equilibrium because the reaction in one direction is too slow, then we cannot measure a meaningful standard state potential.This page titled 22.3: Electrode Potentials is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

22.4: Calculation of Cell Potentials from Electrode Potentials

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The potential of an electrochemical cell is the difference between the electrode potentials of the cathode and the anode\[E_\text{cell} = E_\text{cathode} - E_\text{anode} \label{ecell} \]where \(E_\text{cathode}\) and \(E_\text{anode}\) are both reduction potentials. Given a set of conditions, we can use the Nernst equation to calculate the cell potential, as shown by the following example.Calculate (a) the standard state potential and (b) the potential when [Ag+] = 0.0020 M and [Cd2+] = 0.0050 M, for the following reaction at 25oC.\[\mathrm{Cd}(s)+2 \mathrm{Ag}^{+}(a q)\rightleftharpoons2 \mathrm{Ag}(s)+\mathrm{Cd}^{2+}(a q) \nonumber \]For part (b), calculate the potential twice, once using concentrations and once using activities assuming that the solution's ionic strength is 0.100.(a) In this reaction Cd is oxidized at the anode and Ag+ is reduced at the cathode. Using standard state electrode potentials from Appendix 3, we find that the standard state potential is\[E^{\circ} = E^{\circ}_{\text{Ag}^+/ \text{Ag}} - E^{\circ}_{\text{Cd}^{2+}/ \text{Cd}} = 0.7996 - (-0.4030) = 1.2026 \ \text{V} \nonumber \](b) To calculate the potential when [Ag+] is 0.0020 M and [Cd2+] is 0.0050 M, we use the appropriate relationship for the reaction quotient, Qr, when writing the Nernst equation\[E = E^{\circ} - \frac{0.05916 \ \mathrm{V}}{n} \log \frac{\left[\mathrm{Cd}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}} \nonumber \]\[E=1.2026 \ \mathrm{V}-\frac{0.05916 \ \mathrm{V}}{2} \log \frac{0.050}{(0.020)^{2}}=1.14 \ \mathrm{V} \nonumber \]To calculate the potential using activities, we first calculate the activity coefficients for Cd2+ and Ag+. Following the approach outlined in the appendix in Chapter 35.7 gives\[\log \gamma_{\ce{Cd^{2+}}} = \frac {-0.51 \times (+2)^2 \times \sqrt{0.100}} {1 + 3.3 \times 0.50 \times \sqrt{0.100}} = -0.2078 \nonumber \]\[]log \gamma_{\ce{Ag^{+}}} = \frac {-0.51 \times (+1)^2 \times \sqrt{0.100}} {1 + 3.3 \times 0.25 \times \sqrt{0.100}} = -0.1279 \nonumber \]\[a_{\ce{Cd^{2+}}} = \gamma_{\ce{Cd^{2+}}} \times [\ce{Cd}^{2+}] = 0.6197 \times 0.0050 = 0.003098 \nonumber \]\[a_{\ce{Ag^{+}}} = \gamma_{\ce{Ag^{+}}} \times [\ce{Ag}^{+}] = 7449 \times 0.0020 = 0.00149 \nonumber \]Finally, we substitute activities for concentrations in the Nernst equation to arrive at a potential of\[E=1.2026 \ \mathrm{V}-\frac{0.05916 \ \mathrm{V}}{2} \log \frac{0.003098}{(0.00149)^{2}}=1.11 \ \mathrm{V} \nonumber \]This page titled 22.4: Calculation of Cell Potentials from Electrode Potentials is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

22.5: Currents in Electrochemical Cells

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Most electrochemical techniques rely on either controlling the current and measuring the resulting potential, or controlling the potential and measuring the resulting current; only potentiometry (see Chapter 23) measures a potential under conditions where there is essentially no current. Understanding the relationship between current, i, and potential, E, is important. Although we learned in Sections 22.3 and 22.4 how to calculated electrode potentials and cell potentials using the Nernst equation, the experimentally measured potentials may differ from their thermodynamic values for a variety of reasons that we outline here.The movement of an electrical charge in an electrochemical cell generates a potential, \(E_{ir}\) defined by Ohm's law\[E_{ir} = iR \label{ohm} \]where i is the current and R is the solution's resistance. To account for this, we can include an additional term to the equation for the electrochemical cell's potential\[E_\text{cell} = E_\text{cathode} - E_\text{anode} - E_{ir} = E_\text{Nernst} - iR \label{cellpot} \]where \(E_\text{Nernst}\) is the potential from the Nernst equation. The resulting decease in the potential from its idealized value is called the iR drop.Equation \ref{cellpot} indicates that we expect a linear relationship between an electrochemical cell's potential, \(E_\text{cell}\). When this is not the case, the electrochemical cell is said to be polarized. There are several sources that contribute to polarization, which we consider in this section; first, however, we define ideal polarized and nonpolarized electrodes.An ideal polarized electrode is one in which a change in potential over a fairly wide range has no effect on the current that flows through the electrode, as we see in for the range of potentials defined by the solid green line. Such electrodes are useful because they do not themselves undergo oxidation or reduction—they are electrochemically inert—which makes them a good choice for studying the electrochemical behavior of other species.An ideal nonpolarized electrode is one in which a change in current has no effect on the electrode's potential, as we see in between the limits defined by the solid red line with deviations shown by the dashed red line. Such electrodes are useful because the provided a stable potential against which we can reference the redox potential of other species.The magnitude of polarization when drawing a current is called the overpotential, \(\eta\) and expressed as the difference between the applied potential, E, and the potential from the Nernst equation.\[\eta = E - E_\text{Nernst} \label{overpot} \]The overpotential can be subdivided into a variety of sources, a few of which are discussed below.The reduction of Fe3+ to Fe2+ in an electrochemical cell consumes an electron, which is drawn from the electrode. The oxidation of another species, perhaps the solvent, at a second electrode is the source of this electron. Because the reduction of Fe3+ to Fe2+ consumes one electron, the flow of electrons between the electrodes—in other words, the current—is a measure of the rate at which Fe3+ is reduced.The rate of the reaction \(\text{Fe}^{3+}(aq) \rightleftharpoons \text{ Fe}^{2+}(aq) + e^-\) is the change in the concentration of Fe3+ as a function of time.In order for the reduction of Fe3+ to Fe2+ to take place, Fe3+ must move from the bulk solution into the layer of solution immediately adjacent to the electrode and then diffuse to the electrode's surface; this is called the diffusion layer. Once the reduction takes place, the Fe2+ produced must diffuse away from the electrode's surface and enter into the bulk solution. These two processes are called mass transfer and if we try to change the electrode's potential too quickly, mass transfer may result in concentrations of Fe3+ to Fe2+ at the electrode's surface that are different from that in bulk solution, resulting in concentration polarization.Let's use the reduction of Fe3+ to Fe2+ at the cathode of a galvanic cell to think though how concentration polarization affects the potential we measure. From the Nernst equation we know that\[E = E_{\ce{Fe^{3+}}/\ce{Fe^{2+}}}^{\circ} - \frac {0.05916} {1} \log \frac {[\ce{Fe^{2+}}]}{[\ce{Fe^{3+}}]} = +0.771 - \frac {0.05916} {1} \log \frac {[\ce{Fe^{2+}}]}{[\ce{Fe^{3+}}]} \label{iron1} \]If the mass transfer of Fe3+ from bulk solution to the electrode's surface is slow and if mass transfer of Fe2+ from the electrode's surface to bulk solution is slow, then the concentration of Fe3+ at the electrode's surface is smaller than in bulk solution and the concentration of Fe2+ at the electrode's surface is greater than in the bulk solution. As a result, the ratio \(\frac {[\ce{Fe^{2+}}]}{[\ce{Fe^{3+}}]}\) is greater than that predicted by the bulk concentrations of Fe3+ and Fe2+ and the potential of the cathode is smaller (less positive) than the value +0.771 V predicted by the bulk concentrations of Fe3+ and Fe2+. The resulting potential of the electrochemical cell\[E_\text{cell} = E_\text{cathode} - E_\text{anode} \label{iron2} \]is less positive than that predicted by the bulk concentrations of Fe3+ and Fe2+ due to this concentration polarization.Other kinetic processes can contribute to polarization, including the rate of chemical reactions that take place within the layer of solution near the electrode's surface, the kinetics of reactions in which the electroactive species absorb or desorb from the electrode's surface, and the kinetics of the electron transfer process itself. More details on these are included in later chapters covering specific electrochemical techniques.This page titled 22.5: Currents in Electrochemical Cells is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

22.6: Types of Electroanalytical Methods

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In the next three chapters we will consider a variety of different interfacial electrochemical experiments; that is, experiments in which the redox reaction takes place at the surface of an electrode. Because electrochemistry is such a broad field, let’s use to organize these techniques by the experimental conditions we choose to use (Do we control the potential or the current? How do we change the applied potential or applied current? Do we stir the solution?) and the analytical signal we decide to measure (Current? Potential?).At the first level, we divide electrochemical techniques into static techniques and dynamic techniques. In a static technique we do not allow current to pass through the electrochemical cell and, as a result, the concentrations of all species remain constant. Potentiometry, in which we measure the potential of an electrochemical cell under static conditions, is one of the most important quantitative electrochemical methods and is discussed in Chapter 23.Dynamic techniques, in which we allow current to flow and force a change in the concentration of species in the electrochemical cell, comprise the largest group of interfacial electrochemical techniques. Coulometry, in which we measure current as a function of time, is covered Chapter 24. Voltammetry and amperometry, in which we measure current as a function of a fixed or variable potential, are the subjects of Chapter 25.This page titled 22.6: Types of Electroanalytical Methods is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

23.1: Reference Electrodes

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In potentiometry we measure the difference between the potential of two electrodes. The potential of one electrode—the working or indicator electrode—responds to the analyte’s activity and the other electrode—the counter or reference electrode—has a known, fixed potential. By convention, the reference electrode is the anode; thus, the short hand notation for a potentiometric electrochemical cell isreference electrode || indicator electrodeand the cell potential is\[E_{\mathrm{cell}}=E_{\mathrm{ind}}-E_{\mathrm{ref}} \nonumber \]The ideal reference electrode provides a stable, known potential so that we can attribute any change in Ecell to the analyte’s effect on the indicator electrode’s potential. In addition, the reference electrode should be easy to make and easy to use. Although the standard hydrogen electrode is the reference electrode used to define electrode potentials, it use is not common. Instead, the two reference electrodes discussed in this section find the most applications.A calomel reference electrode is based on the following redox couple between Hg2Cl2 and Hg (calomel is the common name for Hg2Cl2)\[\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)+2 e^{-}\rightleftharpoons2 \mathrm{Hg}(l)+2 \mathrm{Cl}^{-}(a q) \nonumber \]for which the potential is\[E=E_{\mathrm{Hg}_{2} \mathrm{Cl}_{2} / \mathrm{Hg}}^{\mathrm{o}}-\frac{0.05916}{2} \log \left(a_{\text{Cl}^-}\right)^{2}=+0.2682 \mathrm{V}-\frac{0.05916}{2} \log \left(a_{\text{Cl}^-}\right)^{2} \nonumber \]The potential of a calomel electrode, therefore, depends on the activity of Cl– in equilibrium with Hg and Hg2Cl2.As shown in , in a saturated calomel electrode (SCE) the concentration of Cl– is determined by the solubility of KCl. The electrode consists of an inner tube packed with a paste of Hg, Hg2Cl2, and KCl, situated within a second tube that contains a saturated solution of KCl. A small hole connects the two tubes and a porous wick serves as a salt bridge to the solution in which the SCE is immersed. A stopper in the outer tube provides an opening for adding addition saturated KCl. The short hand notation for this cell is\[\mathrm{Hg}(l) | \mathrm{Hg}_{2} \mathrm{Cl}_{2}(s), \mathrm{KCl}(a q, \text { sat'd }) \| \nonumber \]Because the concentration of Cl– is fixed by the solubility of KCl, the potential of an SCE remains constant even if we lose some of the inner solution to evaporation. A significant disadvantage of the SCE is that the solubility of KCl is sensitive to a change in temperature. At higher temperatures the solubility of KCl increases and the electrode’s potential decreases. For example, the potential of the SCE is +0.2444 V at 25oC and +0.2376 V at 35oC. The potential of a calomel electrode that contains an unsaturated solution of KCl is less dependent on the temperature, but its potential changes if the concentration, and thus the activity of Cl–, increases due to evaporation.For example, the potential of a calomel electrode is +0.280 V when the concentration of KCl is 1.00 M and +0.336 V when the concentration of KCl is 0.100 M. If the activity of Cl– is 1.00, the potential is +0.2682 V.Another common reference electrode is the silver/silver chloride electrode, which is based on the reduction of AgCl to Ag.\[\operatorname{AgCl}(s)+e^{-} \rightleftharpoons \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q) \nonumber \]As is the case for the calomel electrode, the activity of Cl– determines the potential of the Ag/AgCl electrode; thus\[E = E_\text{AgCl/Ag}^{\circ}-0.05916 \log a_{\text{Cl}^-} = 0.2223 \text{ V} - 0.05916 \log a_{\text{Cl}^-} \nonumber \]When prepared using a saturated solution of KCl, the electrode's potential is +0.197 V at 25oC. Another common Ag/AgCl electrode uses a solution of 3.5 M KCl and has a potential of +0.205 V at 25oC. As you might expect, the potential of a Ag/AgCl electrode using a saturated solution of KCl is more sensitive to a change in temperature than an electrode that uses an unsaturated solution of KCl.A typical Ag/AgCl electrode is shown in and consists of a silver wire, the end of which is coated with a thin film of AgCl, immersed in a solution that contains the desired concentration of KCl. A porous plug serves as the salt bridge. The electrode’s short hand notation is\[\operatorname{Ag}(s) | \operatorname{Ag} \mathrm{Cl}(s), \mathrm{KCl}\left(a q, a_{\mathrm{Cl}^{-}}=x\right) \| \nonumber \]The standard state reduction potentials in most tables are reported relative to the standard hydrogen electrode’s potential of +0.00 V. Because we rarely use the SHE as a reference electrode, we need to convert an indicator electrode’s potential to its equivalent value when using a different reference electrode. As shown in the following example, this is easy to do.The potential for an Fe3+/Fe2+ half-cell is +0.750 V relative to the standard hydrogen electrode. What is its potential if we use a saturated calomel electrode or a saturated silver/silver chloride electrode?When we use a standard hydrogen electrode the potential of the electrochemical cell is\[E_\text{cell} = E_{\text{Fe}^{3+}/\text{Fe}^{2+}} - E_\text{SHE} = 0.750 \text{ V} -0.000 \text{ V} = 0.750 \text{ V} \nonumber \]We can use the same equation to calculate the potential if we use a saturated calomel electrode\[E_\text{cell} = E_{\text{Fe}^{3+}/\text{Fe}^{2+}} - E_\text{SHE} = 0.750 \text{ V} -0.2444 \text{ V} = 0.506 \text{ V} \nonumber \]or a saturated silver/silver chloride electrode\[E_\text{cell} = E_{\text{Fe}^{3+}/\text{Fe}^{2+}} - E_\text{SHE} = 0.750 \text{ V} -0.197 \text{ V} = 0.553 \text{ V} \nonumber \] provides a pictorial representation of the relationship between these different potentials.This page titled 23.1: Reference Electrodes is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

23.2: Metallic Indicator Electrodes

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In potentiometry, the potential of the indicator electrode is proportional to the analyte’s activity. Two classes of indicator electrodes are used to make potentiometric measurements: metallic electrodes, which are the subject of this section, and ion-selective electrodes, which are covered in the next section.If we place a copper electrode in a solution that contains Cu2+, the electrode’s potential due to the reaction\[\mathrm{Cu}^{2+}(a q)+2 e^{-} \rightleftharpoons \mathrm{Cu}(s) \nonumber \]is determined by the activity of Cu2+.\[E=E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\mathrm{o}}-\frac{0.05916}{2} \log \frac{1}{a_{\mathrm{Cu}^{2+}}}=+0.3419 \mathrm{V}-\frac{0.05916}{2} \log \frac{1}{a_{\mathrm{Cu}^{2+}}} \nonumber \]If copper is the indicator electrode in a potentiometric electrochemical cell that also includes a saturated calomel reference electrode\[\mathrm{SCE} \| \mathrm{Cu}^{2+}\left(a q, a_{\mathrm{Cu^{2+}}}=x\right) | \text{Cu}(s) \nonumber \]then we can use the cell potential to determine an unknown activity of Cu2+ in the indicator electrode’s half-cell\[E_{\text{cell}}= E_{\text { ind }}-E_{\text {SCE }}= +0.3419 \mathrm{V}-\frac{0.05916}{2} \log \frac{1}{a_{\mathrm{Cu}^{2+}}}-0.2224 \mathrm{V} \nonumber \]An indicator electrode in which the metal is in contact with a solution containing its ion is called an electrode of the first kind. In general, if a metal, M, is in a solution of Mn+, the cell potential is\[E_{\mathrm{call}}=K-\frac{0.05916}{n} \log \frac{1}{a_{M^{n+}}}=K+\frac{0.05916}{n} \log a_{M^{n+}} \nonumber \]where K is a constant that includes the standard-state potential for the Mn+/M redox couple and the potential of the reference electrode.For a variety of reasons—including the slow kinetics of electron transfer at the metal–solution interface, the formation of metal oxides on the electrode’s surface, and interfering reactions—electrodes of the first kind are limited to the following metals: Ag, Bi, Cd, Cu, Hg, Pb, Sn, Tl, and Zn.Many of these electrodes, such as Zn, cannot be used in acidic solutions because they are easily oxidized by H+.\[\mathrm{Zn}(s)+2 \mathrm{H}^{+}(a q)\rightleftharpoons \text{ H}_{2}(g)+\mathrm{Zn}^{2+}(a q) \nonumber \]The potential of an electrode of the first kind responds to the activity of Mn+. We also can use this electrode to determine the activity of another species if it is in equilibrium with Mn+. For example, the potential of a Ag electrode in a solution of Ag+ is\[E=0.7996 \mathrm{V}+0.05916 \log a_{\mathrm{Ag}^{+}} \label{second1} \]If we saturate the indicator electrode’s half-cell with AgI, the solubility reaction\[\operatorname{Agl}(s)\rightleftharpoons\operatorname{Ag}^{+}(a q)+\mathrm{I}^{-}(a q) \label{second2} \]determines the concentration of Ag+; thus\[a_{\mathrm{Ag}^{+}}=\frac{K_{\mathrm{sp}, \mathrm{Agl}}}{a_{\text{I}^-}} \label{second3} \]where Ksp,AgI is the solubility product for AgI. Substituting Equation \ref{second3} into Equation \ref{second1}\[E=0.7996 \text{ V}+0.05916 \log \frac{K_{\text{sp, Agl}}}{a_{\text{I}^-}} \label{second4} \]shows that the potential of the silver electrode is a function of the activity of I–. If we incorporate this electrode into a potentiometric electrochemical cell with a saturated calomel electrode\[\mathrm{SCE} \| \mathrm{AgI}(s), \text{ I}^-\left(a q, a_{\text{I}^-}=x\right) | \mathrm{Ag}(\mathrm{s}) \label{second5} \]then the cell potential is\[E_{\mathrm{cell}}=K-0.05916 \log a_{\text{I}^-} \label{second6} \]where K is a constant that includes the standard-state potential for the Ag+/Ag redox couple, the solubility product for AgI, and the reference electrode’s potential.If an electrode of the first kind responds to the activity of an ion in equilibrium with Mn+, we call it an electrode of the second kind. Two common electrodes of the second kind are the calomel and the silver/silver chloride reference electrodes.In an electrode of the second kind we link together a redox reaction and another reaction, such as a solubility reaction. You might wonder if we can link together more than two reactions. The short answer is yes. An electrode of the third kind, for example, links together a redox reaction and two other reactions. Such electrodes are less common and we will not consider them in this text.An electrode of the first kind or the second kind develops a potential as the result of a redox reaction that involves the metallic electrode. An electrode also can serve as a source of electrons or as a sink for electrons in an unrelated redox reaction, in which case we call it a redox electrode. The Pt cathode in \(\PageIndex{1}\) is a redox electrode because its potential is determined by the activity of Fe2+ and Fe3+ in the indicator half-cell. Note that a redox electrode’s potential often responds to the activity of more than one ion, which limits its usefulness for direct potentiometry.. Potentiometric electrochemical cell in which the anode is a metallic electrode of the first kind (Ag) and the cathode is a metallic redox electrode (Pt).This page titled 23.2: Metallic Indicator Electrodes is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

23.3: Membrane Ion-Selective Electrodes

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If metals were the only useful materials for constructing indicator electrodes, then there would be few useful applications of potentiometry. In 1906, Cremer discovered that the potential difference across a thin glass membrane is a function of pH when opposite sides of the membrane are in contact with solutions that have different concentrations of H3O+. The existence of this membrane potential led to the development of a new class of indicator electrodes, which we call ion-selective electrodes (ISEs). In addition to the glass pH electrode, ion-selective electrodes are available for a wide range of ions. It also is possible to construct a membrane electrode for a neutral analyte by using a chemical reaction to generate an ion that is monitored with an ion-selective electrode. The development of new ion-selective membrane electrodes continues to be an active area of research.There are two broad classes of materials that are used as membranes: crystalline solid-state membranes and non-crystalline membranes. Examples of crystalline solid-state membranes are single crystals of LaF3 and polycrystalline AgS. Examples non-crystalline membranes are glass and hydrophobic membranes that hold a liquid ion-exchanger. Each of these are considered below.To be useful, an ion-selective membrane must be structurally stable (a crystalline membrane that is soluble, for example, is not structurally stable), capable of being machined to a suitable size and shape that can be incorporated into the indicator electrode in a potentiometric electrochemical cell, electrically conductive so it is possible to measure the electrochemical cell's potential, and selective toward the analyte. shows a typical potentiometric electrochemical cell equipped with an ion-selective electrode. The short hand notation for this cell is\[\text { ref (sample) }\left\|\mathrm{A}_{\text { samp }}\left(a q, a_{\mathrm{A_{\text{samp}}}}=x\right) | \text{membrane} | \mathrm{A}_{\text { int }}\left(a q, a_{\mathrm{A_{\text{int}}}}=y\right)\right\| \text { ref (internal) } \nonumber \]where the ion-selective membrane separates the two solutions that contain analyte with activities of x and y: the sample solution and the ion-selective electrode’s internal solution. The potential of this electrochemical cell\[E_\text{cell} = E_\text{ref(int)} - E_\text{ref(samp)} + E_\text{mem} \label{membrane1} \]includes the potential of each reference electrode and the difference in potential across the membrane, Emem, which is the membrane's boundary potential. The notations ref(sample) and ref(internal) represent a reference electrode immersed in the sample and a reference electrode immersed in the ion-selective electrode's internal solution. Because the potential of the two reference electrodes are constant, any change in Ecell reflects a change in the membrane’s boundary potential.The analyte’s interaction with the membrane generates a boundary potential if there is a difference in its activity on the membrane’s two sides. Current is carried through the membrane by the movement of either the analyte or an ion already present in the membrane’s matrix. The membrane potential is given by the following Nernst-like equation\[E_{\mathrm{mem}}=E_{\mathrm{asym}}-\frac{R T}{z F} \ln \frac{\left(a_{A}\right)_{\mathrm{int}}}{\left(a_{A}\right)_{\mathrm{samp}}} \label{membrane2} \]where (aA)samp is the analyte’s activity in the sample, (aA)int is the analyte’s activity in the ion-selective electrode’s internal solution, and z is the analyte’s charge. Ideally, Emem is zero when (aA)int = (aA)samp. The term Easym, which is an asymmetry potential, accounts for the fact that Emem usually is not zero under these conditions.For now we simply note that a difference in the analyte’s activity results in the membrane's boundary potential. As we consider different types of ion-selective electrodes, we will explore more specifically the source of the membrane potential.Substituting Equation \ref{membrane2} into Equation \ref{membrane1}, assuming a temperature of 25oC, and rearranging gives\[E_{\mathrm{cell}}=K+\frac{0.05916}{z} \log \left(a_{A}\right)_{\mathrm{samp}} \label{membrane3} \]where K is a constant that includes the potentials of the two reference electrodes, the asymmetry potential, and the analyte's activity in the internal solution. Equation \ref{membrane3} is a general equation and applies to all types of ion-selective electrodes.The membrane's boundary potential results from a chemical interaction between the analyte and active sites on the membrane’s surface. Because the signal depends on a chemical process, most membranes are not selective toward a single analyte. Instead, the membrane's boundary potential is proportional to the concentration of each ion that interacts with the membrane’s active sites. We can rewrite Equation \ref{membrane3} to include the contribution to the potential of an interferent, I \[E_\text{cell} = K + \frac {0.05916} {z_A} \log \left\{ a_A + K_{A,I}(a_I)^{z_A/z_I} \right\} \label{membrane4} \]where zA and zI are the charges of the analyte and the interferent, and KA,I is a selectivity coefficient that accounts for the relative response of the interferent. The selectivity coefficient is defined as\[K_{A,I} = \frac {(a_A)_e} {(a_I)_e^{z_A/z_I}} \label{membrane5} \]where (aA)e and (aI)e are the activities of analyte and the interferent that yield identical cell potentials. When the selectivity coefficient is 1.00, the membrane responds equally to the analyte and the interferent. A membrane shows good selectivity for the analyte when KA,I is significantly less than 1.00.Selectivity coefficients for most commercially available ion-selective electrodes are provided by the manufacturer. If the selectivity coefficient is not known, it is easy to determine its value experimentally by preparing a series of solutions, each of which contains the same activity of interferent, (aI)add, but a different activity of analyte. As shown in , a plot of cell potential versus the log of the analyte’s activity has two distinct linear regions. When the analyte’s activity is significantly larger than KA,I \(\times\) (aI)add, the potential is a linear function of log(aA), as given by Equation \ref{membrane3}. If KA,I \(\times\) (aI)add is significantly larger than the analyte’s activity, however, the cell’s potential remains constant. The activity of analyte and interferent at the intersection of these two linear regions is used to calculate KA,I.Sokalski and co-workers described a method for preparing ion-selective electrodes with significantly improved selectivities [Sokalski, T.; Ceresa, A.; Zwicki, T.; Pretsch, E. J. Am. Chem. Soc. 1997, 119, 11347–11348]. For example, a conventional Pb2+ ISE has a \(\log K_{\text{Pb}^{2+}/\text{Mg}^{2+}}\) of –3.6. If the potential for a solution in which the activity of Pb2+ is \(4.1 \times 10^{-12}\) is identical to that for a solution in which the activity of Mg2+ is 0.01025, what is the value of \(\log K_{\text{Pb}^{2+}/\text{Mg}^{2+}}\) for their ISE?Making appropriate substitutions into Equation \ref{membrane5}, we find that\[K_{\text{Pb}^{2+}/\text{Mg}^{2+}} = \frac {(a_{\text{Pb}^{2+}})_e} {(a_{\text{Mg}^{2+}})_e^{z_{\text{Pb}^{2+}}/z_{\text{Mg}^{2+}}}} = \frac {4.1 \times 10^{-12}} {(0.01025)^{+2/+2}} = 4.0 \times 10^{-10} \nonumber \]The value of \(\log K_{\text{Pb}^{2+}/\text{Mg}^{2+}}\), therefore, is –9.40.The earliest ion-selective electrodes were based on the observation that a thin glass membrane separating two solutions with different levels of acidity develops a measurable difference in potential on opposite sides of the membrane. Incorporating the glass electrode into a potentiometer along with a reference electrode provides a way to measure the potential. Commercial glass membrane pH electrodes often are available in a combination form that includes both the indicator electrode and the reference electrode. The use of a single electrode greatly simplifies the measurement of pH. An example of a typical combination electrode is shown in .The first commercial glass electrodes were manufactured using Corning 015, a glass with a composition that is approximately 22% Na2O, 6% CaO, and 72% SiO2. Membranes fashioned from Corning 015 have an excellent selectivity for hydrogen ions, H+, below a pH of 9; above this pH the membrane becomes more selective for other cations and the measured pH value deviates from its actual value. Replacing Na2O and CaO with Li2O and BaO extends the useful pH range of glass membranes to pH levels greater than 12.When immersed in an aqueous solution for several hours, the outer approximately 10 nm of the glass membrane’s surface becomes hydrated, resulting in the formation of negatively charged sites, —SiO–. Sodium ions, Na+, serve as counter ions. Because H+ binds more strongly to —SiO– than does Na+, they displace the sodium ions on both sides of the membrane.\[\mathrm{H}^{+}+-\mathrm{SiO}^{-} \mathrm{Na}^{+}\rightleftharpoons-\mathrm{SiO}^{-} \mathrm{H}^{+}+\mathrm{Na}^{+} \label{glass1} \]explaining the membrane’s selectivity for H+. The transport of charge across the membrane is carried by the Na+ ions within the glass membrane. The potential of a glass electrode obeys the equation\[E_{\mathrm{cell}}=K+0.05916 \log a_{\mathrm{H}^{+}} \label{glass2} \]As noted above, at sufficiently basic pH values a glass electrode no longer provides an accurate measure of a sample's pH as the membrane becomes more selective for other monovalent cations, such as Na+ and K+.For a Corning 015 glass membrane, the selectivity coefficient KH+/Na+ is \(\approx 10^{-11}\). What is the expected error if we measure the pH of a solution in which the activity of H+ is \(2 \times 10^{-13}\) and the activity of Na+ is 0.05?A solution in which the activity of H+, (aH+)act, is \(2 \times 10^{-13}\) has a pH of 12.7. Because the electrode responds to both H+ and Na+, the apparent activity of H+, (aH+)app, is\[(a_{\text{H}^+})_\text{app} = (a_{\text{H}^+})_\text{act} + (K_{\text{H}^+ / \text{Na}^+} \times a_{\text{Na}^+}) = 2 \times 10^{-13} + (10^{-11} \times 0.05) = 7 \times 10^{-13} \nonumber \]The apparent activity of H+ is equivalent to a pH of 12.2, an error of –0.5 pH units.Glass pH electrodes also show deviations from ideal behavior at pH levels less than 0.5, although the reasons for this are not clear. Still, a glass electrode has a wide dynamic range for measuring pH.Because an ion-selective electrode’s glass membrane is very thin—it is only about 50 μm thick—they must be handled with care to avoid cracks or breakage. Glass electrodes usually are stored in a storage buffer recommended by the manufacturer, which ensures that the membrane’s outer surface remains hydrated. If a glass electrode dries out, it is reconditioned by soaking for several hours in a solution that contains the analyte. The composition of a glass membrane will change over time, which affects the electrode’s performance. The average lifetime for a typical glass electrode is several years.The observation that the Corning 015 glass membrane responds to ions other than H+ led to the development of glass membranes with a greater selectivity for other cations. For example, a glass membrane with a composition of 11% Na2O, 18% Al2O3, and 71% SiO2 is used as an ion-selective electrode for Na+. Other glass ion-selective electrodes have been developed for the analysis of Li+, K+, Rb+, Cs+, \(\text{NH}_4^+\), Ag+, and Tl+. Table \(\PageIndex{1}\) provides several examples.\(K_{\mathrm{Na}^{+} / \mathrm{H}^{+}}=1000\)\(K_{\mathrm{Na}^{+} / \mathrm{K}^{+}}=0.001\)\(K_{\mathrm{Na}^{+} / \mathrm{Li}^{+}}=0.001\)\(K_{\mathrm{Li}^{+} / \mathrm{Na}^{+}}=0.3\)\(K_{\mathrm{Li}^{+} / \mathrm{K}^{+}}=0.001\)A solid-state ion-selective electrode has a membrane that consists of either a polycrystalline inorganic salt or a single crystal of an inorganic salt. We can fashion a polycrystalline solid-state ion-selective electrode by sealing a 1–2 mm thick pellet of AgS—or a mixture of AgS and a second silver salt or another metal sulfide—into the end of a nonconducting plastic cylinder, filling the cylinder with an internal solution that contains the analyte, and placing a reference electrode into the internal solution. shows a typical design.The NaCl in a salt shaker is an example of polycrystalline material because it consists of many small crystals of sodium chloride. The NaCl salt plates used in IR spectroscopy, on the other hand, are an example of a single crystal of sodium chloride.The membrane potential for a Ag2S pellet develops as the result of a difference in the extent of the solubility reaction\[\mathrm{Ag}_{2} \mathrm{S}(s)\rightleftharpoons2 \mathrm{Ag}^{+}(a q)+\mathrm{S}^{2-}(a q) \label{ss1} \]on the membrane’s two sides, with charge carried across the membrane by Ag+ ions. When we use the electrode to monitor the activity of Ag+, the cell potential is\[E_{\text {cell }}=K+0.05916 \log a_{\mathrm{Ag}^{+}} \label{ss2} \]The membrane also responds to the activity of \(\text{S}^{2-}\), with a cell potential of\[E_{\mathrm{cell}}=K-\frac{0.05916}{2} \log a_{\text{S}^{2-}} \label{ss3} \]If we combine an insoluble silver salt, such as AgCl, with the Ag2S, then the membrane potential also responds to the concentration of Cl–, with a cell potential of\[E_{\text {cell }}=K-0.05916 \log a_{\mathrm{Cl}^{-}} \label{ss4} \]By mixing Ag2S with CdS, CuS, or PbS, we can make an ion-selective electrode that responds to the activity of Cd2+, Cu2+, or Pb2+. In this case the cell potential is\[E_{\mathrm{cell}}=K+\frac{0.05916}{2} \ln a_{M^{2+}} \label{ss5} \]where aM2+ is the activity of the metal ion.Table \(\PageIndex{2}\) provides examples of polycrystalline, Ag2S-based solid-state ion-selective electrodes. The selectivity of these ion-selective electrodes depends on the relative solubility of the compounds. A Cl– ISE using a Ag2S/AgCl membrane is more selective for Br– (KCl–/Br– = 102) and for I– (KCl–/I– = 106) because AgBr and AgI are less soluble than AgCl. If the activity of Br– is sufficiently high, AgCl at the membrane/solution interface is replaced by AgBr and the electrode’s response to Cl– decreases substantially. Most of the polycrystalline ion-selective electrodes listed in Table \(\PageIndex{2}\) operate over an extended range of pH levels. The equilibrium between S2– and HS– limits the analysis for S2– to a pH range of 13–14.\(K_{\text{Ag}^+/\text{Cu}^{2+}} = 10^{-6}\)\(K_{\text{Ag}^+/\text{Pb}^{2+}} = 10^{-10}\)Hg2+ interferes\(K_{\text{Cd}^{2+}/\text{Fe}^{2+}} = 200\)\(K_{\text{Cd}^{2+}/\text{Pb}^{2+}} = 6\)Ag+, Hg2+, and Cu2+ must be absent\(K_{\text{Cu}^{2+}/\text{Fe}^{3+}} = 10\)\(K_{\text{Cu}^{2+}/\text{Cu}^{+}} = 10^{-6}\)Ag+ and Hg2+ must be absent\(K_{\text{Pb}^{2+}/\text{Fe}^{3+}} = 1\)\(K_{\text{Pb}^{2+}/\text{Cd}^{2+}} = 1\)Ag+, Hg2+, and Cu2+ must be absent\(K_{\text{Br}^-/\text{I}^{-}} = 5000\)\(K_{\text{Br}^-/\text{Cl}^{-}} = 0.005\)\(K_{\text{Br}^-/\text{OH}^{-}} = 10^{-5}\)S2– must be absent\(K_{\text{Cl}^-/\text{I}^{-}} = 10^{6}\)\(K_{\text{Cl}^-/\text{Br}^{-}} = 100\)\(K_{\text{Cl}^-/\text{OH}^{-}} = 0.01\)S2– must be absent\(K_{\text{I}^-/\text{S}^{2-}} = 30\)\(K_{\text{I}^-/\text{Br}^{-}} = 10^{-4}\)\(K_{\text{I}^-/\text{Cl}^{-}} = 10^{-6}\)\(K_{\text{I}^-/\text{OH}^{-}} = 10^{-7}\)\(K_{\text{SCN}^-/\text{I}^{-}} = 10^{3}\)\(K_{\text{SCN}^-/\text{Br}^{-}} = 100\)\(K_{\text{SCN}^-/\text{Cl}^{-}} = 0.1\)\(K_{\text{SCN}^-/\text{OH}^{-}} = 0.01\)S2– must be absentThe membrane of a F– ion-selective electrode is fashioned from a single crystal of LaF3, which usually is doped with a small amount of EuF2 to enhance the membrane’s conductivity. Because EuF2 provides only two F– ions—compared to the three F– ions in LaF3—each EuF2 produces a vacancy in the crystal’s lattice. Fluoride ions pass through the membrane by moving into adjacent vacancies. As shown in , the LaF3 membrane is sealed into the end of a non-conducting plastic cylinder, which contains a standard solution of F–, typically 0.1 M NaF, and a Ag/AgCl reference electrode.The membrane potential for a F– ISE results from a difference in the solubility of LaF3 on opposite sides of the membrane, with the potential given by\[E_{\mathrm{cell}}=K-0.05916 \log a_{\mathrm{F}^-} \label{ss6} \]One advantage of the F– ion-selective electrode is its freedom from interference. The only significant exception is OH– (KF–/OH– = 0.1), which imposes a maximum pH limit for a successful analysis. Below a pH of 4 the predominate form of fluoride in solution is HF, which does not contribute to the membrane potential. For this reason, an analysis for fluoride is carried out at a pH greater than 4.What is the maximum pH that we can tolerate if we need to analyze a solution in which the activity of F– is \(1 \times 10^{-5}\) with an error of less than 1%?In the presence of OH– the cell potential is\[E_{\mathrm{cell}}=K-0.05916\left\{a_{\mathrm{F}^-}+K_{\mathrm{F}^- / \mathrm{OH}^{-}} \times a_{\mathrm{OH}^-}\right\} \nonumber \]To achieve an error of less than 1%, the term \(K_{\mathrm{F}^- / \mathrm{OH}^{-}} \times a_{\mathrm{OH}^-}\) must be less than 1% of aF–; thus\[K_{\mathrm{F}^- / \mathrm{OH}^-} \times a_{\mathrm{OH}^{-}} \leq 0.01 \times a_{\mathrm{F}^-} \nonumber \]\[0.10 \times a_{\mathrm{OH}^{-}} \leq 0.01 \times\left(1.0 \times 10^{-5}\right) \nonumber \]Solving for aOH– gives the maximum allowable activity for OH– as \(1 \times 10^{-6}\), which corresponds to a pH of less than 8.Unlike a glass membrane ion-selective electrode, a solid-state ISE does not need to be conditioned before it is used, and it may be stored dry. The surface of the electrode is subject to poisoning, as described above for a Cl– ISE in contact with an excessive concentration of Br–. If an electrode is poisoned, it can be returned to its original condition by sanding and polishing the crystalline membrane.Poisoning simply means that the surface has been chemically modified, such as AgBr forming on the surface of a AgCl membrane.Another class of ion-selective electrodes uses a hydrophobic membrane that contains a liquid organic complexing agent that reacts selectively with the analyte. Three types of organic complexing agents have been used: cation exchangers, anion exchangers, and neutral ionophores. A membrane potential exists if the analyte’s activity is different on the two sides of the membrane. Current is carried through the membrane by the analyte.An ionophore is a ligand whose exterior is hydrophobic and whose interior is hydrophilic. The crown ether shown here is one example of a neutral ionophore.One example of a liquid-based ion-selective electrode is that for Ca2+, which uses a porous plastic membrane saturated with the cation exchanger di-(n-decyl) phosphate. As shown in , the membrane is placed at the end of a non-conducting cylindrical tube and is in contact with two reservoirs. The outer reservoir contains di-(n-decyl) phosphate in di-n-octylphenylphosphonate, which soaks into the porous membrane. The inner reservoir contains a standard aqueous solution of Ca2+ and a Ag/AgCl reference electrode. Calcium ion-selective electrodes also are available in which the di-(n-decyl) phosphate is immobilized in a polyvinyl chloride (PVC) membrane that eliminates the need for the outer reservoir.The membrane potential for the Ca2+ ISE develops as the result of a difference in the extent of the complexation reaction\[\mathrm{Ca}^{2+}(a q)+2\left(\mathrm{C}_{10} \mathrm{H}_{21} \mathrm{O}\right)_{2} \mathrm{PO}_{2}^{-}(mem) \rightleftharpoons \mathrm{Ca}\left[\left(\mathrm{C}_{10} \mathrm{H}_{21} \mathrm{O}\right)_{2} \mathrm{PO}_{2}\right]_2 (mem) \label{liq1} \]on the two sides of the membrane, where (mem) indicates a species that is present in the membrane. The cell potential for the Ca2+ ion-selective electrode is\[E_{\mathrm{cell}}=K+\frac{0.05916}{2} \log a_{\mathrm{ca}^{2+}} \label{liq2} \]The selectivity of this electrode for Ca2+ is very good, with only Zn2+ showing greater selectivity.Table \(\PageIndex{3}\) lists the properties of several liquid-based ion-selective electrodes. An electrode using a liquid reservoir can be stored in a dilute solution of analyte and needs no additional conditioning before use. The lifetime of an electrode with a PVC membrane, however, is proportional to its exposure to aqueous solutions. For this reason these electrodes are best stored by covering the membrane with a cap along with a small amount of wetted gauze to maintain a humid environment. Before using the electrode it is conditioned in a solution of analyte for 30–60 minutes.\(K_{\text{Ca}^{2+}/\text{Zn}^{2+}} = 1-5\)\(K_{\text{Ca}^{2+}/\text{Al}^{3+}} = 0.90\)\(K_{\text{Ca}^{2+}/\text{Mn}^{2+}} = 0.38\)\(K_{\text{Ca}^{2+}/\text{Cu}^{2+}} = 0.070\)\(K_{\text{Ca}^{2+}/\text{Mg}^{2+}} = 0.032\)\(K_{\text{K}^{+}/\text{Rb}^{+}} = 1.9\)\(K_{\text{K}^{+}/\text{Cs}^{+}} = 0.38\)\(K_{\text{K}^{+}/\text{Li}^{+}} = 10^{-4}\)\(K_{\text{Li}^{+}/\text{H}^{+}} = 1\)\(K_{\text{Li}^{+}/\text{Na}^{+}} = 0.03\)\(K_{\text{Li}^{+}/\text{K}^{+}} = 0.007\)\(K_{\text{NH}_4^{+}/\text{K}^{+}} = 0.12\)\(K_{\text{NH}_4^{+}/\text{H}^{+}} = 0.016\)\(K_{\text{NH}_4^{+}/\text{Li}^{+}} = 0.0042\)\(K_{\text{NH}_4^{+}/\text{Na}^{+}} = 0.002\)\(K_{\text{ClO}_4^{-}/\text{OH}^{-}} = 1\)\(K_{\text{ClO}_4^{-}/\text{I}^{-}} = 0.012\)\(K_{\text{ClO}_4^{-}/\text{NO}_3^{-}} = 0.0015\)\(K_{\text{ClO}_4^{-}/\text{Br}^{-}} = 5.6 \times 10^{-4}\)\(K_{\text{ClO}_4^{-}/\text{Cl}^{-}} = 2.2 \times 10^{-4}\)\(K_{\text{NO}_3^{-}/\text{Cl}^{-}} = 0.006\)\(K_{\text{NO}_3^{-}/\text{F}^{-}} = 9 \times 10^{-4}\)This page titled 23.3: Membrane Ion-Selective Electrodes is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

23.4: Molecular-Selective Electrode Systems

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The electrodes in Chapter 23.3 are selective toward ions. In this section we consider how we can incorporate an ion-selective electrode into an electrode that responds to neutral species, such as volatile analytes, such as CO2 and NH3, and biochemically important compounds, such as amino acids and urea.A number of membrane electrodes respond to the concentration of a dissolved gas. The basic design of a gas-sensing electrode, as shown in , consists of a thin membrane that separates the sample from an inner solution that contains an ion-selective electrode. The membrane is permeable to the gaseous analyte, but impermeable to nonvolatile components in the sample’s matrix. The gaseous analyte passes through the membrane where it reacts with the inner solution, producing a species whose concentration is monitored by the ion-selective electrode. For example, in a CO2 electrode, CO2 diffuses across the membrane where it reacts in the inner solution to produce H3O+.\[\mathrm{CO}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons\text{ HCO}_{3}^{-}(a q)+\text{ H}_{3} \mathrm{O}^{+}(a q) \label{gas1} \]The change in the activity of H3O+ in the inner solution is monitored with a pH electrode, for which the cell potential, from Chapter 23.3, is\[E_\text{cell} = K + 0.05916 \log a_{\ce{H+}} \label{gas2} \]To find the relationship between the activity of H3O+ in the inner solution and the activity of CO2 in the inner solution we rearrange the equilibrium constant expression for reaction \ref{gas1}; thus\[a_{\mathrm{H}_{3} \mathrm{O}^{+}}=K_{\mathrm{a}} \times \frac{a_{\mathrm{CO}_{2}}}{a_{\mathrm{HCO}_{3}^{-}}} \label{gas3} \]where Ka is the equilibrium constant. If the activity of \(\text{HCO}_3^-\) in the internal solution is sufficiently large, then its activity is not affected by the small amount of CO2 that passes through the membrane. Substituting Equation \ref{gas3} into Equation \ref{gas2} gives\[E_{\mathrm{cell}}=K^{\prime}+0.05916 \log a_{\mathrm{co}_{2}} \label{gas4} \]where K′ is a constant that includes the constant for the pH electrode, the equilibrium constant for reaction \ref{gas1} and the activity of \(\text{HCO}_3^-\) in the inner solution.Table \(\PageIndex{1}\) lists the properties of several gas-sensing electrodes. The composition of the inner solution changes with use, and both the inner solution and the membrane must be replaced periodically. Gas-sensing electrodes are stored in a solution similar to the internal solution to minimize their exposure to atmospheric gases.10 mM NaHCO310 mM NaCl10 mM NH4Cl0.1 M KNO320 mM NaNO20.1 M KNO31 mM NaHSO3pH 5The approach for developing gas-sensing electrodes can be modified to create potentiometric electrodes that respond to a biochemically important species. The most common class of potentiometric biosensors are enzyme electrodes, in which we trap or immobilize an enzyme at the surface of a potentiometric electrode. The analyte’s reaction with the enzyme produces a product whose concentration is monitored by the potentiometric electrode. Potentiometric biosensors also have been designed around other biologically active species, including antibodies, bacterial particles, tissues, and hormone receptors.One example of an enzyme electrode is the urea electrode, which is based on the catalytic hydrolysis of urea by urease\[\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons 2 \mathrm{NH}_{4}^{+}(a q)+\text{ CO}_{3}^{-}(a q) \label{bio1} \] shows one version of the urea electrode, which modifies a gas-sensing NH3 electrode by adding a dialysis membrane that traps a pH 7.0 buffered solution of urease between the dialysis membrane and the gas permeable membrane [(a) Papastathopoulos, D. S.; Rechnitz, G. A. Anal. Chim. Acta 1975, 79, 17–26; (b) Riechel, T. L. J. Chem. Educ. 1984, 61, 640–642]. An NH3 electrode, as shown in Table \(\PageIndex{1}\), uses a gas-permeable membrane and a glass pH electrode. The NH3 diffuses across the membrane where it changes the pH of the internal solution.When immersed in the sample, urea diffuses through the dialysis membrane where it reacts with the enzyme urease to form the ammonium ion, \(\text{NH}_4^+\), which is in equilibrium with NH3.\[\mathrm{NH}_{4}^{+}(a q)+\mathrm{H}_{2} \mathrm{O}(l ) \rightleftharpoons \text{ H}_{3} \mathrm{O}^{+}(a q)+\text{ NH}_{3}(a q) \label{bio2} \]The NH3, in turn, diffuses through the gas permeable membrane where a pH electrode measures the resulting change in pH. The electrode’s response to the concentration of urea is\[E_{\text {cell }}=K-0.05916 \log a_{\text {urea }} \label{bio3} \]Another version of the urea electrode ) immobilizes the enzyme urease in a polymer membrane formed directly on the tip of a glass pH electrode [Tor, R.; Freeman, A. Anal. Chem. 1986, 58, 1042–1046]. In this case the response of the electrode is\[\mathrm{pH}=K a_{\mathrm{urea}} \label{bio4} \]Few potentiometric biosensors are available commercially. As shown in and , however, it is possible to convert an ion-selective electrode or a gas-sensing electrode into a biosensor. Several representative examples are described in Table \(\PageIndex{2}\), and additional examples can be found in this chapter’s additional resources.Source: Complied from Cammann, K. Working With Ion-Selective Electrodes, Springer-Verlag: Berlin, 1977 and Lunte, C. E.; Heineman, W. R. “Electrochemical techniques in Bioanalysis,” in Steckham, E. ed. Topics in Current Chemistry, Vol. 143, Springer-Verlag: Berlin, 1988, p.8.Abbreviations for biologically active phase: E = enzyme; B = bacterial particle; T = tissue.This page titled 23.4: Molecular-Selective Electrode Systems is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

23.5: Instruments for Measuring Cell Potentials

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To measure the potential of an electrochemical cell in a way that draws essentially no current we use a potentiometer. To help us understand how a potentiometer accomplishes this, we will describe the instrument as if the analyst is operating it manually. To do so the analyst observes a change in the current or the potential and manually adjusts the instrument’s settings to maintain the desired experimental conditions. It is important to understand that modern electrochemical instruments provide an automated, electronic means for controlling and measuring current and potential, and that they do so by using very different electronic circuitry than that described here. shows a schematic diagram for a manual potentiometer that consists of a power supply, an electrochemical cell with a working electrode and a counter electrode, an ammeter to measure the current that passes through the electrochemical cell, an adjustable, slide-wire resistor, and a tap key for closing the circuit through the electrochemical cell. Using Ohm’s law, the current in the upper half of the circuit is\[i_{\text {upper}}=\frac{E_{\mathrm{PS}}}{R_{a b}} \label{pot1} \]where EPS is the power supply’s potential, and Rab is the resistance between points a and b of the slide-wire resistor. In a similar manner, the current in the lower half of the circuit is\[i_{\text {lower}}=\frac{E_{\text {cell}}}{R_{c b}} \label{pot2} \]where Ecell is the potential difference between the working electrode and the counter electrode, and Rcb is the resistance between the points c and b of the slide-wire resistor. When iupper = ilower = 0, no current flows through the ammeter and the potential of the electrochemical cell is\[E_{\mathrm{coll}}=\frac{R_{c b}}{R_{a b}} \times E_{\mathrm{PS}} \label{pot3} \]To determine Ecell we briefly press the tap key and observe the current at the ammeter. If the current is not zero, then we adjust the slide wire resistor and remeasure the current, continuing this process until the current is zero. When the current is zero, we use Equation \ref{pot3} to calculate Ecell.Using the tap key to briefly close the circuit through the electrochemical cell minimizes the current that passes through the cell and limits the change in the electrochemical cell’s composition. For example, passing a current of 10–9 A through the electrochemical cell for 1 s changes the concentrations of species in the cell by approximately 10–14 moles.\[10^{-9} \text{ A} = 10^{-9} \text{ C/s} \label{pot4} \]\[10^{-9} \text{ C/s} \times 1 \text{ s} \times \frac {1 \text{ mol}} {96485 \text{C}} = 1.0 \times 10^{-14} \text{ mol} \label{pot5} \]Of course, trying to measure a potential in this way is tedious. Modern potentiometers use operational amplifiers to create a high-impedance voltmeter that measures the potential while drawing a current of less than \(10^{–9}\) A. The relative error, \(E_r\), in the measured potential\[E_r = \frac {R_\text{cell}} {R_\text{meter} + R_\text{cell}} \label{pot6} \]where \(R_\text{cell}\) is the resistance of the solution in the electrochemical cell and \(R_\text{meter}\) is the resistance of the meter. For a solution with a resistance of \(10 \text{M}\Omega\) to achieve a relative error of \(-0.1\%)\) or \(-0.001\) requires an \(R_\text{meter}\) of\[-0.001 = \frac {-10 \text{ M}\Omega} {R_\text{meter} + 10 \text{ M}\Omega} \nonumber \]\[-0.001 \times R_\text{meter} - 0.01 = -10 \text{ M}\Omega \nonumber \]\[-0.001 \times R_\text{meter} = -9.990 \text{ M}\Omega \nonumber \]\[R_\text{meter} = 9990 \text{ M}\Omega \nonumber \]This page titled 23.5: Instruments for Measuring Cell Potentials is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

23.6: Quantitative Potentiometry

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The most important application of potentiometry is determining the concentration of an analyte in solution. Most potentiometric electrodes are selective toward the free, uncomplexed form of the analyte, and do not respond to any of the analyte’s complexed forms. This selectivity provides potentiometric electrodes with a significant advantage over other quantitative methods of analysis if we need to determine the concentration of free ions. For example, calcium is present in urine both as free Ca2+ ions and as protein-bound Ca2+ ions. If we analyze a urine sample using atomic absorption spectroscopy, the signal is propor- tional to the total concentration of Ca2+ because both free and bound calcium are atomized. Analyzing urine with a Ca2+ ISE, however, gives a signal that is a function of only free Ca2+ ions because the protein-bound Ca2+ can not interact with the electrode’s membrane. In this section, we consider several important aspects of quantiative potentiometry.In Chapter 23.3, we showed that the potential of an ion-selective electrode for an ion with a charge of z is\[E_{\mathrm{cell}}=K+\frac{0.05916}{z} \log \left(a_{A}\right)_{\mathrm{samp}} \label{quant1} \]where K is a constant that includes the potentials of the ion-selective electrode's internal and external reference electrodes, any asymmetry potential associated with the ion-selective electrode's membrane, and the analyte's activity in the ion-selective electrode's internal solution. Equation \ref{quant1} is a general equation and applies to all types of ion-selective electrodes. Note that when the analyte is a cation, an increase in the analyte's activity results in results in an increase in the potential; when the analyte is an anion, which makes z a negative number, an increase in the analyte's activity results in a decrease in the potential.As the concentrations of ions in solution often are reported as pX values, where\[\text{pX} = - \log a_\text{X} \label{quant2} \]it is convenient to substitute Equation \ref{quant2} into Equation \ref{quant1}\[E_{\mathrm{cell}}=K - \frac{0.05916}{z} \text{ pA} \label{quant3} \]Note that for a cation, an increase in pA results in a decrease in the potential; when the analyte is an anion, an increase in pA results in an increase in the potential.To use Equation \ref{quant3} we need to determine the value of K, which we can do using one or more external standards or by the method of standard addition, both of which were covered in Chapter 1.5. One complication, of course, is that potential is a function of the analyte's activity instead of its concentration.Equation \ref{quant1} is written in terms of the analyte's activity. When we use a potentiometric electrode, however, our goal is to determine the analyte’s concentration. As we learned in Chapter 22, an ion’s activity is the product of its concentration, [Mn+], and a matrix-dependent activity coefficient, \(\gamma_{Mn^{n+}}\).\[a_{M^{n+}}=\left[M^{n+}\right] \gamma_{M^{n+}} \label{quant4} \]Substituting Equation \ref{quant4} into Equation \ref{quant1} and rearranging, gives\[E_{\mathrm{cell}}=K+\frac{0.05916}{n} \log \gamma_{M^{n+}}+\frac{0.05916}{n} \log \left[M^{n+}\right] \label{quant5} \]We can solve Equation \ref{quant5} for the metal ion’s concentration if we know the value for its activity coefficient. Unfortunately, if we do not know the exact ionic composition of the sample’s matrix—which is the usual situation—then we cannot calculate the value of \(\gamma_{Mn^{n+}}\). There is a solution to this dilemma. If we design our system so that the standards and the samples have an identical matrix, then the value of \(\gamma_{Mn^{n+}}\) remains constant and Equation \ref{quant5} simplifies to\[E_{\mathrm{cell}}=K^{\prime}+\frac{0.05916}{n} \log \left[M^{n+}\right] \label{quanat6} \]where \(K^{\prime}\) includes the activity coefficient.In the absence of interferents, a calibration curve of Ecell versus logaA, where A is the analyte, is a straight-line. A plot of Ecell versus log[A], however, may show curvature at higher concentrations of analyte as a result of a matrix-dependent change in the analyte’s activity coefficient. To maintain a consistent matrix we add a high concentration of an inert electrolyte to all samples and standards. If the concentration of added electrolyte is sufficient, then the difference between the sample’s matrix and the matrix of the standards will not affect the ionic strength and the activity coefficient essentially remains constant. The inert electrolyte added to the sample and the standards is called a total ionic strength adjustment buffer (TISAB).The concentration of Ca2+ in a water sample is determined using the method of external standards. The ionic strength of the samples and the standards is maintained at a nearly constant level by making each solution 0.5 M in KNO3. The measured cell potentials for the external standards are shown in the following table.What is the concentration of Ca2+ in a water sample if its cell potential is found to be –0.084 V?Linear regression gives the calibration curve in , with an equation of\[E_{\mathrm{cell}}=0.027+0.0303 \log \left[\mathrm{Ca}^{2+}\right] \nonumber \]Substituting the sample’s cell potential gives the concentration of Ca2+ as \(2.17 \times 10^{-4}\) M. Note that the slope of the calibration curve, which is 0.0303, is slightly larger than its ideal value of 0.05916/2 = 0.02958; this is not unusual and is one reason for using multiple standards. One reason that it is not unusual to find that the experimental slope deviates from its ideal value of 0.05916/n is that this ideal value assumes that the temperature is 25°C.Another approach to calibrating a potentiometric electrode is the method of standard additions, which was introduced in Chapter 1.5. First, we transfer a sample with a volume of Vsamp and an analyte concentration of Csamp into a beaker and measure the potential, (Ecell)samp. Next, we make a standard addition by adding to the sample a small volume, Vstd, of a standard that contains a known concentration of analyte, Cstd, and measure the potential, (Ecell)std. If Vstd is significantly smaller than Vsamp, then we can safely ignore the change in the sample’s matrix and assume that the analyte’s activity coefficient is constant. Example \(\PageIndex{9}\) demonstrates how we can use a one-point standard addition to determine the concentration of analyte in a sample.The concentration of Ca2+ in a sample of sea water is determined using a Ca ion-selective electrode and a one-point standard addition. A 10.00-mL sample is transferred to a 100-mL volumetric flask and diluted to volume. A 50.00-mL aliquot of the sample is placed in a beaker with the Ca ISE and a reference electrode, and the potential is measured as –0.05290 V. After adding a 1.00-mL aliquot of a \(5.00 \times 10^{-2}\) M standard solution of Ca2+ the potential is –0.04417 V. What is the concentration of Ca2+ in the sample of sea water?To begin, we write the Nernst equation before and after adding the standard addition. The cell potential for the sample is\[\left(E_{\mathrm{cell}}\right)_{\mathrm{samp}}=K+\frac{0.05916}{2} \log C_{\mathrm{samp}} \nonumber \]and that following the standard addition is\[\left(E_{\mathrm{cell}}\right)_{\mathrm{std}}=K+\frac{0.05916}{2} \log \left\{ \frac {V_\text{samp}} {V_\text{tot}}C_\text{samp} + \frac {V_\text{std}} {V_\text{tot}}C_\text{std} \right\} \nonumber \]where Vtot is the total volume (Vsamp + Vstd) after the standard addition. Subtracting the first equation from the second equation gives\[\Delta E = \left(E_{\mathrm{cell}}\right)_{\mathrm{std}} - \left(E_{\mathrm{cell}}\right)_{\mathrm{samp}} = \frac{0.05916}{2} \log \left\{ \frac {V_\text{samp}} {V_\text{tot}}C_\text{samp} + \frac {V_\text{std}} {V_\text{tot}}C_\text{std} \right\} - \frac{0.05916}{2}\log C_\text{samp} \nonumber \]Rearranging this equation leaves us with\[\frac{2 \Delta E_{cell}}{0.05916} = \log \left\{ \frac {V_\text{samp}} {V_\text{tot}} + \frac {V_\text{std}C_\text{std}} {V_\text{tot}C_\text{samp}} \right\} \nonumber \]Substituting known values for \(\Delta E\), Vsamp, Vstd, Vtot and Cstd,\[\begin{array}{l}{\frac{2 \times\{-0.04417-(-0.05290)\}}{0.05916}=} \\ {\log \left\{\frac{50.00 \text{ mL}}{51.00 \text{ mL}}+\frac{(1.00 \text{ mL})\left(5.00 \times 10^{-2} \mathrm{M}\right)}{(51.00 \text{ mL}) C_{\mathrm{samp}}}\right\}} \\ {0.2951=\log \left\{0.9804+\frac{9.804 \times 10^{-4}}{C_{\mathrm{samp}}}\right\}}\end{array} \nonumber \]and taking the inverse log of both sides gives\[1.973=0.9804+\frac{9.804 \times 10^{-4}}{C_{\text {samp }}} \nonumber \]Finally, solving for Csamp gives the concentration of Ca2+ as \(9.88 \times 10^{-4}\) M. Because we diluted the original sample of seawater by a factor of 10, the concentration of Ca2+ in the seawater sample is \(9.88 \times 10^{-3}\) M.With the availability of inexpensive glass pH electrodes and pH meters, the determination of pH is one of the most common quantitative analytical measurements. The potentiometric determination of pH, however, is not without complications, several of which we discuss in this section.One complication is confusion over the meaning of pH [Kristensen, H. B.; Saloman, A.; Kokholm, G. Anal. Chem. 1991, 63, 885A–891A]. The conventional definition of pH in most general chemistry textbooks is given in terms of the concentration of H+\[\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \label{quant7} \]As we now know, when we measure pH it actually is a measure of the activity of H+.\[\mathrm{pH}=-\log a_{\mathrm{H}^{+}} \label{quant8} \]Try this experiment—find several general chemistry textbooks and look up pH in each textbook’s index. Turn to the appropriate pages and see how it is defined. Next, look up activity or activity coefficient in each textbook’s index and see if these terms are indexed.Equation \ref{quant7} only approximates the true pH. If we calculate the pH of 0.1 M HCl using Equation \ref{quant7}, we obtain a value of 1.00; the solution’s actual pH, as defined by Equation \ref{quant8}, is 1.1 [Hawkes, S. J. J. Chem. Educ. 1994, 71, 747–749]. The activity and the concentration of H+ are not the same in 0.1 M HCl because the activity coefficient for H+ is not 1.00 in this matrix. shows a more colorful demonstration of the difference between activity and concentration.A second complication in measuring pH is the uncertainty in the relationship between potential and activity. For a glass membrane electrode, the cell potential, (Ecell)samp, for a sample of unknown pH is\[(E_{\text{cell}})_\text {samp} = K-\frac{R T}{F} \ln \frac{1}{a_{\mathrm{H}^{+}}}=K-\frac{2.303 R T}{F} \mathrm{pH}_{\mathrm{samp}} \label{quant9} \]where K includes the potential of the reference electrode, the asymmetry potential of the glass membrane, and any junction potentials in the electrochemical cell. All the contributions to K are subject to uncertainty, and may change from day-to-day, as well as from electrode-to-electrode. For this reason, before using a pH electrode we calibrate it using a standard buffer of known pH. The cell potential for the standard, (Ecell)std, is\[\left(E_{\text {ccll}}\right)_{\text {std}}=K-\frac{2.303 R T}{F} \mathrm{p} \mathrm{H}_{\mathrm{std}} \label{quant10} \]where pHstd is the standard’s pH. Subtracting Equation \ref{quan10} from Equation \ref{quant9} and solving for pHsamp gives\[\text{pH}_\text{samp} = \text{pH}_\text{std} - \frac{\left\{\left(E_{\text {cell}}\right)_{\text {samp}}-\left(E_{\text {cell}}\right)_{\text {std}}\right\} F}{2.303 R T} \label{quant11} \]which is the operational definition of pH adopted by the International Union of Pure and Applied Chemistry [Covington, A. K.; Bates, R. B.; Durst, R. A. Pure & Appl. Chem. 1985, 57, 531–542].Calibrating a pH electrode presents a third complication because we need a standard with an accurately known activity for H+. Table \(\PageIndex{1}\) provides pH values for several primary standard buffer solutions accepted by the National Institute of Standards and Technology.saturated(at 25oC) KHC4H4O7 (tartrate)0.05 m KH2C6H5O7(citrate)To standardize a pH electrode using two buffers, choose one near a pH of 7 and one that is more acidic or basic depending on your sample’s expected pH. Rinse your pH electrode in deionized water, blot it dry with a laboratory wipe, and place it in the buffer with the pH closest to 7. Swirl the pH electrode and allow it to equilibrate until you obtain a stable reading. Adjust the “Standardize” or “Calibrate” knob until the meter displays the correct pH. Rinse and dry the electrode, and place it in the second buffer. After the electrode equilibrates, adjust the “Slope” or “Temperature” knob until the meter displays the correct pH.Some pH meters can compensate for a change in temperature. To use this feature, place a temperature probe in the sample and connect it to the pH meter. Adjust the “Temperature” knob to the solution’s temperature and calibrate the pH meter using the “Calibrate” and “Slope” controls. As you are using the pH electrode, the pH meter compensates for any change in the sample’s temperature by adjusting the slope of the calibration curve using a Nernstian response of 2.303RT/F.This page titled 23.6: Quantitative Potentiometry is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

24.1: Introduction to Coulometry

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Coulometry is based on an exhaustive electrolysis of the analyte. By exhaustive we mean the analyte is oxidized or reduced completely at the working electrode, or reacts completely with a reagent generated at the working electrode. There are two forms of coulometry: controlled-potential coulometry, in which we apply a constant potential to the electrochemical cell, and controlled-current coulometry, in which we pass a constant current through the electrochemical cell.During an electrolysis, the total charge, Q, in coulombs, that passes through the electrochemical cell is proportional to the absolute amount of analyte by Faraday’s law\[Q=n F N_{A} \label{intro1} \]where n is the number of electrons per mole of analyte, F is Faraday’s constant (96 487 C mol–1), and NA is the moles of analyte. A coulomb is equivalent to an A•sec; thus, for a constant current, i, the total charge is\[Q=i t_{e} \label{intro2} \]where te is the electrolysis time. If the current varies with time, as it does in controlled-potential coulometry, then the total charge is\[Q=\int_{0}^{t_e} i(t) d t \label{intro3} \]In coulometry, we monitor current as a function of time and use either Equation \ref{intro2} or Equation \ref{intro3} to calculate Q. Knowing the total charge, we then use Equation \ref{intro1} to determine the moles of analyte. To obtain an accurate value for NA, all the current must oxidize or reduce the analyte; that is, coulometry requires 100% current efficiency or an accurate measurement of the current efficiency using a standard.Current efficiency is the percentage of current that actually leads to the analyte’s oxidation or reduction.This page titled 24.1: Introduction to Coulometry is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

24.2: Controlled-Potential Coulometry

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The easiest way to ensure 100% current efficiency is to hold the working electrode at a constant potential where the analyte is oxidized or reduced completely and where no potential interfering species are oxidized or reduced. As electrolysis progresses, the analyte’s concentration and the current decrease. The resulting current-versus-time profile for controlled-potential coulometry is shown in . Integrating the area under the curve from t = 0 to t = te gives the total charge. In this section we consider the experimental parameters and instrumentation needed to develop a controlled-potential coulometric method of analysis and its applications.To understand how an appropriate potential for the working electrode is selected, let’s develop a constant-potential coulometric method for Cu2+ based on its reduction to copper metal at a Pt working electrode.\[\mathrm{Cu}^{2+}(a q)+2 e^{-} \rightleftharpoons \mathrm{Cu}(s) \label{cp1} \] shows the three reduction reactions that can take place in an aqueous solution of Cu2+ and their standard state reduction potentials: the reduction of O2 to H2O, the reduction of Cu2+ to Cu, and the reduction of H3O+ to H2. From the diagram we know that reaction \ref{cp1} is favored when the working electrode’s potential is more negative than +0.342 V versus the standard hydrogen electrode. To ensure a 100% current efficiency, however, the potential must be sufficiently more positive than +0.000 V so that the reduction of H3O+ to H2 does not contribute significantly to the total current flowing through the electrochemical cell.We can use the Nernst equation for reaction \ref{cp1} to estimate the minimum potential for quantitatively reducing Cu2+.\[E=E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\mathrm{o}}-\frac{0.05916}{2} \log \frac{1}{\left[\mathrm{Cu}^{2+}\right]} \label{cp2} \]So why are we using the concentration of Cu2+ in Equation \ref{cp2} instead of its activity as we did in Chapter 23 when we considered potentiometry? In potentiometry we used activity because we used Ecell to determine the analyte’s concentration. Here we use the Nernst equation to help us select an appropriate potential. Once we identify a potential, we can adjust its value as needed to ensure a quantitative reduction of Cu2+. In addition, in coulometry the analyte’s concentration is given by the total charge, not the applied potential.If we define a quantitative electrolysis as one in which we reduce 99.99% of Cu2+ to Cu, then the concentration of Cu2+ at te is\[\left[\mathrm{Cu}^{2+}\right]_{t_{e}}=0.0001 \times\left[\mathrm{Cu}^{2+}\right]_{0} \label{cp3} \]where [Cu2+]0 is the initial concentration of Cu2+ in the sample. Substituting Equation \ref{cp3} into Equation \ref{cp2} allows us to calculate the desired potential.\[E=E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}-\frac{0.05916}{2} \log \frac{1}{0.0001 \times\left[\mathrm{Cu}^{2+}\right]} \label{cp4} \]If the initial concentration of Cu2+ is \(1.00 \times 10^{-4}\) M, for example, then the working electrode’s potential must be more negative than +0.105 V to quantitatively reduce Cu2+ to Cu. Note that at this potential H3O+ is not reduced to H2, maintaining 100% current efficiency.Many controlled-potential coulometric methods for Cu2+ use a potential that is negative relative to the standard hydrogen electrode—see, for example, Rechnitz, G. A. Controlled-Potential Analysis, Macmillan: New York, 1963, p.49. Based on you might expect that applying a potential <0.000 V will partially reduce H3O+ to H2, resulting in a current efficiency that is less than 100%. The reason we can use such a negative potential is that the reaction rate for the reduction of H3O+ to H2 is very slow at a Pt electrode. This results in a significant overpotential—the need to apply a potential more positive or a more negative than that predicted by thermodynamics—which shifts Eo for the H3O+/H2 redox couple to a more negative value.In controlled-potential coulometry, as shown in , the current decreases over time. As a result, the rate of electrolysis—recall from Chapter 22 that current is a measure of rate—becomes slower and an exhaustive electrolysis of the analyte may require a long time. Because time is an important consideration when designing an analytical method, we need to consider the factors that affect the analysis time.We can approximate how the current changes as a function of time ) as an exponential decay; thus, the current at time t is\[i_{t}=i_{0} e^{-k t} \label{cp5} \]where i0 is the current at t = 0 and k is a rate constant that is directly proportional to the area of the working electrode and the rate of stirring, and that is inversely proportional to the volume of solution. For an exhaustive electrolysis in which we oxidize or reduce 99.99% of the analyte, the current at the end of the analysis, te, is\[i_{t_{e}} \leq 0.0001 \times i_{0} \label{cp6} \]Substituting Equation \ref{cp6} into Equation \ref{cp5} and solving for te gives the minimum time for an exhaustive electrolysis as\[t_{e}=-\frac{1}{k} \times \ln (0.0001)=\frac{9.21}{k} \label{cp7} \]From this equation we see that a larger value for k reduces the analysis time. For this reason we usually carry out a controlled-potential coulometric analysis in a small volume electrochemical cell, using an electrode with a large surface area, and with a high stirring rate. A quantitative electrolysis typically requires approximately 30–60 min, although shorter or longer times are possible.We can use the three-electrode potentiostat in ) to set and control the potential in controlled-potential coulometry . The potential of the working electrode is measured relative to a constant-potential reference electrode that is connected to the working electrode through a high-impedance potentiometer. To set the working electrode’s potential we adjust the slide wire resistor that is connected to the auxiliary electrode. If the working electrode’s potential begins to drift, we adjust the slide wire resistor to return the potential to its initial value. The current flowing between the auxiliary electrode and the working electrode is measured with an ammeter. Of course, a modern potentionstat uses operational amplifiers to maintain the constant potential without our intervention.The working electrode is usually one of two types: a cylindrical Pt electrode manufactured from platinum-gauze ), or a Hg pool electrode. The large overpotential for the reduction of H3O+ at Hg makes it the electrode of choice for an analyte that requires a negative potential. For example, a potential more negative than –1 V versus the SHE is feasible at a Hg electrode—but not at a Pt electrode—even in a very acidic solution. Because mercury is easy to oxidize, it is less useful if we need to maintain a potential that is positive with respect to the SHE. Platinum is the working electrode of choice when we need to apply a positive potential.The auxiliary electrode, which often is a Pt wire, is separated by a salt bridge from the analytical solution. This is necessary to prevent the electrolysis products generated at the auxiliary electrode from reacting with the analyte and interfering in the analysis. A saturated calomel or Ag/AgCl electrode serves as the reference electrode.The other essential need for controlled-potential coulometry is a means for determining the total charge. One method is to monitor the current as a function of time and determine the area under the curve, as shown in . Modern instruments use electronic integration to monitor charge as a function of time. The total charge at the end of the electrolysis is read directly from a digital readout.If the product of controlled-potential coulometry forms a deposit on the working electrode, then we can use the change in the electrode’s mass as the analytical signal. For example, if we apply a potential that reduces Cu2+ to Cu at a Pt working electrode, the difference in the electrode’s mass before and after electrolysis is a direct measurement of the amount of copper in the sample. An analytical technique that uses mass as a signal a gravimetric technique; thus, we call this electrogravimetry.The majority of controlled-potential coulometric analyses involve the determination of inorganic cations and anions, including trace metals and halides ions. Table \(\PageIndex{1}\) summarizes several of these methods.Source: Rechnitz, G. A. Controlled-Potential Analysis, Macmillan: New York, 1963.Electrolytic reactions are written in terms of the change in the analyte’s oxidation state. The actual species in solution depends on the analyte.The ability to control selectivity by adjusting the working electrode’s potential makes controlled-potential coulometry particularly useful for the analysis of alloys. For example, we can determine the composition of an alloy that contains Ag, Bi, Cd, and Sb by dissolving the sample and placing it in a matrix of 0.2 M H2SO4 along with a Pt working electrode and a Pt counter electrode. If we apply a constant potential of +0.40 V versus the SCE, Ag(I) deposits on the electrode as Ag and the other metal ions remain in solution. When electrolysis is complete, we use the total charge to determine the amount of silver in the alloy. Next, we shift the working electrode’s potential to –0.08 V versus the SCE, depositing Bi on the working electrode. When the coulometric analysis for bismuth is complete, we determine antimony by shifting the working electrode’s potential to –0.33 V versus the SCE, depositing Sb. Finally, we determine cadmium following its electrodeposition on the working electrode at a potential of –0.80 V versus the SCE.We also can use controlled-potential coulometry for the quantitative analysis of organic compounds, although the number of applications is significantly less than that for inorganic analytes. One example is the six-electron reduction of a nitro group, –NO2, to a primary amine, –NH2, at a mercury electrode. Solutions of picric acid—also known as 2,4,6-trinitrophenol, or TNP, a close relative of TNT—is analyzed by reducing it to triaminophenol.Another example is the successive reduction of trichloroacetate to dichloroacetate, and of dichloroacetate to monochloroacetate\[\text{Cl}_3\text{CCOO}^-(aq) + \text{H}_3\text{O}^+(aq) + 2 e^- \rightleftharpoons \text{Cl}_2\text{HCCOO}^-(aq) + \text{Cl}^-(aq) + \text{H}_2\text{O}(l) \nonumber \]\[\text{Cl}_2\text{HCCOO}^-(aq) + \text{ H}_3\text{O}^+(aq) + 2 e^- \rightleftharpoons \text{ ClH}_2\text{CCOO}^-(aq) + \text{ Cl}^-(aq) + \text{H}_2\text{O}(l) \nonumber \]We can analyze a mixture of trichloroacetate and dichloroacetate by selecting an initial potential where only the more easily reduced trichloroacetate reacts. When its electrolysis is complete, we can reduce dichloroacetate by adjusting the potential to a more negative potential. The total charge for the first electrolysis gives the amount of trichloroacetate, and the difference in total charge between the first electrolysis and the second electrolysis gives the amount of dichloroacetate.One useful application of controlled-potential coulometry is determining the number of electrons involved in a redox reaction. To make the determination, we complete a controlled-potential coulometric analysis using a known amount of a pure compound. The total charge at the end of the electrolysis is used to determine the value of n using Faraday’s law. A 0.3619-g sample of tetrachloropicolinic acid, C6HNO2Cl4, is dissolved in distilled water, transferred to a 1000-mL volumetric flask, and diluted to volume. An exhaustive controlled-potential electrolysis of a 10.00-mL portion of this solution at a spongy silver cathode requires 5.374 C of charge. What is the value of n for this reduction reaction?The 10.00-mL portion of sample contains 3.619 mg, or \(1.39 \times 10^{-5}\) mol of tetrachloropicolinic acid. Solving for n gives\[n=\frac{Q}{F N_{A}}=\frac{5.374 \text{ C}}{\left(96478 \text{ C/mol } e^{-}\right)\left(1.39 \times 10^{-5} \text{ mol } \mathrm{C}_{6} \mathrm{HNO}_{2} \mathrm{Cl}_{4}\right)} = 4.01 \text{ mol e}^-/\text{mol } \mathrm{C}_{6} \mathrm{HNO}_{2} \mathrm{Cl}_{4} \nonumber \]Thus, reducing a molecule of tetrachloropicolinic acid requires four electrons. The overall reaction, which results in the selective formation of 3,6-dichloropicolinic acid, isThis page titled 24.2: Controlled-Potential Coulometry is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

24.3: Controlled-Current Coulometry

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A second approach to coulometry is to use a constant current in place of a constant potential, which results in the current-versus-time profile shown in . Controlled-current coulometry has two advantages over controlled-potential coulometry. First, the analysis time is shorter because the current does not decrease over time. A typical analysis time for controlled-current coulometry is less than 10 min, compared to approximately 30–60 min for controlled-potential coulometry. Second, because the total charge is simply the product of current and time, there is no need to integrate the current-time curve in .Using a constant current presents us with two important experimental problems. First, during electrolysis the analyte’s concentration—and, therefore, the current that results from its oxidation or reduction—decreases continuously. To maintain a constant current we must allow the potential to change until another oxidation reaction or reduction reaction occurs at the working electrode. Unless we design the system carefully, this secondary reaction results in a current efficiency that is less than 100%. The second problem is that we need a method to determine when the analyte's electrolysis is complete. In a controlled-potential coulometric analysis we know that electrolysis is complete when the current reaches zero, or when it reaches a constant background or residual current. In a controlled-current coulometric analysis, however, current continues to flow even when the analyte’s electrolysis is complete. A suitable method for determining the reaction’s endpoint, te, is needed.To illustrate why a change in the working electrode’s potential may result in a current efficiency of less than 100%, let’s consider the coulometric analysis for Fe2+ based on its oxidation to Fe3+ at a Pt working electrode in 1 M H2SO4.\[\mathrm{Fe}^{2+}(a q) \rightleftharpoons \text{ Fe}^{3+}(a q)+e^{-} \label{ci1} \] shows the relevant potentials for this system. At the beginning of the analysis, the potential of the working electrode remains nearly constant at a level near its initial value.As the concentration of Fe2+ decreases and the concentration of Fe3+ increases, the working electrode’s potential shifts toward more positive values until the oxidation of H2O begins.\[2 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \text{ O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 e^{-} \label{ci2} \]Because a portion of the total current comes from the oxidation of H2O, the current efficiency for the analysis is less than 100% and we cannot use the equation \(Q = it\) to determine the amount of Fe2+ in the sample.Although we cannot prevent the potential from drifting until another species undergoes oxidation, we can maintain a 100% current efficiency if the product of that secondary oxidation reaction both rapidly and quantitatively reacts with the remaining Fe2+. To accomplish this we add an excess of Ce3+ to the analytical solution. As shown in , when the potential of the working electrode shifts to a more positive potential, Ce3+ begins to oxidize to Ce4+\[\mathrm{Ce}^{3+}(a q) \rightleftharpoons \text{ Ce}^{4+}(a q)+e^{-} \label{ci3} \]The Ce4+ that forms at the working electrode rapidly mixes with the solution where it reacts with any available Fe2+.\[\mathrm{Ce}^{4+}(a q)+\text{ Fe}^{2+}(a q) \rightleftharpoons \text{ Ce}^{3+}(a q)+\text{ Fe}^{3+}(a q) \label{ci4} \]Combining reaction \ref{ci3} and reaction \ref{ci4} shows that the net reaction is the oxidation of Fe2+ to Fe3+\[\mathrm{Fe}^{2+}(a q) \rightleftharpoons \text{ Fe}^{3+}(a q)+e^{-} \label{ci5} \]which maintains a current efficiency of 100%. A species used to maintain 100% current efficiency is called a mediator.Adding a mediator solves the problem of maintaining 100% current efficiency, but it does not solve the problem of determining when the analyte's electrolysis is complete. Using the analysis for Fe2+ in , when the oxidation of Fe2+ is complete current continues to flow from the oxidation of Ce3+, and, eventually, the oxidation of H2O. What we need is a signal that tells us when no more Fe2+ is present in the solution.For our purposes, it is convenient to treat a controlled-current coulometric analysis as a reaction between the analyte, Fe2+, and the mediator, Ce3+, as shown by reaction \ref{ci4}. This reaction is identical to a redox titration; thus, we can use the end points for a redox titration—visual indicators and potentiometric or conductometric measurements—to signal the end of a controlled-current coulometric analysis. For example, ferroin provides a useful visual endpoint for the Ce3+ mediated coulometric analysis for Fe2+, changing color from red to blue when the electrolysis of Fe2+ is complete.We can carry out controlled-current coulometry using the two-electrode galvanostat shown in , which consists of a working electrode and a counter electrode. The working electrode—often a simple Pt electrode—also is called the generator electrode since it is where the mediator reacts to generate the species that reacts with the analyte. If necessary, the counter electrode is isolated from the analytical solution by a salt bridge or a porous frit to prevent its electrolysis products from reacting with the analyte. The current from the power supply through the working electrode is\[i=\frac{E_{\mathrm{PS}}}{R+R_{\mathrm{cell}}} \label{ci6} \]where EPS is the potential of the power supply, R is the resistance of the resistor, and Rcell is the resistance of the electrochemical cell. If R >> Rcell, then the current between the auxiliary and working electrodes\[i=\frac{E_{\mathrm{PS}}}{R} \approx \text{constant} \label{ci7} \]maintains a constant value. To monitor the working electrode’s potential, which changes as the composition of the electrochemical cell changes, we can include an optional reference electrode and a high-impedance potentiometer.Alternatively, we can generate the oxidizing agent or the reducing agent externally, and allow it to flow into the analytical solution. shows one simple method for accomplishing this. A solution that contains the mediator flows into a small-volume electrochemical cell with the products exiting through separate tubes. Depending upon the analyte, the oxidizing agent or the reducing reagent is delivered to the analytical solution. For example, we can generate Ce4+ using an aqueous solution of Ce3+, directing the Ce4+ that forms at the anode to our sample.There are two other crucial needs for controlled-current coulometry: an accurate clock for measuring the electrolysis time, te, and a switch for starting and stopping the electrolysis. An analog clock can record time to the nearest ±0.01 s, but the need to stop and start the electrolysis as we approach the endpoint may result in an overall uncertainty of ±0.1 s. A digital clock allows for a more accurate measurement of time, with an overall uncertainty of ±1 ms. The switch must control both the current and the clock so that we can make an accurate determination of the electrolysis time.A controlled-current coulometric method sometimes is called a coulometric titration because of its similarity to a conventional titration. For example, in the controlled-current coulometric analysis for Fe2+ using a Ce3+ mediator, the oxidation of Fe2+ by Ce4+ (reaction \ref{ci4}) is identical to the reaction in a redox titration.There are other similarities between controlled-current coulometry and titrimetry. If we combine the equation \(Q = nFN_a\) and the equation \(Q = it_e\) and solve for the moles of analyte, NA, we obtain the following equation.\[N_{A}=\frac{i}{n F} \times t_{e} \label{ci8} \]Compare Equation \ref{ci8} to the relationship between the moles of analyte, NA, and the moles of titrant, NT, in a titration\[N_{A}=N_{T}=M_{T} \times V_{T} \label{ci9} \]where MT and VT are the titrant’s molarity and the volume of titrant at the end point. In constant-current coulometry, the current source is equivalent to the titrant and the value of that current is analogous to the titrant’s molarity. Electrolysis time is analogous to the volume of titrant, and te is equivalent to the a titration’s end point. Finally, the switch for starting and stopping the electrolysis serves the same function as a buret’s stopcock.For simplicity, we assumed above that the stoichiometry between the analyte and titrant is 1:1. The assumption, however, is not important and does not effect our observation of the similarity between controlled-current coulometry and a titration.The use of a mediator makes a coulometric titration a more versatile analytical technique than controlled-potential coulometry. For example, the direct oxidation or reduction of a protein at a working electrode is difficult if the protein’s active redox site lies deep within its structure. A coulometric titration of the protein is possible, however, if we use the oxidation or reduction of a mediator to produce a solution species that reacts with the protein. Table \(\PageIndex{1}\) summarizes several controlled-current coulometric methods based on a redox reaction using a mediator.For an analyte that is not easy to oxidize or reduce, we can complete a coulometric titration by coupling a mediator’s oxidation or reduction to an acid–base, precipitation, or complexation reaction that involves the analyte. For example, if we use H2O as a mediator, we can generate H3O+at the anode\[6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons 4 \mathrm{H}_{3} \text{O}^{+}(a q)+\text{ O}_{2}(g)+4 e^{-} \nonumber \]and generate OH– at the cathode.\[2 \mathrm{H}_{2} \mathrm{O}(l)+2 e^{-} \rightleftharpoons 2 \mathrm{OH}^{-}(a q)+\text{ H}_{2}(g) \nonumber \]If we carry out the oxidation or reduction of H2O using the generator cell in , then we can selectively dispense H3O+ or OH– into a solution that contains the analyte. The resulting reaction is identical to that in an acid–base titration. Coulometric acid–base titrations have been used for the analysis of strong and weak acids and bases, in both aqueous and non-aqueous matrices. Table \(\PageIndex{2}\) summarizes several examples of coulometric titrations that involve acid–base, complexation, and precipitation reactions.In comparison to a conventional titration, a coulometric titration has two important advantages. The first advantage is that electrochemically generating a titrant allows us to use a reagent that is unstable. Although we cannot prepare and store a solution of a highly reactive reagent, such as Ag2+ or Mn3+, we can generate them electrochemically and use them in a coulometric titration. Second, because it is relatively easy to measure a small quantity of charge, we can use a coulometric titration to determine an analyte whose concentration is too small for a conventional titration.The following example shows the calculations for a typical coulometric analysis.To determine the purity of a sample of Na2S2O3, a sample is titrated coulometrically using I– as a mediator and \(\text{I}_3^-\) as the titrant. A sample weighing 0.1342 g is transferred to a 100-mL volumetric flask and diluted to volume with distilled water. A 10.00-mL portion is transferred to an electrochemical cell along with 25 mL of 1 M KI, 75 mL of a pH 7.0 phosphate buffer, and several drops of a starch indicator solution. Electrolysis at a constant current of 36.45 mA requires 221.8 s to reach the starch indicator endpoint. Determine the sample’s purity.As shown in Table \(\PageIndex{1}\), the coulometric titration of \(\text{S}_2 \text{O}_3^{2-}\) with \(\text{I}_3^-\) is\[2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q)+\text{ I}_{3}^{-}(a q)\rightleftharpoons \text{ S}_{4} \mathrm{O}_{6}^{2-}(a q)+3 \mathrm{I}^{-}(a q) \nonumber \]The oxidation of \(\text{S}_2 \text{O}_3^{2-}\) to \(\text{S}_4 \text{O}_6^{2-}\) requires one electron per \(\text{S}_2 \text{O}_3^{2-}\) (n = 1). Combining the equations \(Q = nFN_A\) and \(Q = it_e\), and solving for the moles and grams of Na2S2O3 gives\[N_{A} =\frac{i t_{e}}{n F}=\frac{(0.03645 \text{ A})(221.8 \text{ s})}{\left(\frac{1 \text{ mol } e^{-}}{\text{mol Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}}\right)\left(\frac{96487 \text{ C}}{\text{mol } e^{-}}\right)} =8.379 \times 10^{-5} \text{ mol Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \nonumber \]This is the amount of Na2S2O3 in a 10.00-mL portion of a 100-mL sample; thus, there are 0.1325 grams of Na2S2O3 in the original sample. The sample’s purity, therefore, is\[\frac{0.1325 \text{ g} \text{ Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}}{0.1342 \text{ g} \text { sample }} \times 100=98.73 \% \text{ w} / \text{w } \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \nonumber \]Note that for this calcuation, it does not matter whether \(\text{S}_2 \text{O}_3^{2-}\) is oxidized at the working electrode or is oxidized by \(\text{I}_3^-\).This page titled 24.3: Controlled-Current Coulometry is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

25.1: Potential Excitation Signals and Currents in Voltammetry

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In voltammetry we apply a time-dependent potential to an electrochemical cell and measure the resulting current as a function of that potential.As shown in , the potential may consist of (a) a linear scan or (b) a series of pulses. For the linear scan in (a), the direction of the scan can be reversed and repeated for additional cycles. The series of pulses in (b) shows just one of several different pulsed potential excitation signals; we will consider other pulse trains in the section on polarography.The current responses in show the three common types of signals. In (c) and (d) the current is monitored directly as the potential is changed. In (e) a change in current is recorded using the current immediately before and after the application of a potential pulse. The current itself has three components: faradic current from the oxidation or reduction of the analyte, a charging current, and residual currents.Faradic current is the result of oxidation or reduction of the analyte at the working electrode. The ease with which electrons move between the electrode and the species that reacts at the electrode affects the faradiac current. When electron transfer kinetics are fast, the redox reaction is at equilibrium. Under these conditions the redox reaction is electrochemically reversible and the Nernst equation applies. If the electron transfer kinetics are sufficiently slow, the concentration of reactants and products at the electrode surface—and thus the magnitude of the faradaic current—are not what is predicted by the Nernst equation. In this case the system is electrochemically irreversible.In addition to the faradaic current from a redox reaction, the current in an electrochemical cell includes other, nonfaradaic sources. Suppose the charge on an electrode is zero and we suddenly change its potential so that the electrode’s surface acquires a positive charge. Cations near the electrode’s surface will respond to this positive charge by migrating away from the electrode; anions, on the other hand, will migrate toward the electrode. This migration of ions occurs until the electrode’s positive surface charge and the negative charge of the solution near the electrode are equal. Because the movement of ions and the movement of electrons are indistinguishable, the result is a small, short-lived nonfaradaic current that we call the charging current. Every time we change the electrode’s potential, a transient charging current flows.The migration of ions in response to the electrode’s surface charge leads to the formation of a structured electrode-solution interface that we call the electrical double layer, or EDL. When we change an electrode’s potential, the charging current is the result of a restructuring of the EDL. The exact structure of the electrical double layer is not important in the context of this text, but you can consult this chapter’s additional resources for additional information. See Chapter 22.1 for additional details.Even in the absence of analyte, a small, measurable current flows through an electrochemical cell. In addition to the charging current discussed above, the residual current includes a faradaic current from the oxidation or reduction of trace impurities in the sample. Methods for discriminating between the analyte’s faradaic current and the residual current are discussed later in this chapter.This page titled 25.1: Potential Excitation Signals and Currents in Voltammetry is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

25.2: Voltammetric Instrumentation

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Although early voltammetric methods used only two electrodes, a modern voltammeter makes use of a three-electrode potentiostat, such as that shown in . The potential of the working electrode is measured relative to a constant-potential reference electrode that is connected to the working electrode through a high-impedance potentiometer. The auxiliary electrode generally is a platinum wire and the reference electrode usually is a SCE or a Ag/AgCl electrode. We apply a time-dependent potential excitation signal to the working electrode—changing its potential relative to the fixed potential of the reference electrode—and measure the current that flows between the working electrode and the auxiliary electrode. Modern potentiostats include waveform generators that allow us to apply a time-dependent potential profile, such as a series of potential pulses, to the working electrode.For the working electrode we can choose among several different materials, including mercury, platinum, gold, silver, and carbon. The earliest voltammetric techniques used a mercury working electrode. Because mercury is a liquid, the working electrode usually is a drop suspended from the end of a capillary tube. In the hanging mercury drop electrode, or HMDE, we extrude the drop of Hg by rotating a micrometer screw that pushes the mercury from a reservoir through a narrow capillary tube a).In the dropping mercury electrode, or DME, mercury drops form at the end of the capillary tube as a result of gravity b). Unlike the HMDE, the mercury drop of a DME grows continuously—as mercury flows from the reservoir under the influence of gravity—and has a finite lifetime of several seconds. At the end of its lifetime the mercury drop is dislodged, either manually or on its own, and is replaced by a new drop. The static mercury drop electrode, or SMDE, uses a solenoid driven plunger to control the flow of mercury c). Activation of the solenoid momentarily lifts the plunger, allowing mercury to flow through the capillary, forming a single, hanging Hg drop. Repeated activation of the solenoid produces a series of Hg drops. In this way the SMDE may be used as either a HMDE or a DME. There is one additional type of mercury electrode: the mercury film electrode. A solid electrode—typically carbon, platinum, or gold—is placed in a solution of Hg2+ and held at a potential where the reduction of Hg2+ to Hg is favorable, depositing a thin film of mercury on the solid electrode’s surface.Mercury has several advantages as a working electrode. Perhaps its most important advantage is its high overpotential for the reduction of H3O+ to H2, which makes accessible potentials as negative as –1 V versus the SCE in acidic solutions and –2 V versus the SCE in basic solutions ). A species such as Zn2+, which is difficult to reduce at other electrodes without simultaneously reducing H3O+, is easy to reduce at a mercury working electrode. Other advantages include the ability of metals to dissolve in mercury—which results in the formation of an amalgam—and the ability to renew the surface of the electrode by extruding a new drop. One limitation to mercury as a working electrode is the ease with which it is oxidized. Depending on the solvent, a mercury electrode can not be used at potentials more positive than approximately –0.3 V to +0.4 V versus the SCE.Solid electrodes constructed using platinum, gold, silver, or carbon may be used over a range of potentials, including potentials that are negative and positive with respect to the SCE ). For example, the potential window for a Pt electrode extends from approximately +1.2 V to –0.2 V versus the SCE in acidic solutions, and from +0.7 V to –1 V versus the SCE in basic solutions. A solid electrode can replace a mercury electrode for many voltammetric analyses that require negative potentials, and is the electrode of choice at more positive potentials. Except for the carbon paste electrode, a solid electrode is fashioned into a disk and sealed into the end of an inert support with an electrical lead ). The carbon paste electrode is made by filling the cavity at the end of the inert support with a paste that consists of carbon particles and a viscous oil. Solid electrodes are not without problems, the most important of which is the ease with which the electrode’s surface is altered by the adsorption of a solution species or by the formation of an oxide layer. For this reason a solid electrode needs frequent reconditioning, either by applying an appropriate potential or by polishing.A typical arrangement for a voltammetric electrochemical cell is shown in . In addition to the working electrode, the reference electrode, and the auxiliary electrode, the cell also includes a N2-purge line for removing dissolved O2, and an optional stir bar. Electrochemical cells are available in a variety of sizes, allowing the analysis of solution volumes ranging from more than 100 mL to as small as 50 μL.This page titled 25.2: Voltammetric Instrumentation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

25.3: Linear Sweep Voltammetry

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In the simplest voltammetric experiment we apply a linear potential ramp as an excitation signal and record the current that flows in response to the change in potential. Among the experimental variables under our control are the initial potential, the final potential, the scan rate, and whether we choose to stir the solution or leave it unstirred. We call this linear sweep voltammetry.To illustrate how linear sweep voltammetry works, let's consider what happens when we reduce \(\text{Fe(CN)}_6^{3-}\) to \(\text{Fe(CN)}_6^{4-}\) at the working electrode. The relationship between the concentrations of \(\text{Fe(CN)}_6^{3-}\), the concentration of \(\text{Fe(CN)}_6^{4-}\), and the potential is given by the Nernst equation\[E=+0.356 \text{ V}-0.05916 \log \frac{\left[\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\right]_{x=0}}{\left[\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\right]_{x=0}} \label{lsv1} \]where +0.356V is the standard-state potential for the \(\text{Fe(CN)}_6^{3-}\)/\(\text{Fe(CN)}_6^{4-}\) redox couple, and x = 0 indicates that the concentrations of \(\text{Fe(CN)}_6^{3-}\) and \(\text{Fe(CN)}_6^{4-}\) are those at the surface of the working electrode. We use surface concentrations instead of bulk concentrations because the equilibrium position for the redox reaction\[\mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q)+e^{-}\rightleftharpoons\mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q) \label{lsv2} \]is established at the electrode’s surface.Let’s assume we have a solution for which the initial concentration of \(\text{Fe(CN)}_6^{3-}\) is 1.0 mM and that \(\text{Fe(CN)}_6^{4-}\) is absent. shows the relationship between the applied potential and the species that are stable at the electrode's surface.If we apply a potential of +0.530 V to the working electrode, the concentrations of \(\text{Fe(CN)}_6^{3-}\) and \(\text{Fe(CN)}_6^{4-}\) at the surface of the electrode are unaffected, and no faradaic current is observed. If we switch the potential to +0.356 V some of the \(\text{Fe(CN)}_6^{3-}\) at the electrode’s surface is reduced to \(\text{Fe(CN)}_6^{4-}\)until we reach a condition where\[\left[\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\right]_{x=0}=\left[\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\right]_{x=0}=0.50 \text{ mM} \label{lsv3} \]If this is all that happens after we apply the potential, then there would be a brief surge of faradaic current that quickly returns to zero, which is not the most interesting of results (although this is the basis for chronoamperometry, an electrochemical method we will not consider in this text). Although the concentrations of \(\text{Fe(CN)}_6^{3-}\) and \(\text{Fe(CN)}_6^{4-}\) at the electrode surface are 0.50 mM, their concentrations in bulk solution remains unchanged.Because of this difference in concentration, there is a concentration gradient between the electrode’s surface and the bulk solution. This concentration gradient creates a driving force that transports \(\text{Fe(CN)}_6^{4-}\) away from the electrode and that transports \(\text{Fe(CN)}_6^{3-}\) to the electrode ). As the \(\text{Fe(CN)}_6^{3-}\) arrives at the electrode it, too, is reduced to \(\text{Fe(CN)}_6^{4-}\). A faradaic current continues to flow until there is no difference between the concentrations of \(\text{Fe(CN)}_6^{3-}\) and \(\text{Fe(CN)}_6^{4-}\) at the electrode and their concentrations in bulk solution (although this might take a long time!).Although the potential at the working electrode determines if a faradaic current flows, the magnitude of the current is determined by the rate of the resulting oxidation or reduction reaction. Two factors contribute to the rate of the electrochemical reaction: the rate at which the reactants and products are transported to and from the electrode—what we call mass transport—and the rate at which electrons pass between the electrode and the reactants and products in solution.There are three modes of mass transport that affect the rate at which reactants and products move toward or away from the electrode surface: diffusion, migration, and convection. Diffusion occurs whenever the concentration of an ion or a molecule at the surface of the electrode is different from that in bulk solution. If we apply a potential sufficient to completely reduce \(\text{Fe(CN)}_6^{3-}\) at the electrode surface, the result is a concentration gradient similar to that shown in . The region of solution over which diffusion occurs is the diffusion layer. In the absence of other modes of mass transport, the width of the diffusion layer, \(\delta\), increases with time as the \(\text{Fe(CN)}_6^{3-}\) must diffuse from an increasingly greater distance.Convection occurs when we mix the solution, which carries reactants toward the electrode and removes products from the electrode. The most common form of convection is stirring the solution with a stir bar; other methods include rotating the electrode and incorporating the electrode into a flow-cell.The final mode of mass transport is migration, which occurs when a charged particle in solution is attracted to or repelled from an electrode that carries a surface charge. If the electrode carries a positive charge, for example, an anion will move toward the electrode and a cation will move toward the bulk solution. Unlike diffusion and convection, migration affects only the mass transport of charged particles.The movement of material to and from the electrode surface is a complex function of all three modes of mass transport. In the limit where diffusion is the only significant form of mass transport, the current, \(i\), in a voltammetric cell is proportional to the slope of the concentration profile in \[i \propto \frac {\partial C} {\partial x} \label{lsv4} \]where \(C\) is the concentration of \(\text{Fe(CN)}_6^{3-}\) and \(x\) is distance.For Equation \ref{lsv4} to be valid, convection and migration must not interfere with the formation of a diffusion layer. We can eliminate migration by adding a high concentration of an inert supporting electrolyte. Because ions of similar charge are equally attracted to or repelled from the surface of the electrode, each has an equal probability of undergoing migration. A large excess of an inert electrolyte ensures that few reactants or products experience migration. Although it is easy to eliminate convection by not stirring the solution, there are experimental designs where we cannot avoid convection, either because we must stir the solution or because we are using an electrochemical flow cell. Fortunately, as shown in , the dynamics of a fluid moving past an electrode results in a small diffusion layer—typically 1–10 μm in thickness—in which the rate of mass transport by convection drops to zero. shows the linear sweep voltammogram (the center image, which shows the current as a function of time) and eight snapshots of the concentration profiles for the reduction of \(\text{Fe(CN)}_6^{3-}\) to \(\text{Fe(CN)}_6^{4-}\) in an unstirred solution. The initial potential was set to +0.530 V and the final potential was set to +0.182 V with a scan rate of 0.050 V/s.At the initial potential, only \(\text{Fe(CN)}_6^{3-}\) is stable at the electrode surface, and no current flows. After 0.696 s the potential is 0.495 V (image to the left of the linear sweep voltammogram) and, because \(\text{Fe(CN)}_6^{3-}\) remains stable at the electrode surface, no current flows. Moving clockwise around the linear sweep voltammogram, the applied potential becomes smaller and the concentration of \(\text{Fe(CN)}_6^{3-}\) at the electrode surface decreases and the concentration of \(\text{Fe(CN)}_6^{4-}\) increases. Initially the slope of the concentration gradient, and, therefore, the current increases; as the concentration of \(\text{Fe(CN)}_6^{3-}\) at the electrode surface approaches zero, however, the concentration gradient becomes less steep and the current decreases. The result is the linear sweep voltammogram in the center of the diagram.If we run the same experiment as in , but stir the solution, the resulting linear sweep voltammogram and concentration profiles are those in . Stirring the solution, as we saw in creates a diffusion layer whose thickness is independent of time.As a result, instead of the peak current in , the current reaches a steady-state value, which we call the limiting current, \(i_l\). The linear sweep voltammogram also has a characteristic half-wave potential, \(E_{1/2}\), when the current is 50% of the limiting current. shows how the limiting current and half-wave potential are measured.Earlier we noted, in Equation \ref{lsv4}, that the current in linear sweep voltammetry is proportional to the slope of the concentration profile. The current is also a function of other variables, as shown here for the reduction of \(\text{Fe(CN)}_6^{3-}\) to \(\text{Fe(CN)}_6^{4-}\)\[i = \frac{ n F A D \left( \left[ \ce{Fe(CN)6^{3-}} \right]_\text{bulk} - \left[ \ce{Fe(CN)6^{3-}} \right]_\text{x = 0} \right)} {\delta} \label{lsv5} \]where n the number of electrons in the redox reaction, F is Faraday’s constant, A is the area of the electrode, D is the diffusion coefficient for \(\text{Fe(CN)}_6^{3-}\), \(\delta\) is the thickness of the diffusion layer, and \(\left( \left[ \ce{Fe(CN)6^{3-}} \right]_\text{bulk} - \left[ \ce{Fe(CN)6^{3-}} \right]_\text{x = 0} \right)\) is the difference in the concentration of \( \ce{Fe(CN)6^{3-}}\) between the bulk solution and the electrode's surface.Because \(n\), \(F\), \(A\), and \(D\) are constants, and because \(\delta\) is a constant if we stir the solution, we can write Equation \ref{lsv5} as\[i = K_{\ce{Fe(CN)6^{3-}}} \left( \left[ \ce{Fe(CN)6^{3-}} \right]_\text{bulk} - \left[ \ce{Fe(CN)6^{3-}} \right]_\text{x = 0} \right) \label{lsv6} \]where \(K_{\ce{Fe(CN)6^{3-}}}\) is a constant. If we use the limiting current, then \(\left[ \ce{Fe(CN)6^{3-}} \right]_\text{x = 0}\) is zero, and Equation \ref{lsv6} becomes\[i_l = K_{\ce{Fe(CN)6^{3-}}} \left[ \ce{Fe(CN)6^{3-}} \right]_\text{bulk} \label{lsv7} \]A reversible electrochemical reaction is one in which the concentration of the oxidized and reduced species at the electrode surface remain in thermodynamic equilibrium with each other. When this is true, the Nernst equation explains the relationship between the applied potential, their concentration, and the standard state potential.Equation \ref{lsv7} shows us that the limiting current is a measure of the concentration of \(\text{Fe(CN)}_6^{3-}\) in bulk solution, which means we can use the limiting current for quantitative work. also shows that there is a qualitative relationship between the half-wave potential, \(E_{1/2}\), and the limiting current; however, it is not yet clear what the half-wave potential represents.If we solve Equation \ref{lsv7} for \(\left[ \ce{Fe(CN)6^{3-}} \right]_\text{bulk} \) and substitute into Equation \ref{lsv6} and rearrange, we have\[ \left[ \ce{Fe(CN)6^{3-}} \right]_\text{x = 0} = \frac {i_l - i} {K_{\ce{Fe(CN)6^{3-}}}} \label{lsv8} \]If we take the same approach with \(\text{Fe(CN)}_6^{4-}\), which forms at the electrode solution, then we have\[i = -\frac{ n F A D \left( \left[ \ce{Fe(CN)6^{4-}} \right]_\text{bulk} - \left[ \ce{Fe(CN)6^{4-}} \right]_\text{x = 0} \right)} {\delta} = K_{\ce{Fe(CN)6^{4-}}} \left[ \ce{Fe(CN)6^{4-}} \right]_\text{x = 0} \label{lsv9} \]\[ \left[ \ce{Fe(CN)6^{4-}} \right]_\text{x = 0} = \frac {-i} {K_{\ce{Fe(CN)6^{4-}}}} \label{lsv10} \]where the minus sign accounts for the concentration profile having a negative slope. Substituting Equation \ref{lsv9} and Equation \ref{lsv10} into Equation \ref{lsv1}, which is the Nersnt equation, gives\[E = E^{\circ} - 0.05916 \log \frac {-i/K_{\ce{Fe(CN)6^{4-}}}} {(i_l - i)/K_{\ce{Fe(CN)6^{3-}}}} \label{lsv11} \]\[E = E^{\circ} + 0.05916 \log \frac{K_{\ce{Fe(CN)6^{3-}}}}{K_{\ce{Fe(CN)6^{4-}}}} - 0.05916 \log \frac {i} {i_l - i} \label{lsv12} \]When \(i = \frac {i_l - i} {2}\), which is the definition of \(E_{1/2}\), Equation \ref{lsv12} simplifies to\[E_{1/2} = E^{\circ} + 0.05916 \log \frac{K_{\ce{Fe(CN)6^{3-}}}}{K_{\ce{Fe(CN)6^{4-}}}} \label{lsv13} \]The only difference between \(K_{\ce{Fe(CN)6^{3-}}}\) and \(K_{\ce{Fe(CN)6^{4-}}}\) are the diffusion coefficients, \(D\), for \(\ce{Fe(CN)6^{3-}}\) and for \(\ce{Fe(CN)6^{4-}}\). As these values should be similar, we have\[E_{1/2} \approx E^{\circ} \label{lsv14} \]and \(E_{1/2}\) provides an estimate for the standard state reduction potential.When an electrochemical reaction is not reversible, the Nernst equation no longer applies, which means we can no longer assume that the half-wave potential provides an estimate for the standard state reduction potential. The relationship between the limiting current and the concentration of the electroactive species in bulk solution still holds true and quantitative work remains possible.The presence of dissolved oxygen creates a complication as it is capable of undergoing reduction reactions at the electrode's surface that may interfere with the determination of the analyte's limiting current or half-wave potential. For example, O2 is reduced to H2O2 with a standard state potential of +0.695 V\[\ce{O2}(g) + 2\ce{H+}(aq) + 2e^{-} \rightleftharpoons \ce{H2O2}(aq) \label{lsv15} \]and H2O2 subsequently is reduced to H2O at a standard state potential of +1.763 V.\[\ce{H2O2}(aq) + 2\ce{H+}(aq) + 2e^{-} \rightleftharpoons 2 \ce{H2O}(aq) \label{lsv16} \]This is the reason that a typical cell for voltammetry (see includes the ability to pass N2 through the solution to remove dissolved O2. Once the solution is deaerated, N2 is allowed to flow over the solution to prevent O2 from reentering the solution.As we learned in the previous section, the limiting current in linear sweep voltammetry is proportional to the concentration of the species undergoing oxidation or reduction at the electrode surface, which makes it a useful tool for a quantitative analysis. Because we are interested only in the limiting current, most quantitative methods simply hold the potential of the working electrode at a fixed value and measure the limiting current. Because we are measuring the current as a function of time instead of potential, these are called amperometric methods (where ampere is the unit for current). Several examples of amperometic methods are gathered here.One important detector for high-performance liquid chromatography (HPLC) is one in which the mobile phase eluting from the column passes through a small volume electrochemical cell in which the working electrode is held at a potential that will oxidize or reduce the analytes. The resulting current is plotted as function of time to yield the chromatogram. A similar arrangement is used in flow-injection analysis (FIA). See Chapter 28 (HPLC) and Chapter 33 (FIA) for further details.One important application of amperometry is in the construction of chemical sensors. One of the first amperometric sensors was developed in 1956 by L. C. Clark to measure dissolved O2 in blood. shows the sensor’s design, which is similar to a potentiometric membrane electrode. A thin, gas-permeable membrane is stretched across the end of the sensor and is separated from the working electrode and the counter electrode by a thin solution of KCl. The working electrode is a Pt disk cathode, and a Ag ring anode serves as the counter electrode. Although several gases can diffuse across the membrane, including O2, N2, and CO2, only oxygen undergoes reduction at the cathode\[\mathrm{O}_{2}(g)+4 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+4 e^{-}\rightleftharpoons 6 \mathrm{H}_{2} \mathrm{O}(l) \label{lsv17} \]with its concentration at the electrode’s surface quickly reaching zero. The concentration of O2 at the membrane’s inner surface is fixed by its diffusion through the membrane, which creates a limiting current. The result is a steady-state current that is proportional to the concentration of dissolved oxygen. Because the electrode consumes oxygen, the sample is stirred to prevent the depletion of O2 at the membrane’s outer surface.The oxidation of the Ag anode is the other half-reaction.\[\mathrm{Ag}(s)+\text{ Cl}^{-}(a q)\rightleftharpoons \mathrm{AgCl}(s)+e^{-} \nonumber \]Another example of an amperometric sensor is a glucose sensor. In this sensor the single membrane in is replaced with three membranes. The outermost membrane of polycarbonate is permeable to glucose and O2. The second membrane contains an immobilized preparation of glucose oxidase that catalyzes the oxidation of glucose to gluconolactone and hydrogen peroxide.\[\beta-\mathrm{D}-\text {glucose }(a q)+\text{ O}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \text {gluconolactone }(a q)+\text{ H}_{2} \mathrm{O}_{2}(a q) \label{lsv18} \]The hydrogen peroxide diffuses through the innermost membrane of cellulose acetate where it undergoes oxidation at a Pt anode.\[\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \rightleftharpoons \text{ O}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 e^{-} \label{lsv19} \] summarizes the reactions that take place in this amperometric sensor. FAD is the oxidized form of flavin adenine nucleotide—the active site of the enzyme glucose oxidase—and FADH2 is the active site’s reduced form. Note that O2 serves a mediator, carrying electrons to the electrode.By changing the enzyme and mediator, it is easy to extend to the amperometric sensor in to the analysis of other analytes. For example, a CO2 sensor has been developed using an amperometric O2 sensor with a two-layer membrane, one of which contains an immobilized preparation of autotrophic bacteria [Karube, I.; Nomura, Y.; Arikawa, Y. Trends in Anal. Chem. 1995, 14, 295–299]. As CO2 diffuses through the membranes it is converted to O2 by the bacteria, increasing the concentration of O2 at the Pt cathode.This page titled 25.3: Linear Sweep Voltammetry is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

25.4: Cyclic Voltammetry

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In linear sweep voltammetry we scan the potential in one direction, either to more positive potentials or to more negative potentials. In cyclic voltammetry we complete a scan in both directions. a shows a typical potential-excitation signal. In this example, we first scan the potential to more positive values, resulting in the following oxidation reaction for the species R.\[R \rightleftharpoons O+n e^{-} \label{cv1} \]When the potential reaches a predetermined switching potential, we reverse the direction of the scan toward more negative potentials. Because we generated the species O on the forward scan, during the reverse scan it reduces back to R.\[O+n e^{-} \rightleftharpoons R \label{cv2} \]Cyclic voltammetry is carried out in an unstirred solution, which, as shown in b, results in peak currents instead of limiting currents. The voltammogram has separate peaks for the oxidation reaction and for the reduction reaction, each characterized by a peak potential and a peak current.The peak current in cyclic voltammetry is given by the Randles-Sevcik equation\[i_{p}=\left(2.69 \times 10^{5}\right) n^{3 / 2} A D^{1 / 2} \nu^{1 / 2} C_{A} \label{cv3} \]where n is the number of electrons in the redox reaction, A is the area of the working electrode, D is the diffusion coefficient for the electroactive species, \(\nu\) is the scan rate, and CA is the concentration of the electroactive species at the electrode. For a well-behaved system, the anodic and the cathodic peak currents are equal, and the ratio ip,a/ip,c is 1.00. The half-wave potential, E1/2, is midway between the anodic and cathodic peak potentials.\[E_{1 / 2}=\frac{E_{p, a}+E_{p, c}}{2} \label{cv4} \]Scanning the potential in both directions provides an opportunity to explore the electrochemical behavior of species generated at the electrode. This is a distinct advantage of cyclic voltammetry over other voltammetric techniques. shows the cyclic voltammogram for the same redox couple at both a faster and a slower scan rate. At the faster scan rate, \(\PageIndex{2}\)a, we see two peaks. At the slower scan rate in b, however, the peak on the reverse scan disappears. One explanation for this is that the products from the reduction of R on the forward scan have sufficient time to participate in a chemical reaction whose products are not electroactive.This page titled 25.4: Cyclic Voltammetry is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

25.5: Polarography

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The first important voltammetric technique to be developed—polarography—uses the dropping mercury (DME) electrode as the working electrode (see . In polarography, as in linear sweep voltammetry, we vary the potential and measure the current. The change in potential can be in the form of a linear ramp, as was the case for linear sweep voltammetry, or it can involve a series of pulses.As shown in , the current is measured while applying a linear potential ramp.Although polarography takes place in an unstirred solution, we obtain a limiting current instead of a peak current. When a Hg drop separates from the glass capillary and falls to the bottom of the electrochemical cell, it mixes the solution. Each new Hg drop, therefore, grows into a solution whose composition is identical to the bulk solution. The oscillations in the current are a result of the Hg drop’s growth, which leads to a time-dependent change in the area of the working electrode. The limiting current—which also is called the diffusion current—is measured using either the maximum current, imax, or from the average current, iavg. The relationship between the analyte’s concentration, CA, and the limiting current is given by the Ilkovic equations\[i_{\max }=706 n D^{1 / 2} m^{2 / 3} t^{1 / 6} C_{A}=K_{\max } C_{A} \label{pol1} \]\[i_{avg}=607 n D^{1 / 2} m^{2 / 3} t^{1 / 6} C_{A}=K_{\mathrm{avg}} C_{A} \label{pol2} \]where n is the number of electrons in the redox reaction, D is the analyte’s diffusion coefficient, m is the flow rate of Hg, t is the drop’s lifetime and Kmax and Kavg are constants. The half-wave potential, E1/2, provides qualitative information about the redox reaction.Normal polarography has been replaced by various forms of pulse polarography, several examples of which are shown in [see Osteryoung, J. J. Chem. Educ. 1983, 60, 296–298 for a comprehensive review]. Normal pulse polarography a), for example, uses a series of potential pulses characterized by a cycle of time \(\tau\), a pulse-time of tp, a pulse potential of \(\Delta E_\text{p}\), and a change in potential per cycle of \(\Delta E_\text{s}\). Typical experimental conditions for normal pulse polarography are \(\tau \approx 1 \text{ s}\), tp ≈ 50 ms, and \(\Delta E_\text{s} \approx 2 \text{ mV}\). The initial value of \(\Delta E_\text{p} \approx 2 \text{ mV}\), and it increases by ≈ 2 mV with each pulse. The current is sampled at the end of each potential pulse for approximately 17 ms before returning the potential to its initial value. The shape of the resulting voltammogram is similar to , but without the current oscillations. Because we apply the potential for only a small portion of the drop’s lifetime, there is less time for the analyte to undergo oxidation or reduction and a smaller diffusion layer. As a result, the faradaic current in normal pulse polarography is greater than in the polarography, resulting in better sensitivity and smaller detection limits.In differential pulse polarography b) the current is measured twice per cycle: for approximately 17 ms before applying the pulse and for approximately 17 ms at the end of the cycle. The difference in the two currents gives rise to the peak-shaped voltammogram. Typical experimental conditions for differential pulse polarography are \(\tau \approx 1 \text{ s}\), tp ≈ 50 ms, \(\Delta E_\text{p}\) ≈ 50 mV, and \(\Delta E_\text{s}\) ≈ 2 mV.The voltammogram for differential pulse polarography is approximately the first derivative of the voltammogram for normal pulse polarography. To see why this is the case, note that the change in current over a fixed change in potential, \(\Delta i / \Delta E\), approximates the slope of the voltammogram for normal pulse polarography. You may recall that the first derivative of a function returns the slope of the function at each point. The first derivative of a sigmoidal function is a peak-shaped function.Other forms of pulse polarography include staircase polarography c) and square-wave polarography d). One advantage of square-wave polarography is that we can make \(\tau\) very small—perhaps as small as 5 ms, compared to 1 s for other forms of pulse polarography—which significantly decreases analysis time. For example, suppose we need to scan a potential range of 400 mV. If we use normal pulse polarography with a \(\Delta E_\text{s}\) of 2 mV/cycle and a \(\tau\) of 1 s/cycle, then we need 200 s to complete the scan. If we use square-wave polarography with a \(\Delta E_\text{s}\) of 2 mV/cycle and a \(\tau\) of 5 ms/cycle, we can complete the scan in 1 s. At this rate, we can acquire a complete voltammogram using a single drop of Hg!Polarography is used extensively for the analysis of metal ions and inorganic anions, such as \(\text{IO}_3^-\) and \(\text{NO}_3^-\). We also can use polarography to study organic compounds with easily reducible or oxidizable functional groups, such as carbonyls, carboxylic acids, and carbon-carbon double bonds.This page titled 25.5: Polarography is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

25.6: Stripping Methods

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Another important voltammetric technique is stripping voltammetry, which consists of three related techniques: anodic stripping voltammetry, cathodic stripping voltammetry, and adsorptive stripping voltammetry. Because anodic stripping voltammetry is the more widely used of these techniques, we will consider it in greatest detail.Anodic stripping voltammetry consists of two steps ). The first step is a controlled potential electrolysis in which we hold the working electrode—usually a hanging mercury drop or a mercury film electrode—at a cathodic potential sufficient to deposit the metal ion on the electrode. For example, when analyzing Cu2+ the deposition reaction is\[\mathrm{Cu}^{2+}+2 e^{-} \rightleftharpoons \mathrm{Cu}(\mathrm{Hg}) \label{sv1} \]where Cu(Hg) indicates that the copper is amalgamated with the mercury. This step serves as a means of concentrating the analyte by transferring it from the larger volume of the solution to the smaller volume of the electrode. During most of the electrolysis we stir the solution to increase the rate of deposition. Near the end of the deposition time we stop the stirring—eliminating convection as a mode of mass transport—and allow the solution to become quiescent. Typical deposition times of 1–30 min are common, with analytes at lower concentrations requiring longer times.In the second step, we scan the potential anodically—that is, toward a more positive potential. When the working electrode’s potential is sufficiently positive, the analyte is stripped from the electrode, returning to solution in its oxidized form.\[\mathrm{Cu}(\mathrm{Hg})\rightleftharpoons \text{ Cu}^{2+}+2 e^{-} \label{sv2} \]Monitoring the current during the stripping step gives a peak-shaped voltammogram, as shown in . The peak current is proportional to the analyte’s concentration in the solution. Because we are concentrating the analyte in the electrode, detection limits are much smaller than other electrochemical techniques. An improvement of three orders of magnitude—the equivalent of parts per billion instead of parts per million—is routine.Anodic stripping voltammetry is very sensitive to experimental conditions, which we must carefully control to obtain results that are accurate and precise. Key variables include the area of the mercury film or the size of the hanging Hg drop, the deposition time, the rest time, the rate of stirring, and the scan rate during the stripping step. Anodic stripping voltammetry is particularly useful for metals that form amalgams with mercury, several examples of which are listed in Table \(\PageIndex{1}\).The experimental design for cathodic stripping voltammetry is similar to anodic stripping voltammetry with two exceptions. First, the deposition step involves the oxidation of the Hg electrode to \(\text{Hg}_2^{2+}\), which then reacts with the analyte to form an insoluble film at the surface of the electrode. For example, when Cl– is the analyte the deposition step is\[2 \mathrm{Hg}(l)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \text{ Hg}_{2} \mathrm{Cl}_{2}(s)+2 e^{-} \label{sv3} \]Second, stripping is accomplished by scanning cathodically toward a more negative potential, reducing \(\text{Hg}_2^{2+}\) back to Hg and returning the analyte to solution.\[\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)+2 e^{-}\rightleftharpoons 2 \mathrm{Hg}( l)+2 \mathrm{Cl}^{-}(a q) \label{sv4} \]Table \(\PageIndex{1}\) lists several analytes analyzed successfully by cathodic stripping voltammetry.In adsorptive stripping voltammetry, the deposition step occurs without electrolysis. Instead, the analyte adsorbs to the electrode’s surface. During deposition we maintain the electrode at a potential that enhances adsorption. For example, we can adsorb a neutral molecule on a Hg drop if we apply a potential of –0.4 V versus the SCE, a potential where the surface charge of mercury is approximately zero. When deposition is complete, we scan the potential in an anodic or a cathodic direction, depending on whether we are oxidizing or reducing the analyte. Examples of compounds that have been analyzed by absorptive stripping voltammetry also are listed in Table \(\PageIndex{1}\).This page titled 25.6: Stripping Methods is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

25.7: Applications of Voltammetry

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Voltammetry finds use for both quantitative analyses and characterization analyses. Examples of each are highlighted in this section.Voltammetry has been used for the quantitative analysis of a wide variety of samples, including environmental samples, clinical samples, pharmaceutical formulations, steels, gasoline, and oil.The choice of which voltammetric technique to use depends on the sample’s characteristics, including the analyte’s expected concentration and the sample’s location. For example, amperometry is ideally suited for detecting analytes in flow systems, including the in vivo analysis of a patient’s blood or as a selective sensor for the rapid analysis of a single analyte. The portability of amperometric sensors, which are similar to potentiometric sensors, also make them ideal for field studies. Although cyclic voltammetry is used to determine an analyte’s concentration, other methods described in this chapter are better suited for quantitative work.Pulse polarography and stripping voltammetry frequently are interchangeable. The choice of which technique to use often depends on the analyte’s concentration and the desired accuracy and precision. Detection limits for normal pulse polarography generally are on the order of 10–6 M to 10–7 M, and those for differential pulse polarography, staircase, and square wave polarography are between 10–7 M and 10–9 M. Because we concentrate the analyte in stripping voltammetry, the detection limit for many analytes is as little as 10–10 M to 10–12 M. On the other hand, the current in stripping voltammetry is much more sensitive than pulse polarography to changes in experimental conditions, which may lead to poorer precision and accuracy. We also can use pulse polarography to analyze a wider range of inorganic and organic analytes because there is no need to first deposit the analyte at the electrode surface.Stripping voltammetry also suffers from occasional interferences when two metals, such as Cu and Zn, combine to form an intermetallic compound in the mercury amalgam. The deposition potential for Zn is sufficiently negative that any Cu2+ in the sample also deposits into the mercury drop or film, leading to the formation of intermetallic compounds such as CuZn and CuZn2. During the stripping step, zinc in the intermetallic compounds strips at potentials near that of copper, decreasing the current for zinc at its usual potential and increasing the apparent current for copper. It is possible to overcome this problem by adding an element that forms a stronger intermetallic compound with the interfering metal. Thus, adding Ga3+ minimizes the interference of Cu when analyzing for Zn by forming an intermetallic compound of Cu and Ga.In any quantitative analysis we must correct the analyte’s signal for signals that arise from other sources. The total current, itot, in voltammetry consists of two parts: the current from the analyte’s oxidation or reduction, iA, and a background or residual current, ir.\[i_{t o t}=i_{A}+i_{r} \label{app1} \]The residual current, in turn, has two sources. One source is a faradaic current from the oxidation or reduction of trace interferents in the sample, iint. The other source is the charging current, ich, that accompanies a change in the working electrode’s potential.\[i_{r}=i_{\mathrm{int}}+i_{c h} \label{app2} \]We can minimize the faradaic current due to impurities by carefully preparing the sample. For example, one important impurity is dissolved O2, which undergoes a two-step reduction: first to H2O2 at a potential of –0.1 V versus the SCE, and then to H2O at a potential of –0.9 V versus the SCE. Removing dissolved O2 by bubbling an inert gas such as N2 through the sample eliminates this interference. After removing the dissolved O2, maintaining a blanket of N2 over the top of the solution prevents O2 from reentering the solution.There are two methods to compensate for the residual current. One method is to measure the total current at potentials where the analyte’s faradaic current is zero and extrapolate it to other potentials. This is the method shown in in water is determined by differential pulse polarography in 1 M HCl. The initial potential is set to –0.1 V versus the SCE and is scanned toward more negative potentials at a rate of 5 mV/s. Reduction of As(III) to As occurs at a potential of approximately –0.44 V versus the SCE. The peak currents for a set of standard solutions, corrected for the residual current, are shown in the following table.What is the concentration of As(III) in a sample of water if its peak current is 1.37 μA?Linear regression gives the calibration curve shown in , with an equation of\[i_{p}=0.0176+3.01 \times[\mathrm{As}(\mathrm{III})] \nonumber \]Substituting the sample’s peak current into the regression equation gives the concentration of As(III) as 4.49 μM.Voltammetry is a particularly attractive technique for the analysis of samples that contain two or more analytes. Provided that the analytes behave independently, the voltammogram of a multicomponent mixture is a summation of each analyte’s individual voltammograms. As shown in , if the separation between the half-wave potentials or between the peak potentials is sufficient, we can determine the presence of each analyte as if it is the only analyte in the sample. The minimum separation between the half-wave potentials or peak potentials for two analytes depends on several factors, including the type of electrode and the potential-excitation signal. For normal polarography the separation is at least ±0.2–0.3 V, and differential pulse voltammetry requires a minimum separation of ±0.04–0.05 V.If the voltammograms for two analytes are not sufficiently separated, a simultaneous analysis may be possible. An example of this approach is outlined the following example.The differential pulse polarographic analysis of a mixture of indium and cadmium in 0.1 M HCl is complicated by the overlap of their respective voltammograms [Lanza P. J. Chem. Educ. 1990, 67, 704–705]. The peak potential for indium is at –0.557 V and that for cadmium is at –0.597 V. When a 0.800-ppm indium standard is analyzed, \(\Delta i_p\) (in arbitrary units) is 200.5 at –0.557 V and 87.5 at –0.597 V relative to a saturated Ag/AgCl reference electorde. A standard solution of 0.793 ppm cadmium has a \(\Delta i_p\) of 58.5 at –0.557 V and 128.5 at –0.597 V. What is the concentration of indium and cadmium in a sample if \(\Delta i_p\) is 167.0 at a potential of –0.557 V and 99.5 at a potential of –0.597V.The change in current, \(\Delta i_p\), in differential pulse polarography is a linear function of the analyte’s concentration\[\Delta i_{p}=k_{A} C_{A} \nonumber \]where kA is a constant that depends on the analyte and the applied potential, and CA is the analyte’s concentration. To determine the concentrations of indium and cadmium in the sample we must first find the value of kA for each analyte at each potential. For simplicity we will identify the potential of –0.557 V as E1, and that for –0.597 V as E2. The values of kA are\[\begin{aligned} k_{\mathrm{In}, E_{1}} &=\frac{200.5}{0.800 \ \mathrm{ppm}}=250.6 \ \mathrm{ppm}^{-1} \\ k_{\mathrm{In}, E_{2}} &=\frac{87.5}{0.800 \ \mathrm{ppm}}=109.4 \ \mathrm{ppm}^{-1} \\ k_{\mathrm{Cd} E_{1}} &=\frac{58.5}{0.793 \ \mathrm{ppm}}=73.8 \ \mathrm{ppm}^{-1} \\ k_{\mathrm{Cd} E_{2}} &=\frac{128.5}{0.793 \ \mathrm{ppm}}=162.0 \ \mathrm{ppm}^{-1} \end{aligned} \nonumber \]Next, we write simultaneous equations for the current at the two potentials.\[\begin{array}{l}{\Delta i_{E_{1}}=167.0=250.6 \ \mathrm{ppm}^{-1} \times C_{\mathrm{In}}+73.8 \ \mathrm{ppm}^{-1} \times C_{\mathrm{Cd}}} \\ {\triangle i_{E_{2}}=99.5=109.4 \ \mathrm{ppm}^{-1} \times C_{\mathrm{In}}+162.0 \ \mathrm{ppm}^{-1} \times C_{\mathrm{Cd}}}\end{array} \nonumber \]Solving the simultaneous equations, which is left as an exercise, gives the concentration of indium as 0.606 ppm and the concentration of cadmium as 0.205 ppm.Voltammetry is one of several important analytical techniques for the analysis of trace metals in environmental samples, including groundwater, lakes, rivers and streams, seawater, rain, and snow. Detection limits at the parts-per-billion level are routine for many trace metals using differential pulse polarography, with anodic stripping voltammetry providing parts-per-trillion detection limits for some trace metals.One interesting environmental application of anodic stripping voltammetry is the determination of a trace metal’s chemical form within a water sample. Speciation is important because a trace metal’s bioavailability, toxicity, and ease of transport through the environment often depends on its chemical form. For example, a trace metal that is strongly bound to colloidal particles generally is not toxic because it is not available to aquatic lifeforms. Unfortunately, anodic stripping voltammetry can not distinguish a trace metal’s exact chemical form because closely related species, such as Pb2+ and PbCl+, produce a single stripping peak. Instead, trace metals are divided into “operationally defined” categories that have environmental significance.Operationally defined means that an analyte is divided into categories by the specific methods used to isolate it from the sample. There are many examples of operational definitions in the environmental literature. The distribution of trace metals in soils and sediments, for example, often is defined in terms of the reagents used to extract them; thus, you might find an operational definition for Zn2+ in a lake sediment as that extracted using 1.0 M sodium acetate, or that extracted using 1.0 M HCl.Although there are many speciation schemes in the environmental literature, we will consider one proposed by Batley and Florence [see (a) Batley, G. E.; Florence, T. M. Anal. Lett. 1976, 9, 379–388; (b) Batley, G. E.; Florence, T. M. Talanta 1977, 24, 151–158; (c) Batley, G. E.; Florence, T. M. Anal. Chem. 1980, 52, 1962–1963; (d) Florence, T. M., Batley, G. E.; CRC Crit. Rev. Anal. Chem. 1980, 9, 219–296]. This scheme, which is outlined in Table \(\PageIndex{2}\), combines anodic stripping voltammetry with ion-exchange and UV irradiation, dividing soluble trace metals into seven groups. In the first step, anodic stripping voltammetry in a pH 4.8 acetic acid buffer differentiates between labile metals and nonlabile metals. Only labile metals—those present as hydrated ions, weakly bound complexes, or weakly adsorbed on colloidal surfaces—deposit at the electrode and give rise to a signal. Total metal concentration are determined by ASV after digesting the sample in 2 M HNO3 for 5 min, which converts all metals into an ASV-labile form.A Chelex-100 ion-exchange resin further differentiates between strongly bound metals—usually metals bound to inorganic and organic solids, but also those tightly bound to chelating ligands—and more loosely bound metals. Finally, UV radiation differentiates between metals bound to organic phases and inorganic phases. The analysis of seawater samples, for example, suggests that cadmium, copper, and lead are present primarily as labile organic complexes or as labile adsorbates on organic colloids (Group II in Table \(\PageIndex{1}\)).Differential pulse polarography and stripping voltammetry are used to determine trace metals in airborne particulates, incinerator fly ash, rocks, minerals, and sediments. The trace metals, of course, are first brought into solution using a digestion or an extraction.Amperometric sensors also are used to analyze environmental samples. For example, the dissolved O2 sensor described earlier is used to determine the level of dissolved oxygen and the biochemical oxygen demand, or BOD, of waters and wastewaters. The latter test—which is a measure of the amount of oxygen required by aquatic bacteria as they decompose organic matter—is important when evaluating the efficiency of a wastewater treatment plant and for monitoring organic pollution in natural waters. A high BOD suggests that the water has a high concentration of organic matter. Decomposition of this organic matter may seriously deplete the level of dissolved oxygen in the water, adversely affecting aquatic life. Other amperometric sensors are available to monitor anionic surfactants in water, and CO2, H2SO4, and NH3 in atmospheric gases.Differential pulse polarography and stripping voltammetry are used to determine the concentration of trace metals in a variety of clinical samples, including blood, urine, and tissue. The determination of lead in blood is of considerable interest due to concerns about lead poisoning. Because the concentration of lead in blood is so small, anodic stripping voltammetry frequently is the more appropriate technique. The analysis is complicated, however, by the presence of proteins that may adsorb to the mercury electrode, inhibiting either the deposition or stripping of lead. In addition, proteins may prevent the electrodeposition of lead through the formation of stable, nonlabile complexes. Digesting and ashing the blood sample mini- mizes this problem. Differential pulse polarography is useful for the routine quantitative analysis of drugs in biological fluids, at concentrations of less than 10–6 M [Brooks, M. A. “Application of Electrochemistry to Pharmaceutical Analysis,” Chapter 21 in Kissinger, P. T.; Heinemann, W. R., eds. Laboratory Techniques in Electroanalytical Chemistry, Marcel Dekker, Inc.: New York, 1984, pp 539–568.]. Amperometric sensors using enzyme catalysts also have many clinical uses, several examples of which are shown in Table \(\PageIndex{2}\).In addition to environmental samples and clinical samples, differential pulse polarography and stripping voltammetry are used for the analysis of trace metals in other sample, including food, steels and other alloys, gasoline, gunpowder residues, and pharmaceuticals. Voltammetry is an important technique for the quantitative analysis of organics, particularly in the pharmaceutical industry where it is used to determine the concentration of drugs and vitamins in formulations. For example, voltammetric methods are available for the quantitative analysis of vitamin A, niacinamide, and riboflavin. When the compound of interest is not electroactive, it often can be derivatized to an electroactive form. One example is the differential pulse polarographic determination of sulfanilamide, which is converted into an electroactive azo dye by coupling with sulfamic acid and 1-napthol.In the previous section we learned how to use voltammetry to determine an analyte’s concentration in a variety of different samples. We also can use voltammetry to characterize an analyte’s properties, including verifying its electrochemical reversibility, determining the number of electrons transferred during its oxidation or reduction, and determining its equilibrium constant in a coupled chemical reaction.In a characterization application we study the properties of a system. Three examples are described here: determining if a redox reaction is electrochemically reversible, determining the number of electrons involved in the redox reaction, and studying metal-ligand complexation.Earlier in this chapter we derived a relationship between E1/2 and the standard-state potential for a redox couple using the Nernst equation, noting that a redox reaction must be electrochemically reversible. How can we tell if a redox reaction is reversible by looking at its voltammogram? As we learned in Chapter 25.3, for a reversible redox reaction the relationship between potential and current is\[E=E_{½} - \frac{0.05916}{n} \log \frac{i}{i_{l} - i} \label{app3} \]If a reaction is electrochemically reversible, a plot of E versus \(\log \frac{i}{i_l - i}\) is a straight line with a slope of –0.05916/n. In addition, the slope should yield an integer value for n.The following data were obtained from a linear scan hydrodynamic voltammogram of a reversible reduction reaction.The limiting current is 5.15 μA. Show that the reduction reaction is reversible, and determine values for n and for E1/2. shows a plot of E versus \(\log \frac{i}{i_l - i}\). Because the result is a straight-line, we know the reaction is electrochemically reversible under the conditions of the experiment. A linear regression analysis gives the equation for the straight line as\[E=-0.391 \mathrm{V}-0.0300 \log \frac{i}{i_{l}-i} \nonumber \]From Equation \ref{app3}, the slope is equivalent to –0.05916/n; solving for n gives a value of 1.97, or 2 electrons. From Equation \ref{app3} we also know that E1/2 is the y-intercept for a plot of E versus \(\log \frac{i}{i_l - i}\); thus, E1/2 for the data in this example is –0.391 V versus the SCE.We also can use cyclic voltammetry to evaluate electrochemical reversibility by looking at the difference between the peak potentials for the anodic and the cathodic scans. For an electrochemically reversible reaction, the following equation holds true.\[\Delta E_{p}=E_{p, a}-E_{p, c}=\frac{0.05916 \ \mathrm{V}}{n} \label{app4} \]As an example, for a two-electron reduction we expect a \(\Delta E_p\) of approximately 29.6 mV. For an electrochemically irreversible reaction the value of \(\Delta E_p\) is larger than expected.Another important application of voltammetry is determining the equilibrium constant for a solution reaction that is coupled to a redox reaction. The presence of the solution reaction affects the ease of electron transfer in the redox reaction, shifting E1/2 to a more negative or to a more positive potential. Consider, for example, the reduction of O to R\[O+n e^{-} \rightleftharpoons R \label{app5} \]the voltammogram for which is shown in . If we introduce a ligand, L, that forms a strong complex with O, then we also must consider the reaction\[O+p L\rightleftharpoons O L_{p} \label{app6} \]In the presence of the ligand, the overall redox reaction is\[O L_{p}+n e^{-} \rightleftharpoons R+p L \label{app7} \]Because of its stability, the reduction of the OLp complex is less favorable than the reduction of O. As shown in , the resulting voltammogram shifts to a potential that is more negative than that for O. Furthermore, the shift in the voltammogram increases as we increase the ligand’s concentration.We can use this shift in the value of E1/2 to determine both the stoichiometry and the formation constant for a metal-ligand complex. To derive a relationship between the relevant variables we begin with two equations: the Nernst equation for the reduction of O \[E=E_{O / R}^{\circ}-\frac{0.05916}{n} \log \frac{[R]_{x=0}}{[O]_{x=0}} \label{app8} \]and the stability constant, \(\beta_p\) for the metal-ligand complex at the electrode surface.\[\beta_{p} = \frac{\left[O L_p\right]_{x = 0}}{[O]_{x = 0}[L]_{x = 0}^p} \label{app9} \]In the absence of ligand the half-wave potential occurs when [R]x = 0 and [O]x = 0 are equal; thus, from the Nernst equation we have\[\left(E_{1 / 2}\right)_{n c}=E_{O / R}^{\circ} \label{app10} \]where the subscript “nc” signifies that the complex is not present. When ligand is present we must account for its effect on the concentration of O. Solving Equation \ref{app9} for [O]x = 0 and substituting into the Equation \ref{app8} gives\[E=E_{O/R}^{\circ}-\frac{0.05916}{n} \log \frac{[R]_{x=0}[L]_{x=0}^{p} \beta_{p}}{\left[O L_{p}\right]_{x=0}} \label{app11} \]If the formation constant is sufficiently large, such that essentially all O is present as the complex OLp, then \([R]_{x = 0}\) and \([OL_p]_{x = 0}\) are equal at the half-wave potential, and Equation \ref{app11} simplifies to\[\left(E_{1 / 2}\right)_{c} = E_{O/R}^{\circ} - \frac{0.05916}{n} \log{} [L]_{x=0}^{p} \beta_{p} \label{app12} \]where the subscript “c” indicates that the complex is present. Defining \(\Delta E_{1/2}\) as\[\triangle E_{1 / 2}=\left(E_{1 / 2}\right)_{c}-\left(E_{1 / 2}\right)_{n c} \label{app13} \]and substituting Equation \ref{app10} and Equation \ref{app12} and expanding the log term leaves us with the following equation.\[\Delta E_{1 / 2}=-\frac{0.05916}{n} \log \beta_{p}-\frac{0.05916 p}{n} \log {[L]} \label{app14} \]A plot of \(\Delta E_{1/2}\) versus log[L] is a straight-line, with a slope that is a function of the metal-ligand complex’s stoichiometric coefficient, p, and a y-intercept that is a function of its formation constant \(\beta_p\).A voltammogram for the two-electron reduction (n = 2) of a metal, M, has a half-wave potential of –0.226 V versus the SCE. In the presence of an excess of ligand, L, the following half-wave potentials are recorded.Determine the stoichiometry of the metal-ligand complex and its formation constant.We begin by calculating values of \(\Delta E_{1/2}\) using Equation \ref{app13}, obtaining the values in the following table. shows the resulting plot of \(\Delta E_{1/2}\) as a function of log[L]. A linear regression analysis gives the equation for the straight line as\[\triangle E_{1 / 2}=-0.370 \mathrm{V}-0.0601 \log {[L]} \nonumber \]From Equation \ref{app14} we know that the slope is equal to –0.05916p/n. Using the slope and n = 2, we solve for p obtaining a value of 2.03 ≈ 2. The complex’s stoichiometry, therefore, is ML2. We also know, from Equation \ref{app14}, that the y-intercept is equivalent to –(0.05916/n)log\(\beta_p\). Solving for \(\beta_2\) gives a formation constant of \(3.2 \times 10^{12}\).Cyclic voltammetry is one of the most powerful electrochemical techniques for exploring the mechanism of coupled electrochemical and chemical reactions. The treatment of this aspect of cyclic voltammetry is beyond the level of this text, although you can consult this chapter’s additional resources for additional information.This page titled 25.7: Applications of Voltammetry is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

26.1: A General Description of Chromatography

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Instrumental_Analysis_(LibreTexts)/26%3A_Introduction_to_Chromatographic_Separations/26.01%3A_A_General_Description_of_Chromatography

In chromatography we pass a sample-free phase, which we call the mobile phase, over a second sample-free stationary phase that remains fixed in space ). We inject or place the sample into the mobile phase. As the sample moves with the mobile phase, its components partition between the mobile phase and the stationary phase. A component whose distribution ratio favors the stationary phase requires more time to pass through the system. Given sufficient time and sufficient stationary and mobile phase, we can separate solutes even if they have similar distribution ratios.There are many ways in which we can identify a chromatographic separation: by describing the physical state of the mobile phase and the stationary phase; by describing how we bring the stationary phase and the mobile phase into contact with each other; or by describing the chemical or physical interactions between the solute and the stationary phase. Let’s briefly consider how we might use each of these classifications.We can trace the history of chromatography to the turn of the century when the Russian botanist Mikhail Tswett used a column packed with calcium carbonate and a mobile phase of petroleum ether to separate colored pigments from plant extracts. As the sample moved through the column, the plant’s pigments separated into individual colored bands. After effecting the separation, the calcium carbonate was removed from the column, sectioned, and the pigments recovered. Tswett named the technique chromatography, combining the Greek words for “color” and “to write.” There was little interest in Tswett’s technique until Martin and Synge’s pioneering development of a theory of chromatography (see Martin, A. J. P.; Synge, R. L. M. “A New Form of Chromatogram Employing Two Liquid Phases,” Biochem. J. 1941, 35, 1358–1366). Martin and Synge were awarded the 1952 Nobel Prize in Chemistry for this work.The mobile phase is a liquid or a gas, and the stationary phase is a solid or a liquid film coated on a solid substrate. We often name chromatographic techniques by listing the type of mobile phase followed by the type of stationary phase. In gas–liquid chromatography, for example, the mobile phase is a gas and the stationary phase is a liquid film coated on a solid substrate. If a technique’s name includes only one phase, as in gas chromatography, it is the mobile phase.There are two common methods for bringing the mobile phase and the stationary phase into contact. In column chromatography we pack the stationary phase into a narrow column and pass the mobile phase through the column using gravity or by applying pressure. The stationary phase is a solid particle or a thin liquid film coated on either a solid particulate packing material or on the column’s walls.In planar chromatography the stationary phase is coated on a flat surface—typically, a glass, metal, or plastic plate. One end of the plate is placed in a reservoir that contains the mobile phase, which moves through the stationary phase by capillary action. In paper chromatography, for example, paper is the stationary phase.The interaction between the solute and the stationary phase provides a third method for describing a separation ). In adsorption chromatography, solutes separate based on their ability to adsorb to a solid stationary phase. In partition chromatography, the stationary phase is a thin liquid film on a solid support. Separation occurs because there is a difference in the equilibrium partitioning of solutes between the stationary phase and the mobile phase. A stationary phase that consists of a solid support with covalently attached anionic (e.g., \(-\text{SO}_3^-\) ) or cationic (e.g., \(-\text{N(CH}_3)_3^+\)) functional groups is the basis for ion-exchange chromatography in which ionic solutes are attracted to the stationary phase by electrostatic forces. In size-exclusion chromatography the stationary phase is a porous particle or gel, with separation based on the size of the solutes. Larger solutes are unable to penetrate as deeply into the porous stationary phase and pass more quickly through the column.There are other interactions that can serve as the basis of a separation. In affinity chromatography the interaction between an antigen and an antibody, between an enzyme and a substrate, or between a receptor and a ligand forms the basis of a separation.Of the two methods for bringing the stationary phase and the mobile phases into contact, the most important is column chromatography. In this section we develop a general theory that we may apply to any form of column chromatography. provides a simple view of a liquid–solid column chromatography experiment. The sample is introduced as a narrow band at the top of the column. Ideally, the solute’s initial concentration profile is rectangular a). As the sample moves down (or throught) the column, the solutes begin to separate b,c) and the individual solute bands begin to broaden and develop a Gaussian profile b,c). If the strength of each solute’s interaction with the stationary phase is sufficiently different, then the solutes separate into individual bands d and d). . An alternative view of the separation in showing the concentration of each solute as a function of distance down the column.We can follow the progress of the separation by collecting fractions as they elute from the column e,f), or by placing a suitable detector at the end of the column. A plot of the detector’s response as a function of elution time, or as a function of the volume of mobile phase, is known as a chromatogram ), and consists of a peak for each solute.There are many possible detectors that we can use to monitor the separation. Later sections of this chapter describe some of the most popular.This page titled 26.1: A General Description of Chromatography is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

26.2: Migration Rates of Solutes

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Instrumental_Analysis_(LibreTexts)/26%3A_Introduction_to_Chromatographic_Separations/26.02%3A_Migration_Rates_of_Solutes

Our ability to separate two solutes depends on the equilibrium interactions of the solute with the stationary phase and the mobile phase, which effects both the time it takes a solute to travel through the column and the width of the solute's elution profile. In this section we consider the rate at which the solute moves through the column.Let’s assume we can describe a solute’s distribution between the mobile phase and stationary phase using the following equilibrium reaction\[S_{\text{m}} \rightleftharpoons S_{\text{s}} \label{mig1} \]where Sm is the solute in the mobile phase and Ss is the solute in the stationary phase. The equilibrium constant for this reaction is an equilibrium partition coefficient, KD.\[K_{D}=\frac{\left[S_{\mathrm{s}}\right]}{\left[S_\text{m}\right]} \label{mig2} \]This is not a trivial assumption. In this section we are, in effect, treating the solute’s equilibrium between the mobile phase and the stationary phase as if it is identical to the equilibrium in a simple liquid–liquid extraction carried out in a separatory funnel. You might question whether this is a reasonable assumption. There is an important difference between the two experiments that we need to consider. In a liquid–liquid extraction the two phases remain in contact with each other at all times, allowing for a true equilibrium. In chromatography, however, the mobile phase is in constant motion. A solute that moves into the stationary phase from the mobile phase will equilibrate back into a different portion of the mobile phase; this does not describe a true equilibrium.So, we ask again: Can we treat a solute’s distribution between the mobile phase and the stationary phase as an equilibrium process? The answer is yes, if the mobile phase velocity is slow relative to the kinetics of the solute’s movement back and forth between the two phase. In general, this is a reasonable assumption.We can characterize a chromatographic peak’s properties in several ways, two of which are shown in . Retention time, tr, is the time between the sample’s injection and the maximum response for the solute’s peak. A chromatographic peak’s baseline width, w, as shown in , is determined by extending tangent lines from the inflection points on either side of the peak through the baseline. Although usually we report tr and w using units of time, we can report them using units of volume by multiplying each by the mobile phase’s velocity, or report them in linear units by measuring distances with a ruler.For example, a solute’s retention volume,Vr, is \(t_\text{r} \times u\) where u is the mobile phase’s velocity through the column.In addition to the solute’s peak, also shows a small peak that elutes shortly after the sample is injected into the mobile phase. This peak contains all nonretained solutes, which move through the column at the same rate as the mobile phase. The time required to elute the nonretained solutes is called the column’s void time, tm.In the absence of any additional equilibrium reactions in the mobile phase or the stationary phase, KD is equivalent to the distribution ratio, D,\[D=\frac{\left[S_{0}\right]}{\left[S_\text{m}\right]}=\frac{(\operatorname{mol} \text{S})_\text{s} / V_\text{s}}{(\operatorname{mol} \text{S})_\text{m} / V_\text{m}}=K_{D} \label{mig3} \]where Vs and Vm are the volumes of the stationary phase and the mobile phase, respectively.A conservation of mass requires that the total moles of solute remain constant throughout the separation; thus, we know that the following equation is true.\[(\operatorname{mol} \text{S})_{\operatorname{tot}}=(\operatorname{mol} \text{S})_{\mathrm{m}}+(\operatorname{mol} \text{S})_\text{s} \label{mig4} \]Solving Equation \ref{mig4} for the moles of solute in the stationary phase and substituting into Equation \ref{mig2} leaves us with\[D = \frac{\left\{(\text{mol S})_{\text{tot}} - (\text{mol S})_\text{m}\right\} / V_{\mathrm{s}}}{(\text{mol S})_{\mathrm{m}} / V_{\mathrm{m}}} \label{mig5} \]Rearranging this equation and solving for the fraction of solute in the mobile phase, fm, gives\[f_\text{m} = \frac {(\text{mol S})_\text{m}} {(\text{mol S})_\text{tot}} = \frac {V_\text{m}} {DV_\text{s} + V_\text{m}} \label{mig6} \]Because we may not know the exact volumes of the stationary phase and the mobile phase, we simplify Equation \ref{mig6} by dividing both the numerator and the denominator by Vm; thus\[f_\text{m} = \frac {V_\text{m}/V_\text{m}} {DV_\text{s}/V_\text{m} + V_\text{m}/V_\text{m}} = \frac {1} {DV_\text{s}/V_\text{m} + 1} = \frac {1} {1+k} \label{mig7} \]where k\[k=D \times \frac{V_\text{s}}{V_\text{m}} \label{mig8} \]is the solute’s retention factor. Note that the larger the retention factor, the more the distribution ratio favors the stationary phase, leading to a more strongly retained solute and a longer retention time.Other (older) names for the retention factor are capacity factor, capacity ratio, and partition ratio, and it sometimes is given the symbol \(k^{\prime}\). Keep this in mind if you are using other resources. Retention factor is the approved name from the IUPAC Gold Book.We can determine a solute’s retention factor from a chromatogram by measuring the column’s void time, tm, and the solute’s retention time, tr (see ). Solving Equation \ref{mig7} for k, we find that\[k=\frac{1-f_\text{m}}{f_\text{m}} \label{mig9} \]Earlier we defined fm as the fraction of solute in the mobile phase. Assuming a constant mobile phase velocity, we also can define fm as\[f_\text{m}=\frac{\text { time spent in the mobile phase }}{\text { time spent in the stationary phase }}=\frac{t_\text{m}}{t_\text{r}} \label{mig10} \]Substituting back into Equation \ref{mig9} and rearranging leaves us with\[k=\frac{1-\frac{t_{m}}{t_{r}}}{\frac{t_{\mathrm{m}}}{t_{\mathrm{r}}}}=\frac{t_{\mathrm{r}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{t_{\mathrm{r}}^{\prime}}{t_{\mathrm{m}}} \label{mig11} \]where \(t_\text{r}^{\prime}\) is the adjusted retention time.In a chromatographic analysis of low molecular weight acids, butyric acid elutes with a retention time of 7.63 min. The column’s void time is 0.31 min. Calculate the retention factor for butyric acid.\[k_{\mathrm{but}}=\frac{t_{\mathrm{r}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{7.63 \text{ min}-0.31 \text{ min}}{0.31 \text{ min}}=23.6 \nonumber \]Selectivity is a relative measure of the retention of two solutes, which we define using a selectivity factor, \(\alpha\)\[\alpha=\frac{k_{B}}{k_{A}}=\frac{t_{r, B}-t_{\mathrm{m}}}{t_{r, A}-t_{\mathrm{m}}} \label{mig12} \]where solute A has the smaller retention time. When two solutes elute with identical retention time, \(\alpha = 1.00\); for all other conditions \(\alpha > 1.00\).In the chromatographic analysis for low molecular weight acids described in Example \(\PageIndex{1}\), the retention time for isobutyric acid is 5.98 min. What is the selectivity factor for isobutyric acid and butyric acid?First we must calculate the retention factor for isobutyric acid. Using the void time from Example \(\PageIndex{1}\) we have\[k_{\mathrm{iso}}=\frac{t_{\mathrm{r}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{5.98 \text{ min}-0.31 \text{ min}}{0.31 \text{ min}}=18.3 \nonumber \]The selectivity factor, therefore, is\[\alpha=\frac{k_{\text {but }}}{k_{\text {iso }}}=\frac{23.6}{18.3}=1.29 \nonumber \]This page titled 26.2: Migration Rates of Solutes is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

26.3: Zone Broadening and Column Efficiency

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Instrumental_Analysis_(LibreTexts)/26%3A_Introduction_to_Chromatographic_Separations/26.03%3A_Zone_Broadening_and_Column_Efficiency

Suppose we inject a sample that has a single component. At the moment we inject the sample it is a narrow band of finite width. As the sample passes through the column, the width of this band continually increases in a process we call band broadening. Column efficiency is a quantitative measure of the extent of band broadening.When we inject a sample onto a column it has a uniform, or rectangular concentration profile with respect to distance down the column. As the sample passes through the column, the individual solute particles move in and out of the stationary phase, remaining in place when in the stationary phase and moving down the column when in the mobile phase. Because some solute particles will, on average, spend more time in the mobile phase and some will, on average, spend more time in the stationary phase, the original rectangular band increases in width and takes on a Gaussian shape, as we see in .Our treatment of chromatography in this section assumes that a solute elutes as a symmetrical Gaussian peak, such as that shown in . This ideal behavior occurs when the solute’s partition coefficient, KD\[K_{\mathrm{D}}=\frac{[S_\text{s}]}{\left[S_\text{m}\right]} \label{shape1} \]is the same for all concentrations of solute. If this is not the case, then the chromatographic peak has an asymmetric peak shape similar to those shown in . The chromatographic peak in a is an example of peak tailing, which occurs when some sites on the stationary phase retain the solute more strongly than other sites. b, which is an example of peak fronting most often is the result of overloading the column with sample.As shown in a, we can report a peak’s asymmetry by drawing a horizontal line at 10% of the peak’s maximum height and measuring the distance from each side of the peak to a line drawn vertically through the peak’s maximum. The asymmetry factor, T, is defined as\[T=\frac{b}{a} \label{shape2} \]In their original theoretical model of chromatography, Martin and Synge divided the chromatographic column into discrete sections, which they called theoretical plates. Within each theoretical plate Martin and Synge assumed that there is an equilibrium between the solute present in the stationary phase and the solute present in the mobile phase [Martin, A. J. P.; Synge, R. L. M. Biochem. J. 1941, 35, 1358–1366]. They described column efficiency in terms of the number of theoretical plates, N,\[N=\frac{L}{H} \label{eff1} \]where L is the column’s length and H is the height of a theoretical plate. For any given column, the column efficiency improves—and chromatographic peaks become narrower—when there are more theoretical plates.If we assume that a chromatographic peak has a Gaussian profile, then the extent of band broadening is given by the peak’s variance or standard deviation. The height of a theoretical plate is the peak’s variance per unit length of the column\[H=\frac{\sigma^{2}}{L} \label{eff2} \]where the standard deviation, \(\sigma\), has units of distance. Because retention times and peak widths usually are measured in seconds or minutes, it is more convenient to express the standard deviation in units of time, \(\tau\), by dividing \(\sigma\) by the solute’s average linear velocity, \(\overline{u}\), which is equivalent to dividing the distance it travels, L, by its retention time, tr.\[\tau=\frac{\sigma}{\overline{u}}=\frac{\sigma t_{r}}{L} \label{eff3} \]For a Gaussian peak shape, the width at the baseline, w, is four times its standard deviation, \(\tau\).\[w = 4 \tau \label{eff4} \]Combining Equation \ref{eff2}, Equation \ref{eff3}, and Equation \ref{eff4} defines the height of a theoretical plate in terms of the easily measured chromatographic parameters tr and w.\[H=\frac{L w^{2}}{16 t_\text{r}^{2}} \label{eff5} \]Combing Equation \ref{eff5} and Equation \ref{eff1} gives the number of theoretical plates.\[N=16 \frac{t_{\mathrm{r}}^{2}}{w^{2}}=16\left(\frac{t_{\mathrm{r}}}{w}\right)^{2} \label{eff6} \]A chromatographic analysis for the chlorinated pesticide Dieldrin gives a peak with a retention time of 8.68 min and a baseline width of 0.29 min. Calculate the number of theoretical plates? Given that the column is 2.0 m long, what is the height of a theoretical plate in mm?Using Equation \ref{eff6}, the number of theoretical plates is\[N=16 \frac{t_{\mathrm{r}}^{2}}{w^{2}}=16 \times \frac{(8.68 \text{ min})^{2}}{(0.29 \text{ min})^{2}}=14300 \text{ plates} \nonumber \]Solving Equation \ref{eff1} for H gives the average height of a theoretical plate as\[H=\frac{L}{N}=\frac{2.00 \text{ m}}{14300 \text{ plates}} \times \frac{1000 \text{ mm}}{\mathrm{m}}=0.14 \text{ mm} / \mathrm{plate} \nonumber \]It is important to remember that a theoretical plate is an artificial construct and that a chromatographic column does not contain physical plates. In fact, the number of theoretical plates depends on both the properties of the column and the solute. As a result, the number of theoretical plates for a column may vary from solute to solute.The number of theoretical plates for an asymmetric peak shape is approximately\[N \approx \frac{41.7 \times \frac{t_{r}^{2}}{\left(w_{0.1}\right)^{2}}}{T+1.25}=\frac{41.7 \times \frac{t_{r}^{2}}{(a+b)^{2}}}{T+1.25} \label{eff7} \]where w0.1 is the width at 10% of the peak’s height [Foley, J. P.; Dorsey, J. G. Anal. Chem. 1983, 55, 730–737].Asymmetric peaks have fewer theoretical plates, and the more asymmetric the peak the smaller the number of theoretical plates. For example, the following table gives values for N for a solute eluting with a retention time of 10.0 min and a peak width of 1.00 min.Another approach to understanding the broadening of a solute band as it passes through a column is to consider the factors that affect the rate at which a solute moves through the column and how that is affected by the velocity with which the mobile phase moves through the column. We will consider one approach that considers four contributions: variations in path lengths, longitudinal diffusion, mass transfer in the stationary phase, and mass transfer in the mobile phase.As solute molecules pass through the column they travel paths that differ in length. Because of this difference in path length, two solute molecules that enter the column at the same time will exit the column at different times. The result, as shown in , is a broadening of the solute’s profile on the column. The contribution of multiple paths to the height of a theoretical plate, Hp, is\[H_{p}=2 \lambda d_{p} \label{van1} \]where dp is the average diameter of the particulate packing material and \(\lambda\) is a constant that accounts for the consistency of the packing. A smaller range of particle sizes and a more consistent packing produce a smaller value for \(\lambda\). For a column without packing material, Hp is zero and there is no contribution to band broadening from multiple paths.An inconsistent packing creates channels that allow some solute molecules to travel quickly through the column. It also can creates pockets that temporarily trap some solute molecules, slowing their progress through the column. A more uniform packing minimizes these problems.The second contribution to band broadening is the result of the solute’s longitudinal diffusion in the mobile phase. Solute molecules are in constant motion, diffusing from regions of higher solute concentration to regions where the concentration of solute is smaller. The result is an increase in the solute’s band width ). The contribution of longitudinal diffusion to the height of a theoretical plate, Hd, is\[H_{d}=\frac{2 \gamma D_{m}}{u} \label{van2} \]where Dm is the solute’s diffusion coefficient in the mobile phase, u is the mobile phase’s velocity, and \(\gamma\) is a constant related to the efficiency of column packing. Note that the effect of Hd on band broadening is inversely proportional to the mobile phase velocity: a higher velocity provides less time for longitudinal diffusion. Because a solute’s diffusion coefficient is larger in the gas phase than in a liquid phase, longitudinal diffusion is a more serious problem in gas chromatography.As the solute passes through the column it moves between the mobile phase and the stationary phase. We call this movement between phases mass transfer. As shown in , band broadening occurs if the solute’s movement within the mobile phase or within the stationary phase is not fast enough to maintain an equilibrium in its concentration between the two phases. On average, a solute molecule in the mobile phase moves some distance down the column before it passes into the stationary phase. A solute molecule in the stationary phase, on the other hand, takes longer than expected to move back into the mobile phase. The contributions of mass transfer in the stationary phase, Hs, and mass transfer in the mobile phase, Hm, are given by the following equations\[H_{s}=\frac{q k d_{f}^{2}}{(1+k)^{2} D_{s}} u \label{van3} \]\[H_{m}=\frac{f n\left(d_{p}^{2}, d_{c}^{2}\right)}{D_{m}} u \label{van4} \]where df is the thickness of the stationary phase, dc is the diameter of the column, Ds and Dm are the diffusion coefficients for the solute in the stationary phase and the mobile phase, k is the solute’s retention factor, and q is a constant related to the column packing material. Although the exact form of Hm is not known, it is a function of particle size and column diameter. Note that the effect of Hs and Hm on band broadening is directly proportional to the mobile phase velocity because a smaller velocity provides more time for mass transfer.The abbreviation fn in Equation \ref{van4} means “is a function of.”The height of a theoretical plate is a summation of the contributions from each of the terms that affect band broadening.\[H=H_{p}+H_{d}+H_{s}+H_{m} \label{van5} \]An alternative form of this equation is the van Deemter equation\[H=A+\frac{B}{u}+C u \label{van6} \]which emphasizes the importance of the mobile phase’s velocity. In the van Deemter equation, A accounts for the contribution of multiple paths (Hp), B/u accounts for the contribution of longitudinal diffusion (Hd), and Cu accounts for the combined contribution of mass transfer in the stationary phase and in the mobile phase (Hs and Hm).There is some disagreement on the best equation for describing the relationship between plate height and mobile phase velocity [Hawkes, S. J. J. Chem. Educ. 1983, 60, 393–398]. In addition to the van Deemter equation, other equations include\[H=\frac{B}{u}+\left(C_s+C_{m}\right) u \label{van7} \]where Cs and Cm are the mass transfer terms for the stationary phase and the mobile phase and\[H=A u^{1 / 3}+\frac{B}{u}+C u \label{van8} \]All three equations, and others, have been used to characterize chromatographic systems, with no single equation providing the best explanation in every case [Kennedy, R. T.; Jorgenson, J. W. Anal. Chem. 1989, 61, 1128–1135].To increase the number of theoretical plates without increasing the length of the column, we need to decrease one or more of the terms in Equation \ref{van5}. The easiest way to decrease H is to adjust the velocity of the mobile phase. For smaller mobile phase velocities, column efficiency is limited by longitudinal diffusion, and for higher mobile phase velocities efficiency is limited by the two mass transfer terms. As shown in Figure 26.3.6 —which uses the van Deemter equation—the optimum mobile phase velocity is the minimum in a plot of H as a function of u.The remaining parameters that affect the terms in Equation \ref{van5} are functions of the column’s properties and suggest other possible approaches to improving column efficiency. For example, both Hp and Hm are a function of the size of the particles used to pack the column. Decreasing particle size, therefore, is another useful method for improving efficiency.For a more detailed discussion of ways to assess the quality of a column, see Desmet, G.; Caooter, D.; Broeckhaven, K. “Graphical Data Represenation Methods to Assess the Quality of LC Columns,” Anal. Chem. 2015, 87, 8593–8602.Perhaps the most important advancement in chromatography columns is the development of open-tubular, or capillary columns. These columns have very small diameters (dc ≈ 50–500 μm) and contain no packing material (dp = 0). Instead, the capillary column’s interior wall is coated with a thin film of the stationary phase. Plate height is reduced because the contribution to H from Hp (Equation \ref{van1}) disappears and the contribution from Hm (Equation \ref{van4}) becomes smaller. Because the column does not contain any solid packing material, it takes less pressure to move the mobile phase through the column, which allows for longer columns. The combination of a longer column and a smaller height for a theoretical plate increases the number of theoretical plates by approximately \(100 \times\). Capillary columns are not without disadvantages. Because they are much narrower than packed columns, they require a significantly smaller amount of sample, which may be difficult to inject reproducibly. Another approach to improving resolution is to use thin films of stationary phase, which decreases the contribution to H from Hs (Equation \ref{van3}).The smaller the particles, the more pressure is needed to push the mobile phase through the column. As a result, for any form of chromatography there is a practical limit to particle size.This page titled 26.3: Zone Broadening and Column Efficiency is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

26.4: Optimization and Column Performance

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The goal of a chromatographic separation is to take a sample with more than one solute and to separate the solutes such that each elutes by itself. Our ability to separate two solutes from each other—to resolve them—is affected by a number of variables; how we can optimize the separation of two solutes is the subject of this section.The goal of chromatography is to separate a mixture into a series of chromatographic peaks, each of which constitutes a single component of the mixture. The resolution between two chromatographic peaks, RAB, is a quantitative measure of their separation, and is defined as\[R_{A B}=\frac{t_{t, B}-t_{t,A}}{0.5\left(w_{B}+w_{A}\right)}=\frac{2 \Delta t_{r}}{w_{B}+w_{A}} \label{res1} \]where B is the later eluting of the two solutes. As shown in , the separation of two chromatographic peaks improves with an increase in RAB. If the areas under the two peaks are identical—as is the case in —then a resolution of 1.50 corresponds to an overlap of only 0.13% for the two elution profiles. Because resolution is a quantitative measure of a separation’s success, it is a useful way to determine if a change in experimental conditions leads to a better separation.In a chromatographic analysis of lemon oil a peak for limonene has a retention time of 8.36 min with a baseline width of 0.96 min. \(\gamma\)-Terpinene elutes at 9.54 min with a baseline width of 0.64 min. What is the resolution between the two peaks?Using Equation \ref{res1} we find that the resolution is\[R_{A B}=\frac{2 \Delta t_{r}}{w_{B}+w_{A}}=\frac{2(9.54 \text{ min}-8.36 \text{ min})}{0.64 \text{ min}+0.96 \text{ min}}=1.48 \nonumber \]Now that we have defined the solute retention factor, selectivity, and column efficiency we are able to consider how they affect the resolution of two closely eluting peaks. Because the two peaks have similar retention times, it is reasonable to assume that their peak widths are nearly identical. If the number of theoretical plates is the same for all solutes—not strictly true, but not a bad assumption—then the ratio tr/w is a constant. If two solutes have similar retention times, then their peak widths must be similar. Equation \ref{res1}, therefore, becomes\[R_{A B}=\frac{t_{r, B}-t_{r, A}}{0.5\left(w_{B}+w_{A}\right)} \approx \frac{t_{r, B}-t_{r, A}}{0.5\left(2 w_{B}\right)}=\frac{t_{r, B}-t_{r, A}}{w_{B}} \label{res2} \]where B is the later eluting of the two solutes. Solving equation 26.3.8 for wB and substituting into Equation \ref{res2} leaves us with the following result.\[R_{A B}=\frac{\sqrt{N_{B}}}{4} \times \frac{t_{r, B}-t_{r, A}}{t_{r, B}} \label{res3} \]Rearranging equation 26.2.11 provides us with the following equations for the retention times of solutes A and B.\[t_{r, A}=k_{A} t_{\mathrm{m}}+t_{\mathrm{m}} \label{res4} \]\[t_{\mathrm{r}, B}=k_{B} t_{\mathrm{m}}+t_{\mathrm{m}} \label{res5} \]After substituting these equations into Equation \ref{res3} and simplifying, we have\[R_{A B}=\frac{\sqrt{N_{B}}}{4} \times \frac{k_{B}-k_{A}}{1+k_{B}} \label{ref6} \]Finally, we can eliminate solute A’s retention factor by substituting in equation 26.2.12. After rearranging, we end up with the following equation for the resolution between the chromatographic peaks for solutes A and B.\[R_{A B}=\frac{\sqrt{N_{B}}}{4} \times \frac{\alpha-1}{\alpha} \times \frac{k_{B}}{1+k_{B}} \label{res7} \]Although Equation \ref{res7} is useful for considering how a change in N, \(\alpha\), or k qualitatively affects resolution—which suits our purpose here—it is less useful for making accurate quantitative predictions of resolution, particularly for smaller values of N and for larger values of R. For more accurate predictions use the equation\[R_{A B}=\frac{\sqrt{N}}{4} \times(\alpha-1) \times \frac{k_{B}}{1+k_{\mathrm{avg}}} \nonumber \]where kavg is (kA + kB)/2. For a derivation of this equation and for a deeper discussion of resolution in column chromatography, see Foley, J. P. “Resolution Equations for Column Chromatography,” Analyst, 1991, 116, 1275-1279.Equation \ref{res7} contains terms that correspond to column efficiency, selectivity, and the solute retention factor. We can vary these terms, more or less independently, to improve resolution and analysis time. The first term, which is a function of the number of theoretical plates (for Equation \ref{res7}), accounts for the effect of column efficiency. The second term is a function of \(\alpha\) and accounts for the influence of column selectivity. Finally, the third term in both equations is a function of kB and accounts for the effect of solute B’s retention factor. A discussion of how we can use these parameters to improve resolution is the subject of the remainder of this section.In addition to resolution, another important factor in chromatography is the amount of time needed to elute a pair of solutes, which we can approximate using the retention time for solute B.\[t_{r, B}=\frac{16 R_{AB}^{2} H}{u} \times\left(\frac{\alpha}{\alpha-1}\right)^{2} \times \frac{\left(1+k_{B}\right)^{3}}{k_{B}^{2}} \label{res8} \]where u is the mobile phase’s velocity.One of the simplest ways to improve resolution is to adjust the retention factor for solute B. If all other terms in Equation \ref{res7} remain constant, an increase in kB will improve resolution. As shown by the green curve in Figure 26.4.2 , however, the improvement is greatest if the initial value of kB is small. Once kB exceeds a value of approximately 10, a further increase produces only a marginal improvement in resolution. For example, if the original value of kB is 1, increasing its value to 10 gives an 82% improvement in resolution; a further increase to 15 provides a net improvement in resolution of only 87.5%.Any improvement in resolution from increasing the value of kB generally comes at the cost of a longer analysis time. The red curve in Figure 26.4.2 shows the relative change in the retention time for solute B as a function of its retention factor. Note that the minimum retention time is for kB = 2. Increasing kB from 2 to 10, for example, approximately doubles solute B’s retention time.The relationship between retention factor and analysis time in Figure 26.4.2 works to our advantage if a separation produces an acceptable resolution with a large kB. In this case we may be able to decrease kB with little loss in resolution and with a significantly shorter analysis time.To increase kB without changing selectivity, \(\alpha\), any change to the chromatographic conditions must result in a general, nonselective increase in the retention factor for both solutes. In gas chromatography, we can accomplish this by decreasing the column’s temperature. Because a solute’s vapor pressure is smaller at lower temperatures, it spends more time in the stationary phase and takes longer to elute. In liquid chromatography, the easiest way to increase a solute’s retention factor is to use a mobile phase that is a weaker solvent. When the mobile phase has a lower solvent strength, solutes spend proportionally more time in the stationary phase and take longer to elute.A second approach to improving resolution is to adjust the selectivity, \(\alpha\). In fact, for \(\alpha \approx 1\) it usually is not possible to improve resolution by adjusting the solute retention factor, kB, or the column efficiency, N. A change in \(\alpha\) often has a more dramatic effect on resolution than a change in kB. For example, changing \(\alpha\) from 1.1 to 1.5, while holding constant all other terms, improves resolution by 267%. In gas chromatography, we adjust \(\alpha\) by changing the stationary phase; in liquid chromatography, we change the composition of the mobile phase to adjust \(\alpha\).To change \(\alpha\) we need to selectively adjust individual solute retention factors. Figure 26.4.3 shows one possible approach for the liquid chromatographic separation of a mixture of substituted benzoic acids. Because the retention time of a compound’s weak acid form and its weak base form are different, its retention time will vary with the pH of the mobile phase, as shown in Figure 26.4.3 a. The intersections of the curves in Figure 26.4.3 a show pH values where two solutes co-elute. For example, at a pH of 3.8 terephthalic acid and p-hydroxybenzoic acid elute as a single chromatographic peak.Figure 26.4.3 a shows that there are many pH values where some separation is possible. To find the optimum separation, we plot \(\alpha\) for each pair of solutes. The red, green, and orange curves in Figure 26.4.3 b show the variation in \(\alpha\) with pH for the three pairs of solutes that are hardest to separate (for all other pairs of solutes, \(\alpha\) > 2 at all pH levels). The blue shading shows windows of pH values in which at least a partial separation is possible—this figure is sometimes called a window diagram—and the highest point in each window gives the optimum pH within that range. The best overall separation is the highest point in any window, which, for this example, is a pH of 3.5. Because the analysis time at this pH is more than 40 min (Figure 26.4.3 a), choosing a pH between 4.1–4.4 might produce an acceptable separation with a much shorter analysis time.Let’s use benzoic acid, C6H5COOH, to explain why pH can affect a solute’s retention time. The separation uses an aqueous mobile phase and a nonpolar stationary phase. At lower pHs, benzoic acid is predominately in its weak acid form, C6H5COOH, and partitions easily into the nonpolar stationary phase. At more basic pHs, however, benzoic acid is in its weak base form, C6H5COO–. Because it now carries a charge, its solubility in the mobile phase increases and its solubility in the nonpolar stationary phase decreases. As a result, it spends more time in the mobile phase and has a shorter retention time.Although the usual way to adjust pH is to change the concentration of buffering agents, it also is possible to adjust pH by changing the column’s temperature because a solute’s pKa value is pH-dependent; for a review, see Gagliardi, L. G.; Tascon, M.; Castells, C. B. “Effect of Temperature on Acid–Base Equilibria in Separation Techniques: A Review,” Anal. Chim. Acta, 2015, 889, 35–57.A third approach to improve resolution is to adjust the column’s efficiency by increasing the number of theoretical plates, N. If we have values for kB and \(\alpha\), then we can use Equation \ref{res7} to calculate the number of theoretical plates for any resolution. Table 26.4.1 provides some representative values. For example, if \(\alpha\) = 1.05 and kB = 2.0, a resolution of 1.25 requires approximately 24 800 theoretical plates. If our column provides only 12 400 plates, half of what is needed, then a separation is not possible. How can we double the number of theoretical plates? The easiest way is to double the length of the column, although this also doubles the analysis time. A better approach is to cut the height of a theoretical plate, H, in half, providing the desired resolution without changing the analysis time. Even better, if we can decrease H by more than 50%, it may be possible to achieve the desired resolution with an even shorter analysis time by also decreasing kB or \(\alpha\).Adjusting the retention factor to improve the resolution between one pair of solutes may lead to unacceptably long retention times for other solutes. For example, suppose we need to analyze a four-component mixture with baseline resolution and with a run-time of less than 20 min. Our initial choice of conditions gives the chromatogram in Figure 26.4.4 a. Although we successfully separate components 3 and 4 within 15 min, we fail to separate components 1 and 2. Adjusting conditions to improve the resolution for the first two components by increasing k2 provides a good separation of all four components, but the run-time is too long (Figure 26.4.4 b). This problem of finding a single set of acceptable operating conditions is known as the general elution problem.One solution to the general elution problem is to make incremental adjustments to the retention factor as the separation takes place. At the beginning of the separation we set the initial chromatographic conditions to optimize the resolution for early eluting solutes. As the separation progresses, we adjust the chromatographic conditions to decrease the retention factor—and, therefore, to decrease the retention time—for each of the later eluting solutes (Figure 26.4.4 c). In gas chromatography this is accomplished by temperature programming. The column’s initial temperature is selected such that the first solutes to elute are resolved fully. The temperature is then increased, either continuously or in steps, to bring off later eluting components with both an acceptable resolution and a reasonable analysis time. In liquid chromatography the same effect is obtained by increasing the solvent’s eluting strength. This is known as a gradient elution. We will have more to say about each of these in later sections of this chapter.This page titled 26.4: Optimization and Column Performance is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

26.6: Applications of Chromatography

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Although the primary purpose of chromatography is the separation of a complex mixture into its component parts, as outlined here, a chromatographic separation also provides qualitative and quantitative information about our samples. More detailed examples of qualitative and quantitative applications are found in the chapters that follow.As we learned in Section 26.2, solutes migrate through a chromatographic system at rate that is a function of the properties of the mobile phase and the stationary phase. This means that a particular solute will elute with a consistent retention time. If we expect a solute to elute with a retention time of 5.0 min, the presence of a peak at 5.0 min is suggestive of, but not definitive evidence of the solute's presence in our sample; however, the absence of a peak at 5.0 min is strong evidence that the solute is not present in our sample. For a complex mixture, this sort of screening technique is a useful qualitative application of chromatography. As we will see in the chapters that follow, the type of detector used to monitor a chromatographic separation may provide useful qualitative information.A chromatographic separation yields a sequence of peaks that, ideally, represent a single solute. These peak's are characterized by an area that is proportional to the amount of analyte injected into the mobile phase. By injecting a series of standards, a calibration curve of peak area as a function of the analyte's concentration provides a way to determine the analyte's concentration in a sample. Any of the calibration strategies discussed in Chapter 1.5—external standards, standard additions, and internal standards—find use in a quantitative chromatographic analysis. Determining a solute's peak area is relatively straight-forward when using a computer-interfaced instrument with appropriate software. Alternatively, peak height, which is easier to measure, can be used as a substitute, although care must be taken to ensure that the peaks are symmetrical and that peak widths are consistent for the standards and the samples.This page titled 26.6: Applications of Chromatography is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

27.1: Principles of Gas Chromatography

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In Chapter 26 we covered several important elements of chromatography, including the factors that affect the migration of solutes, the factors that contribute to band broadening, and the factors under our control that we can use to optimize the separation of a mixture. Here we consider two topics that apply to a gas chromatographic separation, both of which are a function of the properties of gases.Many of the chromatographic variables in gathered in the tables included in Chapter 26.5—both those that are measured directly, provided by the manufacturer, or given by the operating conditions, and those derived from these variables—are given in terms or retention times for the solutes, \(t_r\), and for the mobile phase, \(t_m\). The product of time and flow rate is a volume\[V_r = t_r \times u \nonumber \]\[V_m = t_r\times u \nonumber \]where \(V_r\) and \(V_m\) are the volume of mobile phase needed to elute a solute and the volume of mobile phase needed to elute a non-retained solute, which allows us to describe the retention in terms of volumes instead of times.Because the volume of gas is a function of pressure, and the pressure drops across the column from an inlet pressure of \(P_i\) to an outlet pressure of \(P_o\), the retention times are particularly sensitive to operating condition. We can, however, correct the retention volumes by accounting for the compressibility of the gas\[V_r^o = j t_r u \nonumber \]\[V_m^o = j t_m u \nonumber \]where \(j\) is a correction factor that accounts for the drop in pressure\[j = \frac {3 \times (P_i/P_o)^2 - 1} {2 \times (P_i/P_o)^3 - 1} \nonumber \]and where \(V_r^o\) and \(V_m^o\) are the corrected retention volumes for the solute and the non-retained solutes, respectively. The solute's corrected retention volume can be further normalized by dividing the adjusted retention volume, \(V_r^o - V_m^o\), by the mass of the stationary phase, \(w\), and by adjusting for the column's temperature, \(T_c\), relative to 273 K\[V_g = \frac {V_r^o - V_m^o} {w} \times \frac {273} {T_c} \nonumber \]yielding the solute's specific retention volume, \(V_g\). This value is reasonably insensitive to operating conditions, which makes it useful for qualitative purposes. In Chapter 26 we considered three factors that affect band broadening—multiple paths, longitudinal diffusion, and mass transfer—expressing the relationship between the height of a theoretical plate, \(H\), as a function of the mobile phase's velocity, \(u\), using the van Deemter equation\[H = A + \frac{B}{u} + Cu \nonumber \]where \(A\) is the contribution from multiple paths, \(B\) is the contribution from longitudinal diffusion, and \(C\) is the contribution from mass transfer. Because solutes have large diffusion coefficients in the gas phase, the term \(B/u\) is often the limiting factor in gas chromatography.This page titled 27.1: Principles of Gas Chromatography is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

27.2: Instruments for Gas Chromatography

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In gas chromatography (GC) we inject the sample, which may be a gas or a liquid, into an gaseous mobile phase (often called the carrier gas). The mobile phase carries the sample through a packed or a capillary column that separates the sample’s components based on their ability to partition between the mobile phase and the stationary phase. Figure 27.2.1 shows an example of a typical gas chromatograph, which consists of several key components: a supply of compressed gas for the mobile phase; a heated injector, which rapidly volatilizes the components in a liquid sample; a column, which is placed within an oven whose temperature we can control during the separation; and a detector to monitor the eluent as it comes off the column. Let’s consider each of these components.The most common mobile phases for gas chromatography are He, Ar, and N2, which have the advantage of being chemically inert toward both the sample and the stationary phase. The choice of carrier gas often is determined by the needs of the instrument’s detector. For a packed column the mobile phase velocity usually is 25–150 mL/min. The typical flow rate for a capillary column is 1–25 mL/min.Three factors determine how we introduce a sample to the gas chromatograph. First, all of the sample’s constituents must be volatile. Second, the analytes must be present at an appropriate concentration. Finally, the physical process of injecting the sample must not degrade the separation. Each of these needs is considered in this section.Not every sample can be injected directly into a gas chromatograph. To move through the column, the sample’s constituents must be sufficiently volatile. A solute of low volatility, for example, may be retained by the column and continue to elute during the analysis of subsequent samples. A nonvolatile solute will condense at the top of the column, degrading the column’s performance.An attractive approach to isolating analytes is a solid-phase microextraction (SPME). In one approach, which is illustrated in Figure 27.2.2 , a fused-silica fiber is placed inside a syringe needle. The fiber, which is coated with a thin film of an adsorbent material, such as polydimethyl siloxane, is lowered into the sample by depressing a plunger and is exposed to the sample for a predetermined time. After withdrawing the fiber into the needle, it is transferred to the gas chromatograph for analysis.Two additional methods for isolating volatile analytes are a purge-and-trap and headspace sampling. In a purge-and-trap, we bubble an inert gas, such as He or N2, through the sample, releasing—or purging—the volatile compounds. These compounds are carried by the purge gas through a trap that contains an absorbent material, such as Tenax, where they are retained. Heating the trap and back-flushing with carrier gas transfers the volatile compounds to the gas chromatograph. In headspace sampling we place the sample in a closed vial with an overlying air space. After allowing time for the volatile analytes to equilibrate between the sample and the overlying air, we use a syringe to extract a portion of the vapor phase and inject it into the gas chromatograph. Alternatively, we can sample the headspace with an SPME.Thermal desorption is a useful method for releasing volatile analytes from solids. We place a portion of the solid in a glass-lined, stainless steel tube. After purging with carrier gas to remove any O2 that might be present, we heat the sample. Volatile analytes are swept from the tube by an inert gas and carried to the GC. Because volatilization is not a rapid process, the volatile analytes often are concentrated at the top of the column by cooling the column inlet below room temperature, a process known as cryogenic focusing. Once volatilization is complete, the column inlet is heated rapidly, releasing the analytes to travel through the column.The reason for removing O2 is to prevent the sample from undergoing an oxidation reaction when it is heated.To analyze a nonvolatile analyte we must convert it to a volatile form. For example, amino acids are not sufficiently volatile to analyze directly by gas chromatography. Reacting an amino acid, such as valine, with 1-butanol and acetyl chloride produces an esterified amino acid. Subsequent treatment with trifluoroacetic acid gives the amino acid’s volatile N-trifluoroacetyl-n-butyl ester derivative.If an analyte’s concentration is too small to give an adequate signal, then we must concentrate the analyte before we inject the sample into the gas chromatograph. A side benefit of many extraction methods is that they often concentrate the analytes. Volatile organic materials isolated from an aqueous sample by a purge-and-trap, for example, are concentrated by as much as \(1000 \times\).If an analyte is too concentrated, it is easy to overload the column, resulting in peak fronting and a poor separation. In addition, the analyte’s concentration may exceed the detector’s linear response. Injecting less sample or diluting the sample with a volatile solvent, such as methylene chloride, are two possible solutions to this problem.In Chapter 26 we examined several explanations for why a solute’s band increases in width as it passes through the column, a process we called band broadening. We also introduce an additional source of band broadening if we fail to inject the sample into the minimum possible volume of mobile phase. There are two principal sources of this precolumn band broadening: injecting the sample into a moving stream of mobile phase and injecting a liquid sample instead of a gaseous sample. The design of a gas chromatograph’s injector helps minimize these problems.An example of a simple injection port for a packed column is shown in Figure 27.2.3 . The top of the column fits within a heated injector block, with carrier gas entering from the bottom. The sample is injected through a rubber septum using a microliter syringe, such as the one shown in in Figure 27.2.4 . Injecting the sample directly into the column minimizes band broadening because it mixes the sample with the smallest possible amount of carrier gas. The injector block is heated to a temperature at least 50oC above the boiling point of the least volatile solute, which ensures a rapid vaporization of the sample’s components.Because a capillary column’s volume is significantly smaller than that for a packed column, it requires a different style of injector to avoid overloading the column with sample. Figure 27.2.5 shows a schematic diagram of a typical split/splitless injector for use with a capillary column.In a split injection we inject the sample through a rubber septum using a microliter syringe. Instead of injecting the sample directly into the column, it is injected into a glass liner where it mixes with the carrier gas. At the split point, a small fraction of the carrier gas and sample enters the capillary column with the remainder exiting through the split vent. By controlling the flow rate of the carrier gas as it enters the injector, and its flow rate through the septum purge and the split vent, we can control the fraction of sample that enters the capillary column, typically 0.1–10%.For example, if the carrier gas flow rate is 50 mL/min, and the flow rates for the septum purge and the split vent are 2 mL/min and 47 mL/min, respectively, then the flow rate through the column is 1 mL/min (= 50 – 2 – 47). The ratio of sample entering the column is 1/50, or 2%.In a splitless injection, which is useful for trace analysis, we close the split vent and allow all the carrier gas that passes through the glass liner to enter the column—this allows virtually all the sample to enters the column. Because the flow rate through the injector is low, significant precolumn band broadening is a problem. Holding the column’s temperature approximately 20–25oC below the solvent’s boiling point allows the solvent to condense at the entry to the capillary column, forming a barrier that traps the solutes. After allowing the solutes to concentrate, the column’s temperature is increased and the separation begins.For samples that decompose easily, an on-column injection may be necessary. In this method the sample is injected directly into the column without heating. The column temperature is then increased, volatilizing the sample with as low a temperature as is practical.Control of the column’s temperature is critical to attaining a good separation when using gas chromatography. For this reason the column is placed inside a thermostated oven (see Figure 27.2.1 ). In an isothermal separation we maintain the column at a constant temperature. To increase the interaction between the solutes and the stationary phase, the temperature usually is set slightly below that of the lowest-boiling solute.One difficulty with an isothermal separation is that a temperature that favors the separation of a low-boiling solute may lead to an unacceptably long retention time for a higher-boiling solute. Temperature programming provides a solution to this problem. At the beginning of the analysis we set the column’s initial temperature below that for the lowest-boiling solute. As the separation progresses, we slowly increase the temperature at either a uniform rate or in a series of steps.The final part of a gas chromatograph is the detector. The ideal detector has several desirable features: a low detection limit, a linear response over a wide range of solute concentrations (which makes quantitative work easier), sensitivity for all solutes or selectivity for a specific class of solutes, and an insensitivity to a change in flow rate or temperature.One of the earliest gas chromatography detectors takes advantage of the mobile phase’s thermal conductivity. As the mobile phase exits the column it passes over a tungsten-rhenium wire filament (see Figure 27.2.6 ). The filament’s electrical resistance depends on its temperature, which, in turn, depends on the thermal conductivity of the mobile phase. Because of its high thermal conductivity, helium is the mobile phase of choice when using a thermal conductivity detector (TCD).Thermal conductivity, as the name suggests, is a measure of how easily a substance conducts heat. A gas with a high thermal conductivity moves heat away from the filament—and, thus, cools the filament—more quickly than does a gas with a low thermal conductivity.When a solute elutes from the column, the thermal conductivity of the mobile phase in the TCD cell decreases and the temperature of the wire filament, and thus it resistance, increases. A reference cell, through which only the mobile phase passes, corrects for any time-dependent variations in flow rate, pressure, or electrical power, all of which affect the filament’s resistance.Because all solutes affect the mobile phase’s thermal conductivity, the thermal conductivity detector is a universal detector. Another advantage is the TCD’s linear response over a concentration range spanning 104–105 orders of magnitude. The detector also is non-destructive, which allows us to recover analytes using a postdetector cold trap. One significant disadvantage of the TCD detector is its poor detection limit for most analytes.The combustion of an organic compound in an H2/air flame results in a flame that contains electrons and organic cations, presumably CHO+. Applying a potential of approximately 300 volts across the flame creates a small current of roughly 10–9 to 10–12 amps. When amplified, this current provides a useful analytical signal. This is the basis of the popular flame ionization detector, a schematic diagram of which is shown in Figure 27.2.7 .Most carbon atoms—except those in carbonyl and carboxylic groups—generate a signal, which makes the FID an almost universal detector for organic compounds. Most inorganic compounds and many gases, such as H2O and CO2, are not detected, which makes the FID detector a useful detector for the analysis of organic analytes in atmospheric and aqueous environmental samples. Advantages of the FID include a detection limit that is approximately two to three orders of magnitude smaller than that for a thermal conductivity detector, and a linear response over 106–107 orders of magnitude in the amount of analyte injected. The sample, of course, is destroyed when using a flame ionization detector.The electron capture detector is an example of a selective detector. As shown in Figure 27.2.8 , the detector consists of a \(\beta\)-emitter, such as 63Ni. The emitted electrons ionize the mobile phase, usually N2, generating a standing current between a pair of electrodes. When a solute with a high affinity for capturing electrons elutes from the column, the current decreases, which serves as the signal. The ECD is highly selective toward solutes with electronegative functional groups, such as halogens and nitro groups, and is relatively insensitive to amines, alcohols, and hydrocarbons. Although its detection limit is excellent, its linear range extends over only about two orders of magnitude.A \(\beta\)-particle is an electron.A mass spectrometer is an instrument that ionizes a gaseous molecule using sufficient energy that the resulting ion breaks apart into smaller ions. Because these ions have different mass-to-charge ratios, it is possible to separate them using a magnetic field or an electrical field. The resulting mass spectrum contains both quantitative and qualitative information about the analyte. Figure 27.2.9 shows a mass spectrum for toluene.Figure 27.2.10 shows a block diagram of a typical gas chromatography-mass spectrometer (GC–MS) instrument. The effluent from the column enters the mass spectrometer’s ion source in a manner that eliminates the majority of the carrier gas. In the ionization chamber the remaining molecules—a mixture of carrier gas, solvent, and solutes—undergo ionization and fragmentation. The mass spectrometer’s mass analyzer separates the ions by their mass-to-charge ratio and a detector counts the ions and displays the mass spectrum.There are several options for monitoring a chromatogram when using a mass spectrometer as the detector. The most common method is to continuously scan the entire mass spectrum and report the total signal for all ions that reach the detector during each scan. This total ion scan provides universal detection for all analytes. We can achieve some degree of selectivity by monitoring one or more specific mass-to-charge ratios, a process called selective-ion monitoring. A mass spectrometer provides excellent detection limits, typically 25 fg to 100 pg, with a linear range of 105 orders of magnitude. Because we continuously record the mass spectrum of the column’s eluent, we can go back and examine the mass spectrum for any time increment. This is a distinct advantage for GC–MS because we can use the mass spectrum to help identify a mixture’s components.Two additional detectors are similar in design to a flame ionization detector. In the flame photometric detector, optical emission from phosphorous and sulfur provides a detector selective for compounds that contain these elements. The thermionic detector responds to compounds that contain nitrogen or phosphorous.A Fourier transform infrared spectrophotometer (FT–IR) also can serve as a detector. In GC–FT–IR, effluent from the column flows through an optical cell constructed from a 10–40 cm Pyrex tube with an internal diameter of 1–3 mm. The cell’s interior surface is coated with a reflecting layer of gold. Multiple reflections of the source radiation as it is transmit- ted through the cell increase the optical path length through the sample. As is the case with GC–MS, an FT–IR detector continuously records the column eluent’s spectrum, which allows us to examine the IR spectrum for any time increment.This page titled 27.2: Instruments for Gas Chromatography is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

27.3: Gas Chromatographic Columns and Stationary Phases

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There are two broad classes of chromatographic columns: packed columns and capillary columns. In general, a packed column can handle larger samples and a capillary column can separate more complex mixtures.Packed columns are constructed from glass, stainless steel, copper, or aluminum, and typically are 2–6 m in length with internal diameters of 2–4 mm. The column is filled with a particulate solid support, with particle diameters ranging from 37–44 μm to 250–354 μm. Figure 27.3.1 shows a typical example of a packed column.The most widely used particulate support is diatomaceous earth, which is composed of the silica skeletons of diatoms. These particles are very porous, with surface areas ranging from 0.5–7.5 m2/g, which provides ample contact between the mobile phase and the stationary phase. When hydrolyzed, the surface of a diatomaceous earth contains silanol groups (–SiOH), that serve as active sites for absorbing solute molecules in gas-solid chromatography (GSC).In gas-liquid chromatography (GLC), we coat the packing material with a liquid mobile phase. To prevent uncoated packing material from adsorbing solutes, which degrades the quality of the separation, surface silanols are deactivated by reacting them with dimethyldichlorosilane and rinsing with an alcohol—typically methanol—before coating the particles with stationary phase.The column in Figure 27.3.1 , for example, has approximately 1800 plates/m, or a total of approximately 3600 theoretical plates.A capillary, or open tubular column is constructed from fused silica and is coated with a protective polymer coating. Columns range from 15–100 m in length with an internal diameter of approximately 150–300 μm. Figure 27.3.2 shows an example of a typical capillary column.Capillary columns are of three principal types. In a wall-coated open tubular column (WCOT) a thin layer of stationary phase, typically 0.25 nm thick, is coated on the capillary’s inner wall. In a porous-layer open tubular column (PLOT), a porous solid support—alumina, silica gel, and molecular sieves are typical examples—is attached to the capillary’s inner wall. A support-coated open tubular column (SCOT) is a PLOT column that includes a liquid stationary phase. Figure 27.3.3 shows the differences between these types of capillary columns.A capillary column provides a significant improvement in separation efficiency because it has more theoretical plates per meter and is longer than a packed column. For example, the capillary column in Figure 27.3.2 has almost 4300 plates/m, or a total of 129 000 theoretical plates. On the other hand, a packed column can handle a larger sample. Because of its smaller diameter, a capillary column requires a smaller sample, typically less than 10–2 μL.Elution order in gas–liquid chromatography depends on two factors: the boiling point of the solutes and the interaction between the solutes and the stationary phase. If a mixture’s components have significantly different boiling points, then the choice of stationary phase is less critical. If two solutes have similar boiling points, then a separation is possible only if the stationary phase selectively interacts with one of the solutes. As a general rule, nonpolar solutes are separated more easily when using a nonpolar stationary phase, and polar solutes are easier to separate when using a polar stationary phase.There are several important criteria for choosing a stationary phase: it must not react with the solutes, it must be thermally stable, it must have a low volatility, and it must have a polarity that is appropriate for the sample’s components. Table 27.3.1 summarizes the properties of several popular stationary phases.trifluoropropylmethyl polysiloxane(50% trifluoropropyl, 50% methyl)Many stationary phases have the general structure shown in Figure 27.3.4 a. A stationary phase of polydimethyl siloxane, in which all the –R groups are methyl groups, –CH3, is nonpolar and often makes a good first choice for a new separation. The order of elution when using polydimethyl siloxane usually follows the boiling points of the solutes, with lower boiling solutes eluting first. Replacing some of the methyl groups with other substituents increases the stationary phase’s polarity and provides greater selectivity. For example, replacing 50% of the –CH3 groups with phenyl groups, –C6H5, produces a slightly polar stationary phase. Increasing polarity is provided by substituting trifluoropropyl, –C3H6CF, and cyanopropyl, –C3H6CN, functional groups, or by using a stationary phase of polyethylene glycol (Figure 27.3.4 b).An important problem with all liquid stationary phases is their tendency to elute, or bleed from the column when it is heated. The temperature limits in Table 27.3.1 minimize this loss of stationary phase. Capillary columns with bonded or cross-linked stationary phases provide superior stability. A bonded stationary phase is attached chemically to the capillary’s silica surface. Cross-linking, which is done after the stationary phase is in the capillary column, links together separate polymer chains to provide greater stability.Another important consideration is the thickness of the stationary phase with thinner films of stationary phase improving separation efficiency, as we learned in Chapter 26.4. The most common thickness is 0.25 μm, although a thicker films is useful for highly volatile solutes, such as gases, because it has a greater capacity for retaining such solutes. Thinner films are used when separating low volatility solutes, such as steroids.A few stationary phases take advantage of chemical selectivity. The most notable are stationary phases that contain chiral functional groups, which are used to separate enantiomers [Hinshaw, J. V. LC .GC 1993, 11, 644–648].This page titled 27.3: Gas Chromatographic Columns and Stationary Phases is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

27.4: Applications of Gas Chromatography

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Gas chromatography is widely used for the analysis of a diverse array of samples in environmental, clinical, pharmaceutical, biochemical, forensic, food science and petrochemical laboratories. Table 27.4.1 provides some representative examples of applications.green house gases (CO2, CH4, NOx) in airpesticides in water, wastewater, and soilvehicle emissionstrihalomethanes in drinking waterdrugsblood alcoholsanalysis of arson accelerantsdetection of explosivesvolatile organics in spices and fragrancestrace organics in whiskeymonomers in latex paintpurity of solventsrefinery gascomposition of gasolineIn a GC analysis the area under the peak is proportional to the amount of analyte injected onto the column. A peak’s area is determined by integration, which usually is handled by the instrument’s computer or by an electronic integrating recorder. If two peak are resolved fully, the determination of their respective areas is straightforward.Before electronic integrating recorders and computers, two methods were used to find the area under a curve. One method used a manual planimeter; as you use the planimeter to trace an object’s perimeter, it records the area. A second approach for finding a peak’s area is the cut-and-weigh method. The chromatogram is recorded on a piece of paper and each peak of interest is cut out and weighed. Assuming the paper is uniform in thickness and density of fibers, the ratio of weights for two peaks is the same as the ratio of areas. Of course, this approach destroys your chromatogram.Overlapping peaks, however, require a choice between one of several options for dividing up the area shared by the two peaks (Figure 27.4.1 ). Which method we use depends on the relative size of the two peaks and their resolution. In some cases, the use of peak heights provides more accurate results [(a) Bicking, M. K. L. Chromatography Online, April 2006; (b) Bicking, M. K. L. Chromatography Online, June 2006].For quantitative work we need to establish a calibration curve that relates the detector’s response to the analyte’s concentration. If the injection volume is identical for every standard and sample, then an external standardization provides both accurate and precise results. Unfortunately,even under the best conditions the relative precision for replicate injections may differ by 5%; often it is substantially worse. For quantitative work that requires high accuracy and precision, the use of internal standards is recommended.Marriott and Carpenter report the following data for five replicate injections of a mixture that contains 1% v/v methyl isobutyl ketone and 1% v/v p-xylene in dichloromethane [Marriott, P. J.; Carpenter, P. D. J. Chem. Educ. 1996, 73, 96–99].Assume that p-xylene (peak 2) is the analyte, and that methyl isobutyl ketone (peak 1) is the internal standard. Determine the 95% confidence interval for a single-point standardization with and without using the internal standard.For a single-point external standardization we ignore the internal standard and determine the relationship between the peak area for p-xylene, A2, and the concentration, C2, of p-xylene.\[A_{2}=k C_{2} \nonumber \]Substituting the known concentration for p-xylene (1% v/v) and the appropriate peak areas, gives the following values for the constant k.\[78112 \quad 135404 \quad 132332 \quad 112889 \quad 91287 \nonumber \]The average value for k is 110 000 with a standard deviation of 25 100 (a relative standard deviation of 22.8%). The 95% confidence interval is\[\mu=\overline{X} \pm \frac{t s}{\sqrt{n}}=111000 \pm \frac{(2.78)}{\sqrt{5}}=111000 \pm 31200 \nonumber \]For an internal standardization, the relationship between the analyte’s peak area, A2, the internal standard’s peak area, A1, and their respective concentrations, C2 and C1, is\[\frac{A_{2}}{A_{1}}=k \frac{C_{2}}{C_{1}} \nonumber \]Substituting in the known concentrations and the appropriate peak areas gives the following values for the constant k.\[1.5917 \quad 1.5776 \quad 1.5728 \quad 1.5749 \quad 1.5724 \nonumber \]The average value for k is 1.5779 with a standard deviation of 0.0080 (a relative standard deviation of 0.507%). The 95% confidence interval is\[\mu=\overline{X} \pm \frac{t s}{\sqrt{n}}=1.5779 \pm \frac{(2.78)(0.0080)}{\sqrt{5}}=1.5779 \pm 0.0099 \nonumber \]Although there is a substantial variation in the individual peak areas for this set of replicate injections, the internal standard compensates for these variations, providing a more accurate and precise calibration.Figure 27.4.2 shows chromatograms for five standards and for one sample. Each standard and sample contains the same concentration of an internal standard, which is 2.50 mg/mL. For the five standards, the concentrations of analyte are 0.20 mg/mL, 0.40 mg/mL, 0.60 mg/mL, 0.80 mg/mL, and 1.00 mg/mL, respectively. Determine the concentration of analyte in the sample by (a) ignoring the internal standards and creating an external standards calibration curve, and by (b) creating an internal standard calibration curve. For each approach, report the analyte’s concentration and the 95% confidence interval. Use peak heights instead of peak areas.The following table summarizes my measurements of the peak heights for each standard and the sample, and their ratio (although your absolute values for peak heights will differ from mine, depending on the size of your monitor or printout, your relative peak height ratios should be similar to mine). shows the calibration curve and the calibration equation when we ignore the internal standard. Substituting the sample’s peak height into the calibration equation gives the analyte’s concentration in the sample as 0.49 mg/mL. The 95% confidence interval is ±0.24 mg/mL. The calibration curve shows quite a bit of scatter in the data because of uncertainty in the injection volumes. shows the calibration curve and the calibration equation when we include the internal standard. Substituting the sample’s peak height ratio into the calibration equation gives the analyte’s concentration in the sample as 0.54 mg/mL. The 95% confidence interval is ±0.04 mg/mL.The data for this exercise were created so that the analyte’s actual concentration is 0.55 mg/mL. Given the resolution of my ruler’s scale, my answer is pretty reasonable. Your measurements may be slightly different, but your answers should be close to the actual values.In addition to a quantitative analysis, we also can use chromatography to identify the components of a mixture. As noted earlier, when we use an FT–IR or a mass spectrometer as the detector we have access to the eluent’s full spectrum for any retention time. By interpreting the spectrum or by searching against a library of spectra, we can identify the analyte responsible for each chromatographic peak.In addition to identifying the component responsible for a particular chromatographic peak, we also can use the saved spectra to evaluate peak purity. If only one component is responsible for a chromatographic peak, then the spectra should be identical throughout the peak’s elution. If a spectrum at the beginning of the peak’s elution is different from a spectrum taken near the end of the peak’s elution, then at least two components are co-eluting.When using a nonspectroscopic detector, such as a flame ionization detector, we must find another approach if we wish to identify the components of a mixture. One approach is to spike a sample with the suspected compound and look for an increase in peak height. We also can compare a peak’s retention time to the retention time for a known compound if we use identical operating conditions.Because a compound’s retention times on two identical columns are not likely to be the same—differences in packing efficiency, for example, will affect a solute’s retention time on a packed column—creating a table of standard retention times is not possible. Kovat’s retention index provides one solution to the problem of matching retention times. Under isothermal conditions, the adjusted retention times for normal alkanes increase logarithmically. Kovat defined the retention index, I, for a normal alkane as 100 times the number of carbon atoms. For example, the retention index is 400 for butane, C4H10, and 500 for pentane, C5H12. To determine the a compound’s retention index, Icpd, we use the following formula\[I_{cpd} = 100 \times \frac {\log t_{r,cpd}^{\prime} - \log t_{r,x}^{\prime}} {\log t_{r, x+1}^{\prime} - \log t_{r,x}^{\prime}} + I_x \label{12.1} \]where \(t_{r,cpd}^{\prime}\) is the compound’s adjusted retention time, \(t_{r,x}^{\prime}\) and \(t_{r,x+1}^{\prime}\) are the adjusted retention times for the normal alkanes that elute immediately before the compound and immediately after the compound, respectively, and Ix is the retention index for the normal alkane that elutes immediately before the compound. A compound’s retention index for a particular set of chromatographic conditions—such as the choice of stationary phase, mobile phase, column type, column length, and temperature—is reasonably consistent from day-to-day and between different columns and instruments.Tables of Kovat’s retention indices are available; see, for example, the NIST Chemistry Webbook. A search for toluene returns 341 values of I for over 20 different stationary phases, and for both packed columns and capillary columns.In a separation of a mixture of hydrocarbons the following adjusted retention times are measured: 2.23 min for propane, 5.71 min for isobutane, and 6.67 min for butane. What is the Kovat’s retention index for each of these hydrocarbons?Kovat’s retention index for a normal alkane is 100 times the number of carbons; thus, for propane, I = 300 and for butane, I = 400. To find Kovat’s retention index for isobutane we use Equation \ref{12.1}.\[I_\text{isobutane} =100 \times \frac{\log (5.71)-\log (2.23)}{\log (6.67)-\log (2.23)}+300=386 \nonumber \]When using a column with the same stationary phase as in Example 27.4.2 , you find that the retention times for propane and butane are 4.78 min and 6.86 min, respectively. What is the expected retention time for isobutane?Because we are using the same column we can assume that isobutane’s retention index of 386 remains unchanged. Using Equation \ref{12.1}, we have\[386=100 \times \frac{\log x-\log (4.78)}{\log (6.86)-\log (4.78)}+300 \nonumber \]where x is the retention time for isobutane. Solving for x, we find that\[0.86=\frac{\log x-\log (4.78)}{\log (6.86)-\log (4.78)} \nonumber \]\[0.135=\log x-0.679 \nonumber \]\[0.814=\log x \nonumber \]\[x=6.52 \nonumber \]the retention time for isobutane is 6.5 min.This page titled 27.4: Applications of Gas Chromatography is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

28.1: Scope of HPLC

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Gas chromatography consists largely of two specific types of interactions, both of which involve the stationary phase: the partitioning of the solute into a polar or a non-polar stationary phase, or the adsorption of the solute onto a solid packing material. The separation of a complex mixture into its component parts is determined primarily by the boiling points of the solutes and differences in the solubility of the solutes in the stationary phase. The properties of the mobile phase, on the other hand, are less important. It is not surprising that there is not much variety in the basic types of gas chromatography.High-performance liquid chromatography consists of much richer group of techniques, both because the separation depends on the ability of the solutes to partition into the stationary phase and to partition into the mobile phase. The range of the possible types of interactions between the solutes and the stationary phase also is greater in HPLC than in GC. In addition to separations based on differences in solubility of the solutes in the stationary phase and the mobile phase (normal and reverse phase partition chromatography) and separations based on the adsorption of solutes on a solid substrate (adsorption chromatography), the separation of ions is possible using ion-exchange resins as stationary phases (ion-exchange chromatography) and the separation of ions by size (size-exclusion chromatography).This page titled 28.1: Scope of HPLC is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

28.2: Column Efficiency in Liquid Chromatography

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In Chapter 26 we considered three factors that affect band broadening—multiple paths, longitudinal diffusion, and mass transfer—expressing the relationship between the height of a theoretical plate, \(H\), as a function of the mobile phase's velocity, \(u\), using the van Deemter equation\[H = A + \frac{B}{u} + Cu \nonumber \]where \(A\) is the contribution from multiple paths, \(B\) is the contribution from longitudinal diffusion, and \(C\) is the contribution from mass transfer. Unlike gas chromatography, where there is little distance between the point of injection and the column, and little distance between the column and the detector, an HPLC instrument must often include additional tubing to connect together the sample injection port and the column, and the column and the detector. Solutes moving through this tubing, which does not include stationary phase, travel with a velocity that is slower at the walls of the tubing and faster at the center of the tubing; the result is additional band broadening. The magnitude of this contribution to band broadening is minimized by keeping the length of connecting tubing as short as possible. by using tubing with a smaller internal diameter, and by using lower flow rates.This page titled 28.2: Column Efficiency in Liquid Chromatography is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

28.3: Instruments for Liquid Chromatography

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In high-performance liquid chromatography (HPLC) we inject the sample, which is in solution form, into a liquid mobile phase. The mobile phase carries the sample through a packed or capillary column that separates the sample’s components based on their ability to partition between the mobile phase and the stationary phase. Figure 28.3.1 shows an example of a typical HPLC instrument, which has several key components: reservoirs that store the mobile phase; a pump for pushing the mobile phase through the system; an injector for introducing the sample; a column for separating the sample into its component parts; and a detector for monitoring the eluent as it comes off the column. Let’s consider each of these components.An HPLC typically includes two columns: an analytical column, which is responsible for the separation, and a guard column that is placed before the analytical column to protect it from contamination.The most common type of HPLC column is a stainless steel tube with an internal diameter between 2.1 mm and 4.6 mm and a length between 30 mm and 300 mm (Figure 28.3.2 ). The column is packed with 3–10 µm porous silica particles with either an irregular or a spherical shape. Typical column efficiencies are 40 000–60 000 theoretical plates/m. A 25-cm column with 50 000 plates/m has 12 500 theoretical plates.Capillary columns use less solvent and, because the sample is diluted to a lesser extent, produce larger signals at the detector. These columns are made from fused silica capillaries with internal diameters from 44–200 μm and lengths of 50–250 mm. Capillary columns packed with 3–5 μm particles have been prepared with column efficiencies of up to 250 000 theoretical plates [Novotony, M. Science, 1989, 246, 51–57].One limitation to a packed capillary column is the back pressure that develops when pumping the mobile phase through the small interstitial spaces between the particulate micron-sized packing material (Figure 28.3.3 ). Because the tubing and the fittings that carry the mobile phase have pressure limits, a higher back pressure requires a lower flow rate and a longer analysis time. Monolithic columns, in which the solid support is a single, porous rod, offer column efficiencies equivalent to a packed capillary column while allowing for faster flow rates. A monolithic column—which usually is similar in size to a conventional packed column, although smaller, capillary columns also are available—is prepared by forming the mono- lithic rod in a mold and covering it with PTFE tubing or a polymer resin. Monolithic rods made of a silica-gel polymer typically have macropores with diameters of approximately 2 μm and mesopores—pores within the macropores—with diameters of approximately 13 nm [Cabrera, K. Chromatography Online, April 1, 2008].Two problems tend to shorten the lifetime of an analytical column. First, solutes that bind irreversibly to the stationary phase degrade the column’s performance by decreasing the amount of stationary phase available for effecting a separation. Second, particulate material injected with the sample may clog the analytical column. To minimize these problems we place a guard column before the analytical column. A guard column usually contains the same particulate packing material and stationary phase as the analytical column, but is significantly shorter and less expensive—a length of 7.5 mm and a cost one-tenth of that for the corresponding analytical column is typical. Because they are intended to be sacrificial, guard columns are replaced regularly. If you look closely at Figure 28.3.1 , you will see the small guard column just above the analytical column.In a gas chromatograph the pressure from a compressed gas cylinder is sufficient to push the mobile phase through the column. Pushing a liquid mobile phase through a column, however, takes a great deal more effort, generating pressures in excess of several hundred atmospheres. In this section we consider the basic plumbing needed to move the mobile phase through the column and to inject the sample into the mobile phase.A typical HPLC includes between 1–4 reservoirs for storing mobile phase solvents. The instrument in Figure 28.3.1 , for example, has two mobile phase reservoirs that are used for an isocratic elution or a gradient elution by drawing solvents from one or both reservoirs.Before using a mobile phase solvent we must remove dissolved gases, such as N2 and O2, and small particulate matter, such as dust. Because there is a large drop in pressure across the column—the pressure at the column’s entrance is as much as several hundred atmospheres, but it is atmospheric pressure at the column’s exit—gases dissolved in the mobile phase are released as gas bubbles that may interfere with the detector’s response. Degassing is accomplished in several ways, but the most common are the use of a vacuum pump or sparging with an inert gas, such as He, which has a low solubility in the mobile phase. Particulate materials, which may clog the HPLC tubing or column, are removed by filtering the solvents.Bubbling an inert gas through the mobile phase releases volatile dissolved gases. This process is called sparging.The mobile phase solvents are pulled from their reservoirs by the action of one or more pumps. Figure 28.3.4 shows a close-up view of the pumps for the instrument in Figure 28.3.1 . The working pump and the equilibrating pump each have a piston whose back and forth movement maintains a constant flow rate of up to several mL/min and provides the high output pressure needed to push the mobile phase through the chromatographic column. In this particular instrument, each pump sends its mobile phase to a mixing chamber where they combine to form the final mobile phase. The relative speed of the two pumps determines the mobile phase’s final composition.The back and forth movement of a reciprocating pump creates a pulsed flow that contributes noise to the chromatogram. To minimize these pulses, each pump in Figure 28.3.4 has two cylinders. During the working cylinder’s forward stoke it fills the equilibrating cylinder and establishes flow through the column. When the working cylinder is on its reverse stroke, the flow is maintained by the piston in the equilibrating cylinder. The result is a pulse-free flow.There are other ways to control the mobile phase’s composition and flow rate. For example, instead of the two pumps in Figure 28.3.4 , we can place a solvent proportioning valve before a single pump. The solvent proportioning value connects two or more solvent reservoirs to the pump and determines how much of each solvent is pulled during each of the pump’s cycles. Another approach for eliminating a pulsed flow is to include a pulse damper between the pump and the column. A pulse damper is a chamber filled with an easily compressed fluid and a flexible diaphragm. During the piston’s forward stroke the fluid in the pulse damper is compressed. When the piston withdraws to refill the pump, pressure from the expanding fluid in the pulse damper maintains the flow rate.The operating pressure within an HPLC is sufficiently high that we cannot inject the sample into the mobile phase by inserting a syringe through a septum, as is possible in gas chromatography. Instead, we inject the sample using a loop injector, a diagram of which is shown in Figure 28.3.5 . In the load position a sample loop—which is available in a variety of sizes ranging from 0.5 μL to 5 mL—is isolated from the mobile phase and open to the atmosphere. The sample loop is filled using a syringe with a capacity several times that of the sample loop, with excess sample exiting through the waste line. After loading the sample, the injector is turned to the inject position, which redirects the mobile phase through the sample loop and onto the column.The instrument in Figure 28.3.1 uses an autosampler to inject samples. Instead of using a syringe to push the sample into the sample loop, the syringe draws sample into the sample loop.Many different types of detectors have been use to monitor HPLC separations, most of which use spectroscopy or electrochemistry to generate a measurable signal.The most popular HPLC detectors take advantage of an analyte’s UV/Vis absorption spectrum. These detectors range from simple designs, in which the analytical wavelength is selected using appropriate filters, to a modified spectrophotometer in which the sample compartment includes a flow cell. Figure 28.3.6 shows the design of a typical flow cell when using a diode array spectrometer as the detector. The flow cell has a volume of 1–10 μL and a path length of 0.2–1 cm.When using a UV/Vis detector the resulting chromatogram is a plot of absorbance as a function of elution time. If the detector is a diode array spectrometer, then we also can display the result as a three-dimensional chromatogram that shows absorbance as a function of wavelength and elution time. One limitation to using absorbance is that the mobile phase cannot absorb at the wavelengths we wish to monitor. Absorbance detectors provide detection limits of as little as 100 pg–1 ng of injected analyte. If an analyte is fluorescent, we can place the flow cell in a spectrofluorimeter. Detection limits are as little as 1–10 pg of injected analyte.Another common group of HPLC detectors are those based on electrochemical measurements such as amperometry, voltammetry, coulometry, and conductivity. Figure 28.3.7 , for example, shows an amperometric flow cell. Effluent from the column passes over the working electrode—held at a constant potential relative to a downstream reference electrode—that completely oxidizes or reduces the analytes. The current flowing between the working electrode and the auxiliary electrode serves as the analytical signal. Detection limits for amperometric electrochemical detection are from 10 pg–1 ng of injected analyte.Several other detectors have been used in HPLC. Measuring a change in the mobile phase’s refractive index is analogous to monitoring the mobile phase’s thermal conductivity in gas chromatography. A refractive index detector is nearly universal, responding to almost all compounds, but has a relatively poor detection limit of 0.1–1 μg of injected analyte. An additional limitation of a refractive index detector is that it cannot be used for a gradient elution unless the mobile phase components have identical refractive indexes.Another useful detector is a mass spectrometer. Figure 28.3.8 shows a block diagram of a typical HPLC–MS instrument. The effluent from the column enters the mass spectrometer’s ion source using an interface the removes most of the mobile phase, an essential need because of the incompatibility between the liquid mobile phase and the mass spectrometer’s high vacuum environment. In the ionization chamber the remaining molecules—a mixture of the mobile phase components and solutes—undergo ionization and fragmentation. The mass spectrometer’s mass analyzer separates the ions by their mass-to-charge ratio (m/z). A detector counts the ions and displays the mass spectrum.There are several options for monitoring the chromatogram when using a mass spectrometer as the detector. The most common method is to continuously scan the entire mass spectrum and report the total signal for all ions reaching the detector during each scan. This total ion scan provides universal detection for all analytes. We can achieve some degree of selectivity by monitoring only specific mass-to-charge ratios, a process called selective-ion monitoring. The advantages of using a mass spectrometer in HPLC are the same as for gas chromatography. Detection limits are very good, typically 0.1–1 ng of injected analyte, with values as low as 1–10 pg for some samples. In addition, a mass spectrometer provides qualitative, structural information that can help to identify the analytes. The interface between the HPLC and the mass spectrometer is technically more difficult than that in a GC–MS because of the incompatibility of a liquid mobile phase with the mass spectrometer’s high vacuum requirement.This page titled 28.3: Instruments for Liquid Chromatography is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

28.4: Partition Chromatography

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Of the many forms of liquid chromatography, partition chromatography is the most common. In partition chromatography, a solute's retention time is determined by the extent to which it moves from the mobile phase into the stationary phase, and from the stationary phase back into the mobile phase. The extent of this equilibrium partitioning is determined by the polarity of the solutes, the stationary phase, and the mobile phase. In normal-phase partition chromatography, the stationary phase is polar and the mobile phase is non-polar (or of low polarity), with more polar solutes taking longer to elute as they are more strongly retained by the polar stationary phase. In reverse-phase partition chromatography, the stationary phase is non-polar and the mobile phase is polar, with more polar solutes eluting more quickly as they are less strongly retained by the stationary phase. Of the two modes, reverse-phase partition chromatography is the more common.In partition chromatography the stationary phase is a liquid film coated on a packing material, typically 3–10 μm porous silica particles. Because the stationary phase may be partially soluble in the mobile phase, it may elute, or bleed from the column over time. To prevent the loss of stationary phase, which shortens the column’s lifetime, it is bound covalently to the silica particles. Bonded stationary phases are created by reacting the silica particles with an organochlorosilane of the general form Si(CH3)2RCl, where R is an alkyl or substituted alkyl group.To prevent unwanted interactions between the solutes and any remaining –SiOH groups, Si(CH3)3Cl is used to convert unreacted sites to \(–\text{SiOSi(CH}_3)_3\); such columns are designated as end-capped.The properties of a stationary phase depend on the organosilane’s alkyl group. If R is a polar functional group, then the stationary phase is polar. Examples of polar stationary phases include those where R contains a cyano (–C2H4CN), a diol (–C3H6OCH2CHOHCH2OH), or an amino (–C3H6NH2) functional group. The most common nonpolar stationary phases use an organochlorosilane where the R group is an n-octyl (C8) or n-octyldecyl (C18) hydrocarbon chain. Most reversed-phase separations are carried out using a buffered aqueous solution as a polar mobile phase, or using other polar solvents, such as methanol and acetonitrile. Because the silica substrate may undergo hydrolysis in basic solutions, the pH of the mobile phase must be less than 7.5.It seems odd that the more common form of liquid chromatography is identified as reverse-phase instead of normal phase. One of the earliest examples of chromatography was Mikhail Tswett’s separation of plant pigments, which used a polar column of calcium carbonate and a nonpolar mobile phase of petroleum ether. The assignment of normal and reversed, therefore, is all about precedence.The elution order of solutes in HPLC is governed by polarity. For a normal-phase separation, a solute of lower polarity spends proportionally less time in the polar stationary phase and elutes before a solute that is more polar. Given a particular stationary phase, retention times in normal-phase HPLC are controlled by adjusting the mobile phase’s properties. For example, if the resolution between two solutes is poor, switching to a less polar mobile phase keeps the solutes on the column for a longer time and provides more opportunity for their separation. In reversed-phase HPLC the order of elution is the opposite that in a normal-phase separation, with more polar solutes eluting first. Increasing the polarity of the mobile phase leads to longer retention times. Shorter retention times require a mobile phase of lower polarity.There are several indices that help in selecting a mobile phase, one of which is the polarity index [Snyder, L. R.; Glajch, J. L.; Kirkland, J. J. Practical HPLC Method Development, Wiley-Inter- science: New York, 1988]. Table 28.4.1 provides values of the polarity index, \(P^{\prime}\), for several common mobile phases, where larger values of \(P^{\prime}\) correspond to more polar solvents. Mixing together two or more mobile phases—assuming they are miscible—creates a mobile phase of intermediate polarity. For example, a binary mobile phase made by combining solvent A and solvent B has a polarity index, \(P_{AB}^{\prime}\), of\[P_{A B}^{\prime}=\Phi_{A} P_{A}^{\prime}+\Phi_{B} P_{B}^{\prime} \label{12.1} \]where \(P_A^{\prime}\) and \(P_B^{\prime}\) are the polarity indices for solvents A and B, and \(\Phi_A\) and \(\Phi_B\) are the volume fractions for the two solvents.A reversed-phase HPLC separation is carried out using a mobile phase of 60% v/v water and 40% v/v methanol. What is the mobile phase’s polarity index?Using Equation \ref{12.1} and the values in Table 28.4.1 , the polarity index for a 60:40 water–methanol mixture is\[P_{A B}^{\prime}=\Phi_\text{water} P_\text{water}^{\prime}+\Phi_\text{methanol} P_\text{methanol}^{\prime} \nonumber \]\[P_{A B}^{\prime}=0.60 \times 10.2+0.40 \times 5.1=8.2 \nonumber \]Suppose you need a mobile phase with a polarity index of 7.5. Explain how you can prepare this mobile phase using methanol and water.If we let x be the fraction of water in the mobile phase, then 1 – x is the fraction of methanol. Substituting these values into Equation \ref{12.1} and solving for x \[7.5=10.2 x+5.1(1-x) \nonumber \]\[7.5=10.2 x+5.1-5.1 x \nonumber \]\[2.4=5.1 x \nonumber \]gives x as 0.47. The mobile phase is 47% v/v water and 53% v/v methanol.As a general rule, a two unit change in the polarity index corresponds to an approximately 10-fold change in a solute’s retention factor. Here is a simple example. If a solute’s retention factor, k, is 22 when using water as a mobile phase (\(P^{\prime}\) = 10.2), then switching to a mobile phase of 60:40 water–methanol (\(P^{\prime}\) = 8.2) decreases k to approximately 2.2. Note that the retention factor becomes smaller because we are switching from a more polar mobile phase to a less polar mobile phase in a reversed-phase separation.Changing the mobile phase’s polarity index changes a solute’s retention factor. As we learned in Chapter 26.4, however, a change in k is not an effective way to improve resolution when the initial value of k is greater than 10. To effect a better separation between two solutes we must improve the selectivity factor, \(\alpha\). There are two common methods for increasing \(\alpha\): adding a reagent to the mobile phase that reacts with the solutes in a secondary equilibrium reaction or switching to a different mobile phase.Taking advantage of a secondary equilibrium reaction is a useful strategy for improving a separation [(a) Foley, J. P. Chromatography, 1987, 7, 118–128; (b) Foley, J. P.; May, W. E. Anal. Chem. 1987, 59, 102–109; (c) Foley, J. P.; May, W. E. Anal. Chem. 1987, 59, 110–115]. Figure 28.4.1 shows the reversed-phase separation of four weak acids—benzoic acid, terephthalic acid, p-aminobenzoic acid, and p-hydroxybenzoic acid—on a nonpolar C18 column using an aqueous buffer of acetic acid and sodium acetate as the mobile phase. The retention times for these weak acids are shorter when using a less acidic mobile phase because each solute is present in an anionic, weak base form that is less soluble in the nonpolar stationary phase. If the mobile phase’s pH is sufficiently acidic, the solutes are present as neutral weak acids that are more soluble in the stationary phase and take longer to elute. Because the weak acid solutes do not have identical pKa values, the pH of the mobile phase has a different effect on each solute’s retention time, allowing us to find the optimum pH for effecting a complete separation of the four solutes.In Example 28.4.1 we learned how to adjust the mobile phase’s polarity by blending together two solvents. A polarity index, however, is just a guide, and binary mobile phase mixtures with identical polarity indices may not resolve equally a pair of solutes. Table 28.4.2 , for example, shows retention times for four weak acids in two mobile phases with nearly identical values for \(P^{\prime}\). Although the order of elution is the same for both mobile phases, each solute’s retention time is affected differently by the choice of organic solvent. If we switch from using acetonitrile to tetrahydrofuran, for example, we find that benzoic acid elutes more quickly and that p-hydroxybenzoic acid elutes more slowly. Although we can resolve fully these two solutes using mobile phase that is 16% v/v acetonitrile, we cannot resolve them if the mobile phase is 10% tetrahydrofuran.16% acetonitrile (CH3CN)84% pH 4.11 aqueous buffer (\(P^{\prime}\) = 9.5)10% tetrahydrofuran (THF)90% pH 4.11 aqueous buffer (\(P^{\prime}\) = 9.6)Key: BA is benzoic acid; PH is p-hydroxybenzoic acid; PA is p-aminobenzoic acid; TP is terephthalic acid Source: Harvey, D. T.; Byerly, S.; Bowman, A.; Tomlin, J. “Optimization of HPLC and GC Separations Using Response Surfaces,” J. Chem. Educ. 1991, 68, 162–168.One strategy for finding the best mobile phase is to use the solvent triangle shown in Figure 28.4.2 , which allows us to explore a broad range of mobile phases with only seven experiments. We begin by adjusting the amount of acetonitrile in the mobile phase to produce the best possible separation within the desired analysis time. Next, we use Table 28.4.3 to estimate the composition of methanol/H2O and tetrahydrofuran/H2O mobile phases that will produce similar analysis times. Four additional mobile phases are prepared using the binary and ternary mobile phases shown in Figure 28.4.2 . When we examine the chromatograms from these seven mobile phases we may find that one or more provides an adequate separation, or we may identify a region within the solvent triangle where a separation is feasible. Figure 28.4.3 shows a resolution map for the reversed-phase separation of benzoic acid, terephthalic acid, p-aminobenzoic acid, and p-hydroxybenzoic acid on a nonpolar C18 column in which the maximum desired analysis time is set to 6 min [Harvey, D. T.; Byerly, S.; Bowman, A.; Tomlin, J. J. Chem. Educ. 1991, 68, 162–168]. The areas in blue, green, and red show mobile phase compositions that do not provide baseline resolution. The unshaded area represents mobile phase compositions where a separation is possible.The choice to start with acetonitrile is arbitrary—we can just as easily choose to begin with methanol or with tetrahydrofuran.A separation using a mobile phase that has a fixed composition is an isocratic elution. One difficulty with an isocratic elution is that an appropriate mobile phase strength for resolving early-eluting solutes may lead to unacceptably long retention times for late-eluting solutes. Optimizing the mobile phase for late-eluting solutes, on the other hand, may provide an inadequate separation of early-eluting solutes. Changing the mobile phase’s composition as the separation progresses is one solution to this problem. For a reversed-phase separation we use an initial mobile phase that is more polar. As the separation progresses, we adjust the composition of mobile phase so that it becomes less polar (see Figure 28.4.4 ). Such separations are called gradient elutions.The availability of different types detectors provides another way to build selectivity into an analysis. Figure 28.4.5 , for example, shows the reverse-phase separation of a mixture of flavonoids using UV/Vis detection at two different wavelengths. In this case, a wavelength of 260 nm increases the method's sensitivity for rutin relative to that for taxifolin.As shown in Figure 28.4.6 , a fluorescence detector provides additional selectivity because only a few of a sample’s components are fluorescent.With a mass spectrometer as a detector, there are several options for monitoring the chromatogram. The most common method is to continuously scan the entire mass spectrum and report the total signal for all ions reaching the detector during each scan. This total ion scan provides universal detection for all analytes. As seen in Figure 28.4.7 , we can achieve some degree of selectivity by monitoring only specific mass-to-charge ratios, a process called selective-ion monitoring.Partition chromatography is used routinely for both qualitative and quantitative analyses of environmental, pharmaceutical, industrial, forensic, clinical, and consumer product samples.Samples in liquid form are injected into the HPLC after a suitable clean-up to remove any particulate materials, or after a suitable extraction to remove matrix interferents. In determining polyaromatic hydrocarbons (PAH) in wastewater, for example, an extraction with CH2Cl2 serves the dual purpose of concentrating the analytes and isolating them from matrix interferents. Solid samples are first dissolved in a suitable solvent or the analytes of interest brought into solution by extraction. For example, an HPLC analysis for the active ingredients and the degradation products in a pharmaceutical tablet often begins by extracting the powdered tablet with a portion of mobile phase. Gas samples are collected by bubbling them through a trap that contains a suitable solvent. Organic isocyanates in industrial atmospheres are collected by bubbling the air through a solution of 1-(2-methoxyphenyl)piperazine in toluene. The reaction between the isocyanates and 1-(2-methoxyphenyl)piperazine both stabilizes them against degradation before the HPLC analysis and converts them to a chemical form that can be monitored by UV absorption.A quantitative HPLC analysis is often easier than a quantitative GC analysis because a fixed volume sample loop provides a more precise and accurate injection. As a result, most quantitative HPLC methods do not need an internal standard and, instead, use external standards and a normal calibration curve.An internal standard is necessary when using HPLC–MS because the interface between the HPLC and the mass spectrometer does not allow for a reproducible transfer of the column’s eluent into the MS’s ionization chamber.The concentration of polynuclear aromatic hydrocarbons (PAH) in soil is determined by first extracting the PAHs with methylene chloride. The extract is diluted, if necessary, and the PAHs separated by HPLC using a UV/Vis or fluorescence detector. Calibration is achieved using one or more external standards. In a typical analysis a 2.013-g sample of dried soil is extracted with 20.00 mL of methylene chloride. After filtering to remove the soil, a 1.00-mL portion of the extract is removed and diluted to 10.00 mL with acetonitrile. Injecting 5 μL of the diluted extract into an HPLC gives a signal of 0.217 (arbitrary units) for the PAH fluoranthene. When 5 μL of a 20.0-ppm fluoranthene standard is analyzed using the same conditions, a signal of 0.258 is measured. Report the parts per million of fluoranthene in the soil.For a single-point external standard, the relationship between the signal, S, and the concentration, C, of fluoranthene is\[S = kC \nonumber \]Substituting in values for the standard’s signal and concentration gives the value of k as\[k=\frac{S}{C}=\frac{0.258}{20.0 \text{ ppm}}=0.0129 \text{ ppm}^{-1} \nonumber \]Using this value for k and the sample’s HPLC signal gives a fluoranthene concentration of\[C=\frac{S}{k}=\frac{0.217}{0.0129 \text{ ppm}^{-1}}=16.8 \text{ ppm} \nonumber \]for the extracted and diluted soil sample. The concentration of fluoranthene in the soil is\[\frac{16.8 \text{ g} / \mathrm{mL} \times \frac{10.00 \text{ mL}}{1.00 \text{ mL}} \times 20.00 \text{ mL}}{2.013 \text{ g} \text { sample }}=1670 \text{ ppm} \text { fluoranthene } \nonumber \]The concentration of caffeine in beverages is determined by a reversed-phase HPLC separation using a mobile phase of 20% acetonitrile and 80% water, and using a nonpolar C8 column. Results for a series of 10-μL injections of caffeine standards are in the following table.What is the concentration of caffeine in a sample if a 10-μL injection gives a peak area of 424195? The data in this problem comes from Kusch, P.; Knupp, G. “Simultaneous Determination of Caffeine in Cola Drinks and Other Beverages by Reversed-Phase HPTLC and Reversed-Phase HPLC,” Chem. Educator, 2003, 8, 201–205.The figure below shows the calibration curve and calibration equation for the set of external standards. Substituting the sample’s peak area into the calibration equation gives the concentration of caffeine in the sample as 94.4 mg/L.This page titled 28.4: Partition Chromatography is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

28.5: Adsorption Chromatography

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In adsorption chromatography (or liquid-solid chromatography, LSC) the column packing also serves as the stationary phase. In Tswett’s original work the stationary phase was finely divided CaCO3, but modern columns employ porous 3–10 μm particles of silica or alumina. Because the stationary phase is polar, the mobile phase usually is a nonpolar or a moderately polar solvent. Typical mobile phases include hexane, isooctane, and methylene chloride. The usual order of elution—from shorter to longer retention times—isolefins < aromatic hydrocarbons < ethers < esters, aldehydes, ketones < alcohols, amines < amide < carboxylic acidsNonpolar stationary phases, such as charcoal-based absorbents, also are used. For most samples, liquid–solid chromatography does not offer any special advantages over liquid–liquid chromatography. One exception is the analysis of isomers, where LSC excels.This page titled 28.5: Adsorption Chromatography is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

28.6: Ion-Exchange Chromatography

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In ion-exchange chromatography (IEC) the stationary phase is a cross-linked polymer resin, usually divinylbenzene cross-linked polystyrene, with covalently attached ionic functional groups (see Figure 28.6.1 and Table 28.6.1 ). The counterions to these fixed charges are mobile and are displaced by ions that compete more favorably for the exchange sites. Ion-exchange resins are divided into four categories: strong acid cation exchangers; weak acid cation exchangers; strong base anion exchangers; and weak base anion exchangers.Figure 28.6.1 . Structures of styrene, divinylbenzene, and a styrene–divinylbenzene co-polymer modified for use as an ion-exchange resin are shown on the left. The ion-exchange sites, indicated by R and shown in blue, are mostly in the para position and are not necessarily bound to all styrene units. The cross-linking is shown in red. The photo on the right shows an example of the polymer beads. These beads are approximately 0.30–0.85 mm in diameter. Resins for use in ion-exchange chromatography typically are 5–11 μm in diameter.\(-\text{SO}_3^-\)\(-\text{CH}_2\text{CH}_2\text{SO}_3^-\)\(-\text{COO}^-\)\(-\text{CH}_2\text{COO}^-\)\(-\text{CH}_2\text{N(CH}_3)_3^+\)\(-\text{CH}_2\text{CH}_2\text{N(CH}_2\text{CH}_3)_3^+\)\(-\text{NH}_4^+\)\(-\text{CH}_2\text{CH}_2\text{NH(CH}_2\text{CH}_3)_3^+\)Strong acid cation exchangers include a sulfonic acid functional group that retains it anionic form—and thus its capacity for ion-exchange—in strongly acidic solutions. The functional groups for a weak acid cation exchanger, on the other hand, are fully protonated at pH levels less then 4 and lose their exchange capacity. The strong base anion exchangers include a quaternary amine, which retains a positive charge even in strongly basic solutions. Weak base anion exchangers remain protonated only at pH levels that are moderately basic. Under more basic conditions a weak base anion exchanger loses a proton and its exchange capacity.The ion-exchange reaction of a monovalent cation, M+, exchange site is\[-\mathrm{SO}_{3}^{-} \mathrm{H}^{+}(s)+\mathrm{M}^{+}(a q)\rightleftharpoons-\mathrm{SO}_{3}^{-} \mathrm{M}^{+}(s)+\mathrm{H}^{+}(a q) \nonumber \]The equilibrium constant for this ion-exchange reaction, which we call the selectivity coefficient, K, is\[K=\frac{\left\{-\mathrm{SO}_{3}^{-} \mathrm{M}^{+}\right\}\left[\mathrm{H}^{+}\right]}{\left\{-\mathrm{SO}_{3}^{-} \mathrm{H}^{+}\right\}\left[\mathrm{M}^{+}\right]} \label{12.1} \]where we use curly brackets, { }, to indicate a surface concentration instead of a solution concentration.We don’t usually think about a solid’s concentration. There is a good reason for this. In most cases, a solid’s concentration is a constant. If you break a piece of chalk into two parts, for example, the mass and the volume of each piece retains the same proportional relationship as in the original piece of chalk. When we consider an ion binding to a reactive site on the solid’s surface, however, the fraction of sites that are bound, and thus the concentration of bound sites, can take on any value between 0 and some maximum value that is proportional to the density of reactive sites.Rearranging Equation \ref{12.1} shows us that the distribution ratio, D, for the exchange reaction\[D=\frac{\text { amount of } \mathrm{M}^{+} \text { in the stationary phase }}{\text { amount of } \mathrm{M}^{+} \text { in the mobile phase }} \nonumber \]\[D=\frac{\left\{-\mathrm{SO}_{3}^{-} \mathrm{M}^{+}\right\}}{\left[\mathrm{M}^{+}\right]}=K \times \frac{\left\{-\mathrm{SO}_{3}^{-} \mathrm{H}^{+}\right\}}{\left[\mathrm{H}^{+}\right]} \label{12.2} \]is a function of the concentration of H+ and, therefore, the pH of the mobile phase.An ion-exchange resin’s selectivity is somewhat dependent on whether it includes strong or weak exchange sites and on the extent of cross-linking. The latter is particularly important as it controls the resin’s permeability, and, therefore, the accessibility of exchange sites. An approximate order of selectivity for a typical strong acid cation exchange resin, in order of decreasing D, isAl3+ > Ba2+ > Pb2+ > Ca2+ > Ni2+ > Cd2+ > Cu2+ > Co2+ > Zn2+ > Mg2+ > Ag+ > K+ > \(\text{NH}_4^+\) > Na+ > H+ > Li+Note that highly charged cations bind more strongly than cations of lower charge, and that for cations of similar charge, those with a smaller hydrated radius, or that are more polarizable, bind more strongly. For a strong base anion exchanger the general elution order is\(\text{SO}_4^{2-}\) > I– > \(\text{HSO}_4^-\) > \(\text{NO}_3^-\) > Br– > \(\text{NO}_2^-\) > Cl– > \(\text{HCO}_3^-\) > CH3COO– > OH– > F–Anions of higher charge and of smaller hydrated radius bind more strongly than anions with a lower charge and a larger hydrated radius.The mobile phase in IEC usually is an aqueous buffer, the pH and ionic composition of which determines a solute’s retention time. Gradient elutions are possible in which the mobile phase’s ionic strength or pH is changed with time. For example, an IEC separation of cations might use a dilute solution of HCl as the mobile phase. Increasing the concentration of HCl speeds the elution rate for more strongly retained cations because the higher concentration of H+ allows it to compete more successfully for the ion-exchange sites.From Equation \ref{12.2}, a cation’s distribution ratio, D, becomes smaller when the concentration of H+ in the mobile phase increases.An ion-exchange resin is incorporated into an HPLC column either as 5–11 μm porous polymer beads or by coating the resin on porous silica particles. Columns typically are 250 mm in length with internal diameters ranging from 2–5 mm.Measuring the conductivity of the mobile phase as it elutes from the column serves as a universal detector for cationic and anionic analytes. Because the mobile phase contains a high concentration of ions—a mobile phase of dilute HCl, for example, contains significant concentrations of H+ and Cl– ions—we need a method for detecting the analytes in the presence of a significant background conductivity.To minimize the mobile phase’s contribution to conductivity, an ion-suppressor column is placed between the analytical column and the detector. This column selectively removes mobile phase ions without removing solute ions. For example, in cation-exchange chromatography using a dilute solution of HCl as the mobile phase, the suppressor column contains a strong base anion-exchange resin. The exchange reaction\[\mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)+\mathrm{Resin}^{+} \mathrm{OH}^{-}(s)\rightleftharpoons\operatorname{Resin}^{+} \mathrm{Cl}^{-}(s)+\mathrm{H}_{2} \mathrm{O}(l ) \nonumber \]replaces the mobile phase ions H+ and Cl– with H2O. A similar process is used in anion-exchange chromatography where the suppressor column contains a cation-exchange resin. If the mobile phase is a solution of Na2CO3, the exchange reaction\[2 \mathrm{Na}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q)+2 \operatorname{Resin}^{-} \mathrm{H}^{+}(s)\rightleftharpoons2 \operatorname{Resin}^{-} \mathrm{Na}^{+}(s)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \nonumber \]replaces a strong electrolyte, Na2CO3, with a weak electrolyte, H2CO3.Ion-suppression is necessary when the mobile phase contains a high concentration of ions. Single-column ion chromatography, in which an ion-suppressor column is not needed, is possible if the concentration of ions in the mobile phase is small. Typically the stationary phase is a resin with a low capacity for ion-exchange and the mobile phase is a very dilute solution of methane sulfonic acid for cationic analytes, or potassium benzoate or potassium hydrogen phthalate for anionic analytes. Because the background conductivity is sufficiently small, it is possible to monitor a change in conductivity as the analytes elute from the column.A UV/Vis absorbance detector can be used if the analytes absorb ultraviolet or visible radiation. Alternatively, we can detect indirectly analytes that do not absorb in the UV/Vis if the mobile phase contains a UV/Vis absorbing species. In this case, when a solute band passes through the detector, a decrease in absorbance is measured at the detector.Ion-exchange chromatography is an important technique for the analysis of anions and cations in water. For example, an ion-exchange chromatographic analysis for the anions F–, Cl–, Br–, \(\text{NO}_2^-\), \(\text{NO}_3^-\), \(\text{PO}_4^{3-}\), and \(\text{SO}_4^{2-}\) takes approximately 15 minutes (Figure 28.6.2 ). A complete analysis of the same set of anions by a combination of potentiometry and spectrophotometry requires 1–2 days. Ion-exchange chromatography also is used for the analysis of proteins, amino acids, sugars, nucleotides, pharmaceuticals, consumer products, and clinical samples.This page titled 28.6: Ion-Exchange Chromatography is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

28.7: Size-Exclusion Chromatography

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We have considered two classes of micron-sized stationary phases in this chapter: silica particles and cross-linked polymer resin beads. Both materials are porous, with pore sizes ranging from approximately 5–400 nm for silica particles, and from 5 nm to 100 μm for divinylbenzene cross-linked polystyrene resins. In size-exclusion chromatography—which also is known by the terms molecular-exclusion or gel permeation chromatography—the separation of solutes depends upon their ability to enter into the pores of the stationary phase. Smaller solutes spend proportionally more time within the pores and take longer to elute from the column.A stationary phase’s size selectivity extends over a finite range. All solutes significantly smaller than the pores move through the column’s entire volume and elute simultaneously, with a retention volume, Vr, of\[V_{r}=V_{i}+V_{o} \label{12.3} \]where Vi is the volume of mobile phase occupying the stationary phase’s pore space and Vo is volume of mobile phase in the remainder of the column. The largest solute for which Equation \ref{12.3} holds is the column’s inclusion limit, or permeation limit. Those solutes too large to enter the pores elute simultaneously with an retention volume of\[V_{r} = V_{o} \label{12.4} \]Equation \ref{12.4} defines the column’s exclusion limit.For a solute whose size is between the inclusion limit and the exclusion limit, the amount of time it spends in the stationary phase’s pores is proportional to its size. The retention volume for these solutes is\[V_{r}=DV_{i}+V_{o} \label{12.5} \]where D is the solute’s distribution ratio, which ranges from 0 at the exclusion limit to 1 at the inclusion limit. Equation \ref{12.5} assumes that size-exclusion is the only interaction between the solute and the stationary phase that affects the separation. For this reason, stationary phases using silica particles are deactivated as described earlier, and polymer resins are synthesized without exchange sites.Size-exclusion chromatography provides a rapid means for separating larger molecules, including polymers and biomolecules. A stationary phase for proteins that consists of particles with 30 nm pores has an inclusion limit of 7500 g/mol and an exclusion limit of \(1.2 \times 10^6\) g/mol. Mixtures of proteins that span a wider range of molecular weights are separated by joining together in series several columns with different inclusion and exclusion limits.Another important application of size-exclusion chromatography is the estimation of a solute’s molecular weight (MW). Calibration curves are prepared using a series of standards of known molecular weight and measuring each standard’s retention volume. As shown in Figure 28.7.1 , a plot of log(MW) versus Vr is roughly linear between the exclusion limit and the inclusion limit. Because a solute’s retention volume is influenced by both its size and its shape, a reasonably accurate estimation of molecular weight is possible only if the standards are chosen carefully to minimize the effect of shape.Size-exclusion chromatography is carried out using conventional HPLC instrumentation, replacing the HPLC column with an appropriate size-exclusion column. A UV/Vis detector is the most common means for obtaining the chromatogram.This page titled 28.7: Size-Exclusion Chromatography is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

29.1: Properties of Supercritical Fluids

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As shown in Figure 29.1.1 , a supercritical fluid is a species held at a temperature and a pressure that exceeds its critical point. Under these conditions the species is neither a gas nor a liquid. Instead, it is a supercritical fluid.Some properties of a supercritical fluid, as shown in Table 29.1.1 , are similar to a gas; other properties, however, are similar to a liquid. The viscosity of a supercritical fluid, for example, is similar to a gas, which means we can move a supercritical fluid through a capillary column or a packed column without the need for high pressures. The density of a supercritical fluid, on the other hand, is much closer to that of a liquid, which explains why supercritical fluids are good solvents.The most commonly used supercritical fluid is CO2. Its low critical temperature of 31.1oC and its low critical pressure of 72.9 atm are relatively easy to achieve and maintain. Although supercritical CO2 is a good solvent for nonpolar organics, it is less useful for polar solutes. The addition of an organic modifier, such as methanol, improves the mobile phase’s elution strength. Other common mobile phases and their critical temperatures and pressures are listed in Table 29.1.2 .This page titled 29.1: Properties of Supercritical Fluids is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

29.2: Supercritical Fluid Chromatography

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The instrumentation for supercritical fluid chromatography essentially is the same as that for a standard HPLC. The only important additions are a heated oven for the column and a pressure restrictor downstream from the column to maintain the critical pressure. Gradient elutions are accomplished by changing the applied pressure over time. The resulting change in the mobile phase’s density affects its solvent strength. Detection is accomplished using standard GC detectors or HPLC detectors. Analysis time and resolution, although not as good as in GC, usually are better than in conventional HPLC. Supercritical fluid chromatography has many applications in the analysis of polymers, fossil fuels, waxes, drugs, and food products.This page titled 29.2: Supercritical Fluid Chromatography is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

3.1: Operational Amplifiers

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An operational amplifier (or op amp, for short) is an electrical circuit that has a variety of uses, a few of which we consider in this section: how to amplify and measure the signal from a transducer (detector), and how to perform mathematical operations on signals. In this section we will provide a basic overview of operational amplifiers without worrying about the specific internal details of its electrical circuit.An excellent resource for this section and other sections in this chapter is Principles of Electronic Instrumentation by A. James Diefenderfer and published by W. B. Saunders Company, 1972. provides a symbolic representation of an operational amplifier. The large triangular shape is the operational amplifier, which is an extensive circuit whose exact design is not of interest to us; thus, the simple shape. The operational amplifier has two voltage inputs that are identified as \(v_{-}\) and as \(v_{+}\) and labeled as \(-\) and \(+\) on the op amp. The difference between \(v_{-}\) and \(v_{+}\) is defined as \(v_{s}\). The operational amplifier also has a single voltage output that is identified as \(v_{out}\). All voltages are measured relative to a circuit common of 0 V, represented by the small triangle at the bottom of the figure, that provides a shared reference; the circuit common is understood to be present even when it is not shown. Not included in this figure are the connections to a power supply, which are necessary for its operation.The minus sign and the plus sign that appear as labels on the op amp in do not mean that one input has a positive value and that the other input has a negative value. Instead, an input to the lead with a negative sign is inverted: if \(v_{-}\) is a negative DC voltage, then the output voltage, \(v_{out}\), is a positive DC voltage, and if \(v_{-}\) is a positive DC voltage, then the output voltage, \(v_{out}\), is a negative DC voltage. For an AC input to \(v_{-}\), the output is 180° out-of-phase, which implies a reversal in sign. The other input to the op amp is noninverting, which means that applying a positive voltage to \(v_{+}\) results in a positive signal at \(v_{out}\).The ideal operational amplifier has several important properties that derive from its internal circuitry. The first of these properties is that the op amp's gain, \(A\), which is defined as the ratio of the output voltage to the input voltage\[A_{op} = - \frac{v_{out}}{v_{s}} = - \frac{v_{out}}{v_{-} - v_{+}} \label{prop1} \]is very large, typically on the order of 104 – 106. We need to be careful when we use the term gain as there can be a significant difference between the gain of the operational amplifier and the gain of the circuit that contains the operational amplifier. The gain of the operational amplifier, which is what we mean by Equation \ref{prop1}, is called the open-loop gain. The gain of a circuit that contains an operational amplifier is called a closed-loop gain. Where there is ambiguity, we will be careful to refer to the op amp's gain, \(A_{op}\), or to the circuit's gain, \(A_c\), as these are more descriptive.A second property of an operational amplifier is that regardless of the specific values of \(v_{-}\) and \(v_{+}\), the op amp's internal circuitry is designed such that the current between the two inputs is effectively zero; in essence, the impedence, \(Z\), between the two inputs is so large that from Ohm's law, \(V = I \times Z\), the current between these two inputs is \(I \approx 0\). A large input impedence means we can connect our op amp to a high voltage source and know that it will draw a small current instead of overloading the circuit that includes the op amp.A third property of an operational amplifier is that its output impedence is very small, which means we can draw a current from the circuit that meets our needs—this current is drawn from the op amp's power supply—even if the current into the op amp is zero. For example, if the circuit's gain is small, we can use the operational amplifier to provide a large gain in current.This page titled 3.1: Operational Amplifiers is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

3.2: Operational Amplifier Circuits

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In the last section we noted that an operational amplifier magnifies the difference between two voltage inputs\[A_{op} = - \frac{v_{out}}{v_{-} - v_{+}} \label{prop1} \]where the gain, \(A_{op}\), is typically between 104 and 106. To better control the gain—that is, to make the gain something we can adjust to meet our needs—the operational amplifier is incorporated into a circuit that allows for feedback between the output and the inputs. In this section, we examine two feedback circuits. is an example of an operational amplifier circuit with a negative feedback loop that consists of a resistor, \(R_f\), that connects the op amp's output to its input at a summing point, \(S\). Because the feedback loop is connected to the op amp's inverting input, the effect is called negative feedback.We can analyze this circuit using the laws of electricity from Chapter 2. Let's begin by rearranging Equation \ref{prop1} to solve for \(v_{out}\)\[v_{out} = - A_{op} \times (v_{-} - v_{+}) \label{negfb0} \]and then expand the right side of this equation\[v_{out} = -A_{op} \times (v_{-} - v_{+}) = -A_{op} \times v_{-} + A_{op} \times v_{+} \label{negfb1} \]and solve for \(v_{-}\)\[v_{-} = v_{+} - \frac{v_{out}}{A_{op}} \label{negfb2} \]Because the op amp's gain \(A_{op}\) is so large—recall that it is typically in the range 104 and 106—we can simplify Equation \ref{negfb2} to\[v_{-} \approx v_{+} \label{negfb3} \]One consequence of Equation \ref{negfb3} is that for this circuit \(v_{-} \approx 0 \text{ V}\) as it is at the circuit common.From Kirchoff's laws, we know that the total current that enters the summing point must equal the total current that leaves the summing point, or\[I_{in} = I_s + I_f \label{negfb4} \]where \(I_s\) is the current between the op amp's two inputs. As we noted in Chapter 3.1, an operational amplifier's internal circuitry is designed such at \(I_s \approx 0\); thus\[I_{in} = I_f \label{negfb5} \]Substituting in Ohm's law (\(V = I \times R\)) gives\[\frac{v_{in} - v_{-}}{R_{in}} = \frac{v_{-} - v_{out}}{R_f} \label{negfb6} \]From Equation \ref{negfb3}, we know that \(v_{-} \approx 0\), which allows us to simplify Equation \ref{negfb6} to\[\frac{v_{in}}{R_{in}} = -\frac{v_{out}}{R_f} \label{negfb7} \]Rearranging, we find that the gain for the circuit, \(A_c\), is\[A_c = \frac{v_{out}}{v_{in}} = - \frac{R_f}{R_{in}} \label{negfb8} \]Equation \ref{negfb8} shows us that circuit in returns a voltage, \(v_{out}\), that has the opposite sign of \(v_{in}\) with a gain for the circuit that depends on only the relative values of the two resistors, \(R_f\) and \(R_{in}\). shows another operational amplifier with a feedback loop. In this case the input to the op amp, \(v_{in}\), is made to the noninverting lead and the output is feed back into the op amp's inverting lead.From Kirchoff's voltage law, we know that the op amp's output voltage is equal to the sum of the input voltage and the difference, \(v_s\) between the voltage applied to the op amp's two leads; thus\[V_{out} = v_{in} + v_{s} \label{follow1} \]The op amp's gain, \(A_{op}\), is defined in terms of \(v_s\) and \(v_{out}\)\[- A_{op} = \frac{v_{out}}{v_{s}} \label{follow2} \]where the minus sign is due to the change in sign between the output voltage and the voltage applied to the inverting lead. Substituting Equation \ref{follow2} into Equation \ref{follow1} gives\[V_{in} - \frac{V_{out}}{A_{op}} = V_{out} \label{follow3} \]Because the operational amplifier's gain—which is not the same thing as the circuit's gain—is large, Equation \ref{follow3} becomes\[v_{in} = v_{out} \label{follow4} \]Our analysis of this circuit shows that it returns the original voltage without any gain. It does, however, allow us to draw that voltage from the circuit with more current than the original voltage source might be able to handle.This page titled 3.2: Operational Amplifier Circuits is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

3.3: Amplification and Measurement of Signals

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In Chapter 1.3 we identified the basic components of an instrument as a probe that interacts with the sample, an input transducer that converts the sample's chemical and/or physical properties into an electrical signal, a signal processor that converts the electrical signal into a form that an output transducer can convert into a numerical or a visual output that we can understand. We can represent this as a sequence of actions that take place within the instrument\[\text{probe} \rightarrow \text{sample} \rightarrow \text{input transducer} \rightarrow \text{raw data} \rightarrow\text{signal processor} \rightarrow \text{output transducer} \nonumber \]creating the following general flow of information\[\text{chemical and/or physical information} \rightarrow \text{electrical information} \rightarrow \text{numerical or visual response} \nonumber \]As suggested above, information is encoded in two broad ways: as electrical information (such as currents and potentials) and as information in other, non-electrical forms (such as chemical and physical properties). In this section we will consider how we can measure electrical signals.In Chapter 7 we will introduce the phototube, seen here in , as a transducer for converting photons of light into an electrical current that we can measure. A photon of light strikes a photoemissive cathode and ejects an electron that is then drawn to an anode that is held a positive potential. The resulting current is our analytical signal. As each photon generates a single electron, the resulting current is small and needs amplifying if it is to be useful to us. Operational amplifiers provide a way to accomplish this amplification. shows a simple electrical circuit that we can use to amplify and measure a small current. If you compare this to the inverting amplifier circuit in Chapter 3.2, you will see that we are replacing the input voltage and input resistor with the input current, \(I_x\), that we wish to measure.From Kirchoff's current law, we know that at the summing point, \(S\), the current from the transducer is equal to the sum of the current through the feedback loop, \(I_f\), and the current to the op amp's inverting input, \(I_s\).\[I_x = I_s + I_f \label{iv1} \]As we learned in the last section, \(I_s \approx 0\), which means \(I_x \approx I_f\). From Ohm's law, we have\[V_{out} = - I_f \times R_f = -I_x \times R_f \label{iv2} \]Rearranging to solve for \(I_x\)\[I_x = - \frac{V_{out}}{R_f} = k V_{out} \label{iv3} \]shows us that there is a linear relationship between the voltage we measure from the circuit's output and the current that enters the circuit. By choosing to make \(R_f\) large, a small current is converted into a voltage that is easy to measure. In addition, we know from Chapter 2 that the error in measuring the current from the transducer, \(E_x\), is\[E_x = - \frac{R_m}{R_m + R_l} \times 100 \label{iv4} \]where \(R_m\) is the resistance of the measuring circuit and \(R_l\) is the resistance of the source, which generally is large. The resistance of the measuring circuit is\[R_{m} = \frac{R_r}{A_{op}} \label{iv5} \]If we choose \(R_f\) such that it is similar in magnitude to the op amp's gain, \(A_{op}\), then \(R_m\) is small and the relative error is small as well.From Chapter 2 we know that the error in measuring voltage, \(E_x\), is a function of the resistance of the measuring circuit, \(R_m\), and the resistance of the source, \(R_x\).\[E_x = \frac{V_m - V_x}{V_x} \times 100 = - \frac{R_m}{R_m + R_x} \times 100 \label{volt1} \]To maintain a small measurement error requires that \(R_x << R_m\). This creates a complication when the voltage source has a high internal resistance, as is the case, for example, when we measure pH using a glass electrode where the internal resistance is on the order of \(10^7 - 10^8 \Omega\) (see Chapter 23 for details about glass electrodes). The inverting amplifier circuit discussed in Chapter 3.2 has a resistance of perhaps \(10^5 \Omega\). To increase \(R_m\), the voltage we wish to measure, \(V_x\), is first run through a voltage follower circuit, where the internal resistance is on the order of \(10^{12} \Omega\), and the output is then run through the inverting amplifier, as seen in . The result is an amplified output voltage measured under conditions where the relative error is small.In Chapter 13 we will cover molecular absorption spectroscopy in which we measure the absorbance of sample relative to the absorbance of a reference. A difference amplifier, such as that shown in , allows us to amplify and measure the difference between two voltages. In this circuit, the two voltages, \(v_1\) and \(v_2\), are fed into the op amp's two inputs, \(v_{-}\) and \(v_{+}\), passing through identical resistors. A feedback loop with a resistor connects \(v_1\) to \(v_{out}\) and a resistor identical to that in the feedback loop connects \(v_2\) to the circuit common.We can use Ohm's law to define the currents \(I_1\) and \(I_f\) as\[I_1 = \frac{v_1 - v_{-}}{R_i} \label{comp1} \]\[I_f = \frac{v_{-} - v_{out}}{R_f} \label{comp2} \]By now you should see that the currents \(I_1\) and \(I_f\) are approximately the same because the op amp's high internal impedence prevents current from flowing into the op amp. Combining Equation \ref{comp1} and Equation \ref{comp2} gives\[\frac{v_1 - v_{-}}{R_i} = \frac{v_{-} - v_{out}}{R_f} \label{comp3} \]which we can solve for the voltage at the op amp's inverting input\[R_f v_1 - R_f v_{-} = R_i v_{-} - R_i V_{out} \label{comp4} \]\[R_i v_{-} + R_f v_{-} = R_f v_1 + R_i v_{out} \label{comp5} \]\[v_{-} = \frac{R_f v_1 + R_i v_{out}}{R_i + R_f} \label{comp6} \]The input to the op amp's noniverting lead is the output of a voltage divider (see Chapter 2) acting on \(v_2\)\[v_{+} = v_2 \times \left( \frac{R_f}{R_i + R_f} \right) \label{comp7} \]The feedback loop works to ensure that the inputs to \(v_{-}\) and to \(v_{+}\) are identical; thus\[\frac{R_f v_1 + R_i v_{out}}{R_i + R_f} = v_2 \times \left( \frac{R_f}{R_i + R_f} \right) \label{comp8} \]which we can simply to\[V_1 R_f + V_{out} R_i = V_2 R_f \label{comp9} \]\[V_{out} = \frac{R_f}{R_i} \times (V_2 - V_1) \label{comp10} \]The output voltage from the circuit is equal to the difference between the two input voltages, but amplified by the ratio of the resistance of \(R_f\) to \(R_i\).This page titled 3.3: Amplification and Measurement of Signals is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

3.4: Mathematical Operations Using Operational Amplifiers

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The circuit for comparing two voltages is an example of using an operational amplifier to complete a mathematical operation. In this section we will examine several additional examples of mathematical operations completed using operational amplifiers.The inverting amplifier that we considered earlier, and that is reproduced here in , returns an output voltage, \(v_{out}\) that multiplies the input voltage by an amount that depends on the ratio of the resistors \(R_{in}\) and \(R_f\).\[v_{out} = - v_{in} \times \frac{R_f}{R_{in}} \label{math1} \]Multiplication takes place when \(R_f > R_{in}\) and division takes place when \(R_{in} > R_f\). Note that there is a reversal in the sign of the voltage. shows an operational amplifier circuit that adds together four separate input voltages. From our earlier analysis of circuits, you should see that\[I_f = I_1 + I_2 + I_3 + I_4 \label{math2} \]We can replace \(I_f\) in this equation using Ohm's law; thus\[v_{out} = - R_f \times \left( \frac{v_1}{R_1} + \frac{v_2}{R_2} + \frac{v_3}{R_3} + \frac{v_4}{R_4} \right) \label{math3} \]If all five of the resistors are identical, then \(v_{out}\) is a simple summation of the four input voltages.\[v_{out} = - (v_1 + v_2 + v_3 + v_4) \label{math4} \]If we choose \(R_f\) such that it is \(0.25 \times R_1\) and set \(R_1 = R_2 = R_3 = R_4\), then the output voltage is the average of the input voltages\[v_{out} = - \frac{v_1 + v_2 + v_3 + v_4}{4} \label{math5} \]The voltage comparator covered in the last section subtracts one voltage from another. When more than two voltages are involved, then we can adapt the voltage adder circuit in to include subtraction by first running the voltage we wish to subtract through the inverting amplifier introduced in Chapter 3.2. shows this where \(v_{out} = - (v_4 + v_3 + v_2 - v_1)\). shows an operational amplifier circuit that we can use to integrate a time-dependent signal. The circuit has a feedback loop, but it is built around a capacitor instead of a resistor because it stores charge over time. The circuit also has two switches that allow us to use the circuit over a specific period of time. When the hold switch is open, the input voltage cannot enter the circuit. Closing the hold switch sets \(t = t_0\). As long as the reset switch is open, current moves through the feedback loop. Opening the hold switch sets \(t = t_f\), where \(t_f\) is the total elapsed time. When this cycle is over, closing the reset switch drains the capacitor so that it is ready for its next use.As we have seen several times, the current into the summing point, \(I_{in}\) is equal to the current in the feedback loop.\[I_{in} = I_f \label{int1} \]From Chapter 2, we know that the current in the feedback loop is \(I_f = - C_f \frac{d v_{out}}{dt}\) and, we know from Ohm's law that \(i_{in} = \frac{V_{in}}{R_{in}}\). Substituting both relationships into Equation \ref{int1} gives\[\frac{V_{in}}{R_{in}} = - C_f \frac{d v_{out}}{dt} \label{int2} \]Rearranging this equation\[d v_{out} = -\frac{v_{in}}{R_i C_f} dt \label{int3} \]and integrating over time gives\[\int_{v_{out,1}}^{v_{out,2}} d v_{out} = -\frac{1}{R_{in} C_f} \int_{t_1}^{t_2} v_{in} dt \label{int4} \]If we begin the integration having previously discharged the capacitor and define \(t_1\) as the the moment we close the hold switch and define \(t_2\) as the moment we reopen the hold switch, then Equation \ref{int4} becomes\[v_{out} = -\frac{1}{R_{in} C_f} \int_{0}^{t} v_{in} dt \label{int5} \]and the output voltage is the integral of the input voltage multiplied by \((-R_{in} C_f)^{-1}\).Reversing the capacitor and the resistor in the circuit in coverts the circuit from one that returns the integral of the input signal, into one that returns the derivative of the input signal; shows the resulting circuit.For this circuit we have \(I_{in} = C \times \frac{d v_{in}}{dt}\) and \(I_f = - \frac{V_{out}}{R}\). Given that \(I_{in} = I_f\), we are left with\[- \frac{V_{out}}{R} = C \times \frac{d v_{in}}{dt} \label{deriv1} \]\[v_{out} = - R C \times \frac{d v_{in}}{dt} \label{deriv2} \]This page titled 3.4: Mathematical Operations Using Operational Amplifiers is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

30.1: An Overview of Electrophoresis

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Electrophoresis is a class of separation techniques in which we separate analytes by their ability to move through a conductive medium—usually an aqueous buffer—in response to an applied electric field. In the absence of other effects, cations migrate toward the electric field’s negatively charged cathode. Cations with larger charge-to-size ratios—which favors ions of greater charge and of smaller size—migrate at a faster rate than larger cations with smaller charges. Anions migrate toward the positively charged anode and neutral species do not experience the electrical field and remain stationary.As we will see shortly, under normal conditions even neutral species and anions migrate toward the cathode.There are several forms of electrophoresis. In slab gel electrophoresis the conducting buffer is retained within a porous gel of agarose or polyacrylamide. Slabs are formed by pouring the gel between two glass plates separated by spacers. Typical thicknesses are 0.25–1 mm. Gel electrophoresis is an important technique in biochemistry where it frequently is used to separate DNA fragments and proteins. Although it is a powerful tool for the qualitative analysis of complex mixtures, it is less useful for quantitative work.In capillary electrophoresis the conducting buffer is retained within a capillary tube with an inner diameter that typically is 25–75 μm. The sample is injected into one end of the capillary tube, and as it migrates through the capillary the sample’s components separate and elute from the column at different times. The resulting electropherogram looks similar to a GC or an HPLC chromatogram, and provides both qualitative and quantitative information.This page titled 30.1: An Overview of Electrophoresis is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

30.2: Capillary Electrophoresis

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In capillary electrophoresis we inject the sample into a buffered solution retained within a capillary tube. When an electric field is applied across the capillary tube, the sample’s components migrate as the result of two types of actions: electrophoretic mobility and electroosmotic mobility. Electrophoretic mobility is the solute’s response to the applied electrical field in which cations move toward the negatively charged cathode, anions move toward the positively charged anode, and neutral species remain stationary. The other contribution to a solute’s migration is electroosmotic flow, which occurs when the buffer moves through the capillary in response to the applied electrical field. Under normal conditions the buffer moves toward the cathode, sweeping most solutes, including the anions and neutral species, toward the negatively charged cathode.The velocity with which a solute moves in response to the applied electric field is called its electrophoretic velocity, \(\nu_{ep}\); it is defined as\[\nu_{ep}=\mu_{ep} E \label{12.1} \]where \(\mu_{ep}\) is the solute’s electrophoretic mobility, and E is the magnitude of the applied electrical field. A solute’s electrophoretic mobility is defined as\[\mu_{ep}=\frac{q}{6 \pi \eta r} \label{12.2} \]where q is the solute’s charge, \(\eta\) is the buffer’s viscosity, and r is the solute’s radius. Using Equation \ref{12.1} and Equation \ref{12.2} we can make several important conclusions about a solute’s electrophoretic velocity. Electrophoretic mobility and, therefore, electrophoretic velocity, increases for more highly charged solutes and for solutes of smaller size. Because q is positive for a cation and negative for an anion, these species migrate in opposite directions. A neutral species, for which q is zero, has an electrophoretic velocity of zero.When an electric field is applied to a capillary filled with an aqueous buffer we expect the buffer’s ions to migrate in response to their electrophoretic mobility. Because the solvent, H2O, is neutral we might reasonably expect it to remain stationary. What we observe under normal conditions, however, is that the buffer moves toward the cathode. This phenomenon is called the electroosmotic flow.Electroosmotic flow occurs because the walls of the capillary tubing carry a charge. The surface of a silica capillary contains large numbers of silanol groups (–SiOH). At a pH level greater than approximately 2 or 3, the silanol groups ionize to form negatively charged silanate ions (–SiO–). Cations from the buffer are attracted to the silanate ions. As shown in Figure 30.2.1 , some of these cations bind tightly to the silanate ions, forming a fixed layer. Because the cations in the fixed layer only partially neutralize the negative charge on the capillary walls, the solution adjacent to the fixed layer—which is called the diffuse layer—contains more cations than anions. Together these two layers are known as the double layer. Cations in the diffuse layer migrate toward the cathode. Because these cations are solvated, the solution also is pulled along, producing the electroosmotic flow.The anions in the diffuse layer, which also are solvated, try to move toward the anode. Because there are more cations than anions, however, the cations win out and the electroosmotic flow moves in the direction of the cathode.The rate at which the buffer moves through the capillary, what we call its electroosmotic flow velocity, \(\nu_{eof}\), is a function of the applied electric field, E, and the buffer’s electroosmotic mobility, \(\mu_{eof}\).\[\nu_{eof}=\mu_{e o f} E \label{12.3} \]Electroosmotic mobility is defined as\[\mu_{eof}=\frac{\varepsilon \zeta}{4 \pi \eta} \label{12.4} \]where \(\epsilon\) is the buffer dielectric constant, \(\zeta\) is the zeta potential, and \(\eta\) is the buffer’s viscosity.The zeta potential—the potential of the diffuse layer at a finite distance from the capillary wall—plays an important role in determining the electroosmotic flow velocity. Two factors determine the zeta potential’s value. First, the zeta potential is directly proportional to the charge on the capillary walls, with a greater density of silanate ions corresponding to a larger zeta potential. Below a pH of 2 there are few silanate ions and the zeta potential and the electroosmotic flow velocity approach zero. As the pH increases, both the zeta potential and the electroosmotic flow velocity increase. Second, the zeta potential is directly proportional to the thickness of the double layer. Increasing the buffer’s ionic strength provides a higher concentration of cations, which decreases the thickness of the double layer and decreases the electroosmotic flow.The definition of zeta potential given here admittedly is a bit fuzzy. For a more detailed explanation see Delgado, A. V.; González-Caballero, F.; Hunter, R. J.; Koopal, L. K.; Lyklema, J. “Measurement and Interpretation of Electrokinetic Phenomena,” Pure. Appl. Chem. 2005, 77, 1753–1805. Although this is a very technical report, Sections 1.3–1.5 provide a good introduction to the difficulty of defining the zeta potential and of measuring its value.The electroosmotic flow profile is very different from that of a fluid moving under forced pressure. Figure 30.2.2 compares the electroosmotic flow profile with the hydrodynamic flow profile in gas chromatography and liquid chromatography. The uniform, flat profile for electroosmosis helps minimize band broadening in capillary electrophoresis, improving separation efficiency.A solute’s total velocity, \(\nu_{tot}\), as it moves through the capillary is the sum of its electrophoretic velocity and the electroosmotic flow velocity.\[\nu_{t o t}=\nu_{ep}+\nu_{eof} \nonumber \]As shown in Figure 30.2.3 , under normal conditions the following general relationships hold true.\[(\nu_{tot})_{cations} > \nu_{eof} \nonumber \]\[(\nu_{tot})_{neutrals} = \nu_{eof} \nonumber \]\[(\nu_{tot})_{anions} < \nu_{eof} \nonumber \]Cations elute first in an order that corresponds to their electrophoretic mobilities, with small, highly charged cations eluting before larger cations of lower charge. Neutral species elute as a single band with an elution rate equal to the electroosmotic flow velocity. Finally, anions are the last components to elute, with smaller, highly charged anions having the longest elution time.Another way to express a solute’s velocity is to divide the distance it travels by the elapsed time\[\nu_{tot}=\frac{l}{t_{m}} \label{12.5} \]where l is the distance between the point of injection and the detector, and tm is the solute’s migration time. To understand the experimental variables that affect migration time, we begin by noting that\[\nu_{tot} = \mu_{tot}E = (\mu_{ep} + \mu_{eof})E \label{12.6} \]Combining Equation \ref{12.5} and Equation \ref{12.6} and solving for tm leaves us with\[t_{\mathrm{m}}=\frac{l}{\left(\mu_{ep}+\mu_{eof}\right) E} \label{12.7} \]The magnitude of the electrical field is\[E=\frac{V}{L} \label{12.8} \]where V is the applied potential and L is the length of the capillary tube. Finally, substituting Equation \ref{12.8} into Equation \ref{12.7} leaves us with the following equation for a solute’s migration time.\[t_{\mathrm{m}}=\frac{lL}{\left(\mu_{ep}+\mu_{eof}\right) V} \label{12.9} \]To decrease a solute’s migration time—which shortens the analysis time—we can apply a higher voltage or use a shorter capillary tube. We can also shorten the migration time by increasing the electroosmotic flow, although this decreases resolution.As we learned in Chapter 26.3, the efficiency of a separation is given by the number of theoretical plates, N. In capillary electrophoresis the number of theoretic plates is\[N=\frac{l^{2}}{2 D t_{m}}=\frac{\left(\mu_{e p}+\mu_{eof}\right) E l}{2 D L} \label{12.10} \]where D is the solute’s diffusion coefficient. From Equation \ref{12.10}, the efficiency of a capillary electrophoretic separation increases with higher voltages. Increasing the electroosmotic flow velocity improves efficiency, but at the expense of resolution. Two additional observations deserve comment. First, solutes with larger electrophoretic mobilities—in the same direction as the electroosmotic flow—have greater efficiencies; thus, smaller, more highly charged cations are not only the first solutes to elute, but do so with greater efficiency. Second, efficiency in capillary electrophoresis is independent of the capillary’s length. Theoretical plate counts of approximately 100 000–200 000 are not unusual.It is possible to design an electrophoretic experiment so that anions elute before cations—more about this later—in which smaller, more highly charged anions elute with greater efficiencies.In chromatography we defined the selectivity between two solutes as the ratio of their retention factors. In capillary electrophoresis the analogous expression for selectivity is\[\alpha=\frac{\mu_{ep, 1}}{\mu_{ep, 2}} \nonumber \]where \(\mu_{ep,1}\) and \(\mu_{ep,2}\) are the electrophoretic mobilities for the two solutes, chosen such that \(\alpha \ge 1\). We can often improve selectivity by adjusting the pH of the buffer solution. For example, \(\text{NH}_4^+\) is a weak acid with a pKa of 9.75. At a pH of 9.75 the concentrations of \(\text{NH}_4^+\) and NH3 are equal. Decreasing the pH below 9.75 increases its electrophoretic mobility because a greater fraction of the solute is present as the cation \(\text{NH}_4^+\). On the other hand, raising the pH above 9.75 increases the proportion of neutral NH3, decreasing its electrophoretic mobility.The resolution between two solutes is\[R = \frac {0.177(\mu_{ep,2} - \mu_{ep,1})\sqrt{V}} {\sqrt{D(\mu_{avg} + \mu_{eof})}} \label{12.11} \]where \(\mu_{avg}\) is the average electrophoretic mobility for the two solutes. Increasing the applied voltage and decreasing the electroosmotic flow velocity improves resolution. The latter effect is particularly important. Although increasing electroosmotic flow improves analysis time and efficiency, it decreases resolution.The basic instrumentation for capillary electrophoresis is shown in Figure 30.2.4 and includes a power supply for applying the electric field, anode and cathode compartments that contain reservoirs of the buffer solution, a sample vial that contains the sample, the capillary tube, and a detector. Each part of the instrument receives further consideration in this section.Figure 30.2.5 shows a cross-section of a typical capillary tube. Most capillary tubes are made from fused silica coated with a 15–35 μm layer of polyimide to give it mechanical strength. The inner diameter is typically 25–75 μm, which is smaller than the internal diameter of a capillary GC column, with an outer diameter of 200–375 μm.The capillary column’s narrow opening and the thickness of its walls are important. When an electric field is applied to the buffer solution, current flows through the capillary. This current leads to the release of heat, which we call Joule heating. The amount of heat released is proportional to the capillary’s radius and to the magnitude of the electrical field. Joule heating is a problem because it changes the buffer’s viscosity, with the solution at the center of the capillary being less viscous than that near the capillary walls. Because a solute’s electrophoretic mobility depends on its viscosity (see Equation \ref{12.2}), solute species in the center of the capillary migrate at a faster rate than those near the capillary walls. The result is an additional source of band broadening that degrades the separation. Capillaries with smaller inner diameters generate less Joule heating, and capillaries with larger outer diameters are more effective at dissipating the heat. Placing the capillary tube inside a thermostated jacket is another method for minimizing the effect of Joule heating; in this case a smaller outer diameter allows for a more rapid dissipation of thermal energy.There are two common methods for injecting a sample into a capillary electrophoresis column: hydrodynamic injection and electrokinetic injection. In both methods the capillary tube is filled with the buffer solution. One end of the capillary tube is placed in the destination reservoir and the other end is placed in the sample vial.Hydrodynamic injection uses pressure to force a small portion of sample into the capillary tubing. A difference in pressure is applied across the capillary either by pressurizing the sample vial or by applying a vacuum to the destination reservoir. The volume of sample injected, in liters, is given by the following equation\[V_{\text {inj}}=\frac{\Delta P d^{4} \pi t}{128 \eta L} \times 10^{3} \label{12.12} \]where \(\Delta P\) is the difference in pressure across the capillary in pascals, d is the capillary’s inner diameter in meters, t is the amount of time the pressure is applied in seconds, \(\eta\) is the buffer’s viscosity in kg m–1 s–1, and L is the length of the capillary tubing in meters. The factor of 103 changes the units from cubic meters to liters.For a hydrodynamic injection we move the capillary from the source reservoir to the sample. The anode remains in the source reservoir. A hydrodynamic injection also is possible if we raise the sample vial above the destination reservoir and briefly insert the filled capillary.In a hydrodynamic injection we apply a pressure difference of \(2.5 \times 10^3\) Pa (a \(\Delta P \approx 0.02 \text{ atm}\)) for 2 s to a 75-cm long capillary tube with an internal diameter of 50 μm. Assuming the buffer’s viscosity is 10–3 kg m–1 s–1, what volume and length of sample did we inject?Making appropriate substitutions into Equation \ref{12.12} gives the sample’s volume as\[V_{inj}=\frac{\left(2.5 \times 10^{3} \text{ kg} \text{ m}^{-1} \text{ s}^{-2}\right)\left(50 \times 10^{-6} \text{ m}\right)^{4}(3.14)(2 \text{ s})}{\left(0.001 \text{ kg} \text{ m}^{-1} \text{ s}^{-1}\right)(0.75 \text{ m})} \times 10^{3} \mathrm{L} / \mathrm{m}^{3} \nonumber \]\[V_{inj} = 1 \times 10^{-9} \text{ L} = 1 \text{ nL} \nonumber \]Because the interior of the capillary is cylindrical, the length of the sample, l, is easy to calculate using the equation for the volume of a cylinder; thus\[l=\frac{V_{\text {inj}}}{\pi r^{2}}=\frac{\left(1 \times 10^{-9} \text{ L}\right)\left(10^{-3} \text{ m}^{3} / \mathrm{L}\right)}{(3.14)\left(25 \times 10^{-6} \text{ m}\right)^{2}}=5 \times 10^{-4} \text{ m}=0.5 \text{ mm} \nonumber \]Suppose you need to limit your injection to less than 0.20% of the capillary’s length. Using the information from Example 30.2.1 , what is the maximum injection time for a hydrodynamic injection?The capillary is 75 cm long, which means that 0.20% of that sample’s maximum length is 0.15 cm. To convert this to the maximum volume of sample we use the equation for the volume of a cylinder.\[V_{i n j}=l \pi r^{2}=(0.15 \text{ cm})(3.14)\left(25 \times 10^{-4} \text{ cm}\right)^{2}=2.94 \times 10^{-6} \text{ cm}^{3} \nonumber \]Given that 1 cm3 is equivalent to 1 mL, the maximum volume is \(2.94 \times 10^{-6}\) mL or \(2.94 \times 10^{-9}\) L. To find the maximum injection time, we first solve Equation \ref{12.12} for t \[t=\frac{128 V_{inj} \eta L}{P d^{4} \pi} \times 10^{-3} \text{ m}^{3} / \mathrm{L} \nonumber \]and then make appropriate substitutions.\[t=\frac{\left(2.94 \times 10^{-9} \text{ L}\right)\left(0.001 \text{ kg } \text{ m}^{-1} \text{ s}^{-1}\right)(0.75 \text{ m})}{\left(2.5 \times 10^{3} \text{ kg } \mathrm{m}^{-1} \text{ s}^{-2}\right)\left(50 \times 10^{-6} \text{ m}\right)^{4}(3.14)} \times \frac{10^{-3} \text{ m}^{3}}{\mathrm{L}} = 5.8 \text{ s} \nonumber \]The maximum injection time, therefore, is 5.8 s.In an electrokinetic injection we place both the capillary and the anode into the sample and briefly apply an potential. The volume of injected sample is the product of the capillary’s cross sectional area and the length of the capillary occupied by the sample. In turn, this length is the product of the solute’s velocity (see Equation \ref{12.6}) and time; thus\[V_{inj} = \pi r^2 L = \pi r^2 (\mu_{ep} + \mu_{eof})E^{\prime}t \label{12.13} \]where r is the capillary’s radius, L is the capillary’s length, and \(E^{\prime}\) is the effective electric field in the sample. An important consequence of Equation \ref{12.13} is that an electrokinetic injection is biased toward solutes with larger electrophoretic mobilities. If two solutes have equal concentrations in a sample, we inject a larger volume—and thus more moles—of the solute with the larger \(\mu_{ep}\).The electric field in the sample is different that the electric field in the rest of the capillary because the sample and the buffer have different ionic compositions. In general, the sample’s ionic strength is smaller, which makes its conductivity smaller. The effective electric field is\[E^{\prime} = E \times \frac {\chi_\text{buffer}} {\chi_\text{sample}}\nonumber \]where \(\chi_\text{buffer}\) and \(\chi_{sample}\) are the conductivities of the buffer and the sample, respectively.When an analyte’s concentration is too small to detect reliably, it maybe possible to inject it in a manner that increases its concentration. This method of injection is called stacking. Stacking is accomplished by placing the sample in a solution whose ionic strength is significantly less than that of the buffer in the capillary tube. Because the sample plug has a lower concentration of buffer ions, the effective field strength across the sample plug, \(E^{\prime}\), is larger than that in the rest of the capillary.We know from Equation \ref{12.1} that electrophoretic velocity is directly proportional to the electrical field. As a result, the cations in the sample plug migrate toward the cathode with a greater velocity, and the anions migrate more slowly—neutral species are unaffected and move with the electroosmotic flow. When the ions reach their respective boundaries between the sample plug and the buffer, the electrical field decreases and the electrophoretic velocity of the cations decreases and that for the anions increases. As shown in Figure 30.2.6 , the result is a stacking of cations and anions into separate, smaller sampling zones. Over time, the buffer within the capillary becomes more homogeneous and the separation proceeds without additional stacking.Migration in electrophoresis occurs in response to an applied electric field. The ability to apply a large electric field is important because higher voltages lead to shorter analysis times (Equation \ref{12.9}), more efficient separations (Equation \ref{12.10}), and better resolution (Equation \ref{12.11}). Because narrow bored capillary tubes dissipate Joule heating so efficiently, voltages of up to 40 kV are possible.Because of the high voltages, be sure to follow your instrument’s safety guidelines.Most of the detectors used in HPLC also find use in capillary electrophoresis. Among the more common detectors are those based on the absorption of UV/Vis radiation, fluorescence, conductivity, amperometry, and mass spectrometry. Whenever possible, detection is done “on-column” before the solutes elute from the capillary tube and additional band broadening occurs.UV/Vis detectors are among the most popular. Because absorbance is directly proportional to path length, the capillary tubing’s small diameter leads to signals that are smaller than those obtained in HPLC. Several approaches have been used to increase the pathlength, including a Z-shaped sample cell and multiple reflections (see Figure 30.2.7 ). Detection limits are about 10–7 M.Better detection limits are obtained using fluorescence, particularly when using a laser as an excitation source. When using fluorescence detection a small portion of the capillary’s protective coating is removed and the laser beam is focused on the inner portion of the capillary tubing. Emission is measured at an angle of 90o to the laser. Because the laser provides an intense source of radiation that can be focused to a narrow spot, detection limits are as low as 10–16 M.Solutes that do not absorb UV/Vis radiation or that do not undergo fluorescence can be detected by other detectors. Table 30.2.1 provides a list of detectors for capillary electrophoresis along with some of their important characteristics.universal (total ion)selective (single ion)This page titled 30.2: Capillary Electrophoresis is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

30.3: Applications of Capillary Electrophoresis

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There are several different forms of capillary electrophoresis, each of which has its particular advantages. Several of these methods are described briefly in this section.The simplest form of capillary electrophoresis is capillary zone electrophoresis. In CZE we fill the capillary tube with a buffer and, after loading the sample, place the ends of the capillary tube in reservoirs that contain additional buffer. Usually the end of the capillary containing the sample is the anode and solutes migrate toward the cathode at a velocity determined by their respective electrophoretic mobilities and the electroosmotic flow. Cations elute first, with smaller, more highly charged cations eluting before larger cations with smaller charges. Neutral species elute as a single band. Anions are the last species to elute, with smaller, more negatively charged anions being the last to elute.We can reverse the direction of electroosmotic flow by adding an alkylammonium salt to the buffer solution. As shown in Figure 30.3.1 , the positively charged end of the alkyl ammonium ions bind to the negatively charged silanate ions on the capillary’s walls. The tail of the alkyl ammonium ion is hydrophobic and associates with the tail of another alkyl ammonium ion. The result is a layer of positive charges that attract anions in the buffer. The migration of these solvated anions toward the anode reverses the electroosmotic flow’s direction. The order of elution is exactly opposite that observed under normal conditions.Coating the capillary’s walls with a nonionic reagent eliminates the electroosmotic flow. In this form of CZE the cations migrate from the anode to the cathode. Anions elute into the source reservoir and neutral species remain stationary.Capillary zone electrophoresis provides effective separations of charged species, including inorganic anions and cations, organic acids and amines, and large biomolecules such as proteins. For example, CZE was used to separate a mixture of 36 inorganic and organic ions in less than three minutes [Jones, W. R.; Jandik, P. J. Chromatog. 1992, 608, 385–393]. A mixture of neutral species, of course, can not be resolved.One limitation to CZE is its inability to separate neutral species. Micellar electrokinetic capillary chromatography overcomes this limitation by adding a surfactant, such as sodium dodecylsulfate (Figure 30.3.2 a) to the buffer solution. Sodium dodecylsulfate, or SDS, consists of a long-chain hydrophobic tail and a negatively charged ionic functional group at its head. When the concentration of SDS is sufficiently large a micelle forms. A micelle consists of a spherical agglomeration of 40–100 surfactant molecules in which the hydrocarbon tails point inward and the negatively charged heads point outward (Figure 30.3.2 b).Because micelles have a negative charge, they migrate toward the cathode with a velocity less than the electroosmotic flow velocity. Neutral species partition themselves between the micelles and the buffer solution in a manner similar to the partitioning of solutes between the two liquid phases in HPLC. Because there is a partitioning between two phases, we include the descriptive term chromatography in the techniques name. Note that in MEKC both phases are mobile.The elution order for neutral species in MEKC depends on the extent to which each species partitions into the micelles. Hydrophilic neutrals are insoluble in the micelle’s hydrophobic inner environment and elute as a single band, as they would in CZE. Neutral solutes that are extremely hy- drophobic are completely soluble in the micelle, eluting with the micelles as a single band. Those neutral species that exist in a partition equilibrium between the buffer and the micelles elute between the completely hydro- philic and completely hydrophobic neutral species. Those neutral species that favor the buffer elute before those favoring the micelles. Micellar electrokinetic chromatography is used to separate a wide variety of samples, including mixtures of pharmaceutical compounds, vitamins, and explosives.In capillary gel electrophoresis the capillary tubing is filled with a polymeric gel. Because the gel is porous, a solute migrates through the gel with a velocity determined both by its electrophoretic mobility and by its size. The ability to effect a separation using size is helpful when the solutes have similar electrophoretic mobilities. For example, fragments of DNA of varying length have similar charge-to-size ratios, making their separation by CZE difficult. Because the DNA fragments are of different size, a CGE separation is possible.The capillary used for CGE usually is treated to eliminate electroosmotic flow to prevent the gel from extruding from the capillary tubing. Samples are injected electrokinetically because the gel provides too much resistance for hydrodynamic sampling. The primary application of CGE is the separation of large biomolecules, including DNA fragments, proteins, and oligonucleotides.Another approach to separating neutral species is capillary electrochromatography. In CEC the capillary tubing is packed with 1.5–3 μm particles coated with a bonded stationary phase. Neutral species separate based on their ability to partition between the stationary phase and the buffer, which is moving as a result of the electroosmotic flow; Figure 30.3.3 provides a representative example for the separation of a mixture of hydrocarbons. A CEC separation is similar to the analogous HPLC separation, but without the need for high pressure pumps. Efficiency in CEC is better than in HPLC, and analysis times are shorter.This page titled 30.3: Applications of Capillary Electrophoresis is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

31.1: Thermogravimetry

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One method for determining the products of a thermal decomposition is to monitor the sample’s mass as a function of temperature, a process called a thermogravimetric analysis (TGA) or thermogravimetry. Figure 31.1.1 shows a typical thermogram in which each change in mass—each “step” in the thermogram—represents the loss of a volatile product. As the following example illustrates, we can use a thermogram to identify a compound’s decomposition reactions.The thermogram in Figure 31.1.1 shows the mass of a sample of calcium oxalate monohydrate, CaC2O4•H2O, as a function of temperature. The original sample of 17.61 mg was heated from room temperature to 1000oC at a rate of 20oC per minute. For each step in the thermogram, identify the volatilization product and the solid residue that remains.From 100–250oC the sample loses 17.61 mg – 15.44 mg, or 2.17 mg, which is\[\frac{2.17 \ \mathrm{mg}}{17.61 \ \mathrm{mg}} \times 100=12.3 \% \nonumber \]of the sample’s original mass. In terms of CaC2O4•H2O, this corresponds to a decrease in the molar mass of\[0.123 \times 146.11 \ \mathrm{g} / \mathrm{mol}=18.0 \ \mathrm{g} / \mathrm{mol} \nonumber \]The product’s molar mass and the temperature range for the decomposition, suggest that this is a loss of H2O(g), leaving a residue of CaC2O4.The loss of 3.38 mg from 350–550oC is a 19.2% decrease in the sample’s original mass, or a decrease in the molar mass of\[0.192 \times 146.11 \ \mathrm{g} / \mathrm{mol}=28.1 \ \mathrm{g} / \mathrm{mol} \nonumber \]which is consistent with the loss of CO(g) and a residue of CaCO3.Finally, the loss of 5.30 mg from 600-800oC is a 30.1% decrease in the sample’s original mass, or a decrease in molar mass of\[0.301 \times 146.11 \ \mathrm{g} / \mathrm{mol}=44.0 \ \mathrm{g} / \mathrm{mol} \nonumber \]This loss in molar mass is consistent with the release of CO2(g), leaving a final residue of CaO. The three decomposition reactions are\[\begin{array}{c}{\mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}(s) \rightarrow \ \mathrm{CaC}_{2} \mathrm{O}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)} \\ {\mathrm{CaC}_{2} \mathrm{O}_{4}(s) \rightarrow \ \mathrm{CaCO}_{3}(s)+\mathrm{CO}(g)} \\ {\mathrm{CaCO}_{3}(s) \rightarrow \ \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)}\end{array} \nonumber \]Identifying the products of a thermal decomposition provides information that we can use to develop an analytical procedure. For example, the thermogram in Figure 31.1.1 shows that we must heat a precipitate of CaC2O4•H2O to a temperature between 250 and 400oC if we wish to isolate and weigh CaC2O4. Alternatively, heating the sample to 1000oC allows us to isolate and weigh CaO.Under the same conditions as Figure 31.1.1 , the thermogram for a 22.16 mg sample of MgC2O4•H2O shows two steps: a loss of 3.06 mg from 100–250oC and a loss of 12.24 mg from 350–550oC. For each step, identify the volatilization product and the solid residue that remains. Using your results from this exercise and the results from Example 31.1.1 , explain how you can use thermogravimetry to analyze a mixture that contains CaC2O4•H2O and MgC2O4•H2O. You may assume that other components in the sample are inert and thermally stable below 1000oC.From 100–250oC the sample loses 13.8% of its mass, or a loss of\[0.138 \times 130.34 \ \mathrm{g} / \mathrm{mol}=18.0 \ \mathrm{g} / \mathrm{mol} \nonumber \]which is consistent with the loss of H2O(g) and a residue of MgC2O4.From 350–550oC the sample loses 55.23% of its original mass, or a loss of\[0.5523 \times 130.34 \ \mathrm{g} / \mathrm{mol}=71.99 \ \mathrm{g} / \mathrm{mol} \nonumber \]This weight loss is consistent with the simultaneous loss of CO(g) and CO2(g), leaving a residue of MgO.We can analyze the mixture by heating a portion of the sample to 300oC, 600oC, and 1000oC, recording the mass at each temperature. The loss of mass between 600oC and 1000oC, \(\Delta m_2\), is due to the loss of CO2(g) from the decomposition of CaCO3 to CaO, and is proportional to the mass of CaC2O4•H2O in the sample.\[\mathrm{g} \ \mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}=\Delta m_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{CO}_{2}}{44.01 \ \mathrm{g} \ \mathrm{CO}_{2}} \times \frac{146.11 \ \mathrm{g} \ \mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}}{\mathrm{mol} \ \mathrm{CO}_{2}} \nonumber \]The change in mass between 300oC and 600oC, \(\Delta m_1\), is due to the loss of CO(g) from CaC2O4•H2O and the loss of CO(g) and CO2(g) from MgC2O4•H2O. Because we already know the amount of CaC2O4•H2O in the sample, we can calculate its contribution to \(\Delta m_1\).\[\left(\Delta m_{1}\right)_{\mathrm{Ca}}=\mathrm{g} \ \mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}=\Delta m_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{CO}}{146.11 \ \mathrm{g} \ \mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}} \times \frac{28.01 \ \mathrm{g} \ \mathrm{CO}}{\mathrm{mol} \ \mathrm{CO}} \nonumber \]The change in mass between 300oC and 600oC due to the decomposition of MgC2O4•H2O\[\left(m_{1}\right)_{\mathrm{Mg}}=\Delta m_{1}-\left(\Delta m_{1}\right)_{\mathrm{Ca}} \nonumber \]provides the mass of MgC2O4•H2O in the sample.\[\mathrm{g} \ \mathrm{MgC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}=\left(\Delta m_{1}\right)_{\mathrm{Mg}} \times \frac{1 \ \mathrm{mol}\left(\mathrm{CO} \ + \ \mathrm{CO}_{2}\right)}{130.35 \ \mathrm{g} \ \mathrm{MgC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}} \times \frac{78.02 \ \mathrm{g} \ \left(\mathrm{CO} \ + \ \mathrm{CO}_{2}\right)}{\mathrm{mol}\ \left(\mathrm{CO} \ + \ \mathrm{CO}_{2}\right)} \nonumber \]In a thermogravimetric analysis, the sample is placed on a small balance pan attached to one arm of an electromagnetic balance (Figure 31.1.2 ). The sample is lowered into an electric furnace and the furnace’s temperature is increased at a fixed rate of a few degrees per minute while monitoring continuously the sample’s weight. The instrument usually includes a gas line for purging the volatile decomposition products out of the furnace, and a heat exchanger to dissipate the heat emitted by the furnace.Perhaps the most important application gravimetry is exploring a compound's thermal stability, as illustrated in and Exercise \(\PageIndex{1}\) for calcium oxalate hydrate. TGA is particularly useful for studying the thermal stability of polymers.This page titled 31.1: Thermogravimetry is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

31.2: Differential Thermal Analysis and Differential Scanning Calorimetry

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Differential thermal analysis (DTA) and differential scanning calorimetry (DSC) are similar methods in which the response of a sample and a reference to a change in temperature. In DTA the temperature applied to the sample is increased linearly and the difference between the temperature of the reference material, \(T_{ref}\), and the temperature of the sample, \(T_{samp}\), is recorded as function of the sample's temperature\[\Delta T = T_{ref} - T_{samp} \nonumber \]When the sample undergoes an exothermic process, such as a crystallization or a chemical reaction, the temperature of the sample increases more than does the temperature of the reference, resulting in a more negative value for \(\Delta T\). For an endothermic process, such as melting of a crystalline material or the loss of waters of hydration, the sample's temperature lags behind that for the reference materials, resulting in a more positive value for \(\Delta T\). shows the general shape of DTA curve with negative peaks signaling an endothermic process and positive peaks signaling an exothermic process. Changes in \(\Delta T\) that are not peaks, but shifts in the baseline—as seen at the far left of the curve in —are the result of a simple phase transition for which \(\Delta H = 0\).In DSC the temperature applied to the sample is increased linearly and the relative amount of heat needed to maintain the sample and the reference at the same temperature is measured. For an endothermic process, more heat flows into the sample and for an exothermic process, less heat flows into the sample. The result is a DSC curve that looks similar to that for DTA (see ). shows the basic components of a heat-flux differential scanning calorimeter. The sample and the reference materials are sealed within small aluminum pans and placed on separate platforms within the sample chamber. The two platforms are connected by a metal disk that provides a low resistance path for moving heat between the sample and the reference to maintain a \(\Delta T\) of zero between the two. Another instrumental design for differential scanning calorimetry, which is called power compensation DSC, places the sample and the reference in separated heating chambers and measures the difference in the power applied to the two chambers needed to maintain a \(\Delta T\) of zero.Integrating a peak in DSC or DTA to determine its its area, \(A\), gives a signal that is proportional to \(\Delta H\)\[\Delta H = K \times A \nonumber \]where the calibration constant, \(k\), is determined using an established reference material. Both DSC and DTA find applications in the study of polymers, liquid crystals, and pharmaceutical compounds.This page titled 31.2: Differential Thermal Analysis and Differential Scanning Calorimetry is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

32.1: Radioactive Isotopes

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Atoms that have the same number of protons but a different number of neutrons are isotopes. To identify an isotope we use the notation \({}_Z^A E\), where E is the element’s atomic symbol, Z is the element’s atomic number, and A is the element’s atomic mass number. Although an element’s different isotopes have the same chemical properties, their nuclear properties are not identical. The most important difference between isotopes is their stability. The nuclear configuration of a stable isotope remains constant with time. Unstable isotopes, however, disintegrate spontaneously, emitting radioactive decay particles as they transform into a more stable form.An element’s atomic number, Z, is equal to the number of protons and its atomic mass, A, is equal to the sum of the number of protons and neutrons. We represent an isotope of carbon-13 as \(_{6}^{13} \text{C}\) because carbon has six protons and seven neutrons. Sometimes we omit Z from this notation—identifying the element and the atomic number is repetitive because all isotopes of carbon have six protons and any atom that has six protons is an isotope of carbon. Thus, 13C and C–13 are alternative notations for this isotope of carbon.The most important types of radioactive particles are alpha particles, beta particles, gamma rays, and X-rays. An alpha particle, \(\alpha\), is equivalent to a helium nucleus, \({}_2^4 \text{He}\). When an atom emits an alpha particle, the product is a new atom whose atomic number and atomic mass number are, respectively, 2 and 4 less than its unstable parent. The decay of uranium to thorium is one example of alpha emission.\[_{92}^{238} \text{U} \longrightarrow _{90}^{234} \text{Th}+\alpha \nonumber \]A beta particle, \(\beta\), comes in one of two forms. A negatron, \(_{-1}^0 \beta\), is produced when a neutron changes into a proton, increasing the atomic number by one, as shown here for lead.\[_{82}^{214} \mathrm{Pb} \longrightarrow_{83}^{214} \mathrm{Bi} + _{-1}^{0} \beta \nonumber \]The conversion of a proton to a neutron results in the emission of a positron, \(_{1}^0 \beta\).\[_{15}^{30} \mathrm{P} \longrightarrow_{14}^{30} \mathrm{Si} + _{1}^{0} \beta \nonumber \]A negatron, which is the more common type of beta particle, is equivalent to an electron.The emission of an alpha or a beta particle often produces an isotope in an unstable, high energy state. This excess energy is released as a gamma ray, \(\gamma\), or as an X-ray. Gamma ray and X-ray emission may also occur without the release of an alpha particle or a beta particle.A radioactive isotope’s rate of decay, or activity, follows first-order kinetics\[A-{t} = -\frac{d N}{d t}=\lambda N \label{13.1} \]where A is the isotope’s activity, N is the number of radioactive atoms present in the sample at time t, and \(\lambda\) is the isotope’s decay constant. Activity is expressed as the number of disintegrations per unit time.As with any first-order process, we can rewrite Equation \ref{13.1} in an integrated form.\[N_{t}=N_{0} e^{-\lambda t} \label{13.2} \]Substituting Equation \ref{13.2} into Equation \ref{13.1} gives\[A_{t} = \lambda N_{0} e^{-\lambda t}=A_{0} e^{-\lambda t} \label{13.3} \]If we measure a sample’s activity at time t we can determine the sample’s initial activity, A0, or the number of radioactive atoms originally present in the sample, N0.An important characteristic property of a radioactive isotope is its half-life, t1/2, which is the amount of time required for half of the radioactive atoms to disintegrate. For first-order kinetics the half-life is\[t_{1 / 2}=\frac{0.693}{\lambda} \label{13.4} \]Because the half-life is independent of the number of radioactive atoms, it remains constant throughout the decay process. For example, if 50% of the radioactive atoms remain after one half-life, then 25% remain after two half-lives, and 12.5% remain after three half-lives.Suppose we begin with an N0 of 1200 atoms During the first half-life, 600 atoms disintegrate and 600 remain. During the second half-life, 300 of the 600 remaining atoms disintegrate, leaving 300 atoms or 25% of the original 1200 atoms. Of the 300 remaining atoms, only 150 remain after the third half-life, or 12.5% of the original 1200 atoms.Kinetic information about a radioactive isotope usually is given in terms of its half-life because it provides a more intuitive sense of the isotope’s stability. Knowing, for example, that the decay constant for \(_{38}^{90}\text{Sr}\) is 0.0247 yr–1 does not give an immediate sense of how fast it disintegrates. On the other hand, knowing that its half-life is 28.1 yr makes it clear that the concentration of \(_{38}^{90}\text{Sr}\) in a sample remains essentially constant over a short period of time.Radioactivity does not follow a normal distribution because the possible outcomes are not continuous; that is, a sample can emit 1 or 2 or 3 alpha particles (or some other integer value) in a fixed intervale, but it cannot emit 2.59 alpha particles during that same interval. A Poisson distribution provides the probability that a given number of events will occur in a fixed interval in time or space if the event has a known average rate and if each new event is independent of the preceding event. Mathematically a Poisson distribution is defined by the equation\[P(X, \lambda) = \frac {e^{-\lambda} \lambda^X} {X !} \nonumber \]where \(P(X, \lambda)\) is the probability that an event happens X times given the event’s average rate, \(\lambda\). The Poisson distribution has a theoretical mean, \(\mu\), and a theoretical variance, \(\sigma^2\), that are each equal to \(\lambda\).For a more detailed discussion of the distribution of data, including normal distributions and Poisson distributions, see Appendix 1.The accuracy and precision of radiochemical methods generally are within the range of 1–5%. We can improve the precision—which is limited by the random nature of radioactive decay—by counting the emission of radioactive particles for as long a time as is practical. If the number of counts, M, is reasonably large (M ≥ 100), and the counting period is significantly less than the isotope’s half-life, then the percent relative standard deviation for the activity, \((\sigma_A)_{rel}\), is approximately\[\left(\sigma_{A}\right)_{\mathrm{rel}}=\frac{1}{\sqrt{M}} \times 100 \nonumber \]For example, if we determine the activity by counting 10 000 radioactive particles, then the relative standard deviation is 1%. A radiochemical method’s sensitivity is inversely proportional to \((\sigma_A)_{rel}\), which means we can improve the sensitivity by counting more particles.The concentration of a long-lived radioactive isotope remains essentially constant during the period of analysis. As shown in Example 32.1.1 , we can use the sample’s activity to calculate the number of radioactive particles in the sample.The activity in a 10.00-mL sample of wastewater that contains \(_{38}^{90}\text{Sr}\) is \(9.07 \times 10^6\) disintegrations/s. What is the molar concentration of \(_{38}^{90}\text{Sr}\) in the sample? The half-life for \(_{38}^{90}\text{Sr}\) is 28.1 yr.Solving Equation \ref{13.4} for \(\lambda\), substituting into Equation \ref{13.1}, and solving for N gives\[N=\frac{A \times t_{1 / 2}}{0.693} \nonumber \]Before we can determine the number of atoms of \(_{38}^{90}\text{Sr}\) in the sample we must express its activity and its half-life using the same units. Converting the half-life to seconds gives t1/2 as \(8.86 \times 10^8\) s; thus, there are\[\frac{\left(9.07 \times 10^{6} \text { disintegrations/s }\right)\left(8.86 \times 10^{8} \text{ s}\right)}{0.693} = 1.16 \times 10^{16} \text{ atoms} _{38}^{90}\text{Sr} \nonumber \]The concentration of \(_{38}^{90}\text{Sr}\) in the sample is\[\frac{1.16 \times 10^{16} \text { atoms } _{38}^{90} \text{Sr}}{\left(6.022 \times 10^{23} \text { atoms/mol }\right)(0.01000 \mathrm{L})} = 1.93 \times 10^{-6} \text{ M } _{38}^{90}\text{Sr} \nonumber \]The direct analysis of a short-lived radioactive isotope using the method outlined in Example 32.1.1 is less useful because it provides only a transient measure of the isotope’s concentration. Instead, we can measure its activity after an elapsed time, t, and use Equation \ref{13.3} to calculate N0.One example of a characterization application is the determination of a sample’s age based on the decay of a radioactive isotope naturally present in the sample. The most common example is carbon-14 dating, which is used to determine the age of natural organic materials. As cosmic rays pass through the upper atmosphere, some \(_7^{14}\text{N}\) atoms in the atmosphere capture high energy neutrons, converting them into \(_6^{14}\text{C}\). The \(_6^{14}\text{C}\) then migrates into the lower atmosphere where it oxidizes to form C-14 labeled CO2. Animals and plants subsequently incorporate this labeled CO2 into their tissues. Because this is a steady-state process, all plants and animals have the same ratio of \(_6^{14}\text{C}\) to \(_6^{12}\text{C}\) in their tissues. When an organism dies, the radioactive decay of \(_6^{14}\text{C}\) to \(_7^{14}\text{N}\) by \(_{-1}^0 \beta\) emission (t = 5730 years) leads to predictable reduction in the \(_6^{14}\text{C}\) to \(_6^{12}\text{C}\) ratio. We can use the change in this ratio to date samples that are as much as 30000 years old, although the precision of the analysis is best when the sample’s age is less than 7000 years. The accuracy of carbon-14 dating depends upon our assumption that the natural \(_6^{14}\text{C}\) to \(_6^{12}\text{C}\) ratio in the atmosphere is constant over time. Some variation in the ratio has occurred as the result of the increased consumption of fossil fuels and the production of \(_6^{14}\text{C}\) during the testing of nuclear weapons. A calibration curve prepared using samples of known age—examples of samples include tree rings, deep ocean sediments, coral samples, and cave deposits—limits this source of uncertainty.There is no need to prepare a calibration curve for each analysis. Instead, there is a universal calibration curve known as IntCal. The calibration curve from 2013, IntCal13, is described in the following paper: Reimer, P. J., et. al. “IntCal13 and Marine 13 Radiocarbon Age Calibration Curve 0–50,000 Years Cal BP,” Radiocarbon 2013, 55, 1869–1887. This calibration spans 50 000 years before the present (BP).To determine the age of a fabric sample, the relative ratio of \(_6^{14}\text{C}\) to \(_6^{12}\text{C}\) was measured yielding a result of 80.9% of that found in modern fibers. How old is the fabric?Equation \ref{13.3} and Equation \ref{13.4} provide us with a method to convert a change in the ratio of \(_6^{14}\text{C}\) to \(_6^{12}\text{C}\) to the fabric’s age. Letting A0 be the ratio of \(_6^{14}\text{C}\) to \(_6^{12}\text{C}\) in modern fibers, we assign it a value of 1.00. The ratio of \(_6^{14}\text{C}\) to \(_6^{12}\text{C}\) in the sample, A, is 0.809. Solving gives\[t=\ln \frac{A_{0}}{A} \times \frac{t_{1 / 2}}{0.693}=\ln \frac{1.00}{0.809} \times \frac{5730 \text { yr }}{0.693}=1750 \text { yr } \nonumber \]Other isotopes can be used to determine a sample’s age. The age of rocks, for example, has been determined from the ratio of the number of \(_{92}^{238}\text{U}\) to the number of stable \(_{82}^{206}\text{Pb}\) atoms produced by radioactive decay. For rocks that do not contain uranium, dating is accomplished by comparing the ratio of radioactive \(_{19}^{40}\text{K}\) to the stable \(_{18}^{40}\text{Ar}\). Another example is the dating of sediments collected from lakes by measuring the amount of \(_{82}^{210}\text{Pb}\) that is present.This page titled 32.1: Radioactive Isotopes is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

32.2: Instrumentation

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Alpha particles, beta particles, gamma rays, and X-rays are measured by using the particle’s energy to produce an amplified pulse of electrical current in a detector. These pulses are counted to give the rate of disintegration. There are three common types of detectors: gas-filled detectors, scintillation counters, and semiconductor detectors. A gas-filled detector consists of a tube that contains an inert gas, such as Ar. When a radioactive particle enters the tube it ionizes the inert gas, producing an Ar+/e– ion-pair. Movement of the electron toward the anode and of the Ar+ toward the cathode generates a measurable electrical current. A Geiger counter is one example of a gas-filled detector. A scintillation counter uses a fluorescent material to convert radioactive particles into easy to measure photons. For example, one solid-state scintillation counter consists of a NaI crystal that contains 0.2% TlI, which produces several thousand photons for each radioactive particle. Finally, in a semiconductor detector, adsorption of a single radioactive particle promotes thousands of electrons to the semiconductor’s conduction band, increasing conductivity.This page titled 32.2: Instrumentation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

32.3: Neutron Activation Methods

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Few analytes are naturally radioactive. For many analytes, however, we can induce radioactivity by irradiating the sample with neutrons in a process called neutron activation analysis (NAA). The radioactive element formed by neutron activation decays to a stable isotope by emitting a gamma ray, and, possibly, other nuclear particles. The rate of gamma-ray emission is proportional to the analyte’s initial concentration in the sample. For example, if we place a sample containing non-radioactive \(_{13}^{27}\text{Al}\) in a nuclear reactor and irradiate it with neutrons, the following nuclear reaction takes place.\[_{13}^{27} \mathrm{Al}+_{0}^{1} \mathrm{n} \longrightarrow_{13}^{28} \mathrm{Al} \nonumber \]The radioactive isotope of 28Al has a characteristic decay process that includes the release of a beta particle and a gamma ray.\[_{13}^{28} \mathrm{Al} \longrightarrow_{14}^{28} \mathrm{Al} + _{-1}^{0} \beta + \gamma \nonumber \]When irradiation is complete, we remove the sample from the nuclear reactor, allow any short-lived radioactive interferences to decay into the background, and measure the rate of gamma-ray emission.The initial activity at the end of irradiation depends on the number of atoms that are present. This, in turn, is a equal to the difference between the rate of formation for \(_{13}^{28}\text{Al}\) and its rate of disintegration\[\frac {dN_{_{13}^{28} \text{Al}}} {dt} = \Phi \sigma N_{_{13}^{27} \text{Al}} - \lambda N_{_{13}^{28} \text{Al}} \label{13.5} \]where \(\Phi\) is the neutron flux and \(\sigma\) is the reaction cross-section, or probability that a \(_{13}^{27}\text{Al}\) nucleus captures a neutron. Integrating Equation \ref{13.5} over the time of irradiation, ti, and multiplying by \(\lambda\) gives the initial activity, A0, at the end of irradiation as\[A_0 = \lambda N_{_{13}^{28}\text{Al}} = \Phi \sigma N_{_{13}^{27}\text{Al}} (1-e^{-kt}) \nonumber \]If we know the values for A0, \(\Phi\), \(\sigma\), \(\lambda\), and ti, then we can calculate the number of atoms of \(_{13}^{27}\text{Al}\) initially present in the sample.A simpler approach is to use one or more external standards. Letting \((A_0)_x\) and \((A_0)_s\) represent the analyte’s initial activity in an unknown and in an external standard, and letting \(w_x\) and \(w_s\) represent the analyte’s weight in the unknown and in the external standard, we obtain the following pair of equations\[\left(A_{0}\right)_{x}=k w_{x} \label{13.6} \]\[\left(A_{0}\right)_{s}=k w_{s} \label{13.7} \]that we can solve to determine the analyte’s mass in the sample.As noted earlier, gamma ray emission is measured following a period during which we allow short-lived interferents to decay into the background. As shown in Figure 32.3.1 , we determine the sample’s or the standard’s initial activity by extrapolating a curve of activity versus time back to t = 0. Alternatively, if we irradiate the sample and the standard at the same time, and if we measure their activities at the same time, then we can substitute these activities for (A0)x and (A0)s. This is the strategy used in the following example.The concentration of Mn in steel is determined by a neutron activation analysis using the method of external standards. A 1.000-g sample of an unknown steel sample and a 0.950-g sample of a standard steel known to contain 0.463% w/w Mn are irradiated with neutrons for 10 h in a nuclear reactor. After a 40-min delay the gamma ray emission is 2542 cpm (counts per minute) for the unknown and 1984 cpm for the external standard. What is the %w/w Mn in the unknown steel sample?Combining equations \ref{13.6} and \ref{13.7} gives\[w_{x}=\frac{A_{x}}{A_{s}} \times w_{s} \nonumber \]The weight of Mn in the external standard is\[w_{s}=\frac{0.00463 \text{ g } \text{Mn}}{\text{ g } \text { steel }} \times 0.950 \text{ g} \text { steel }=0.00440 \text{ g} \text{ Mn} \nonumber \]Substituting into the above equation gives\[w_{x}=\frac{2542 \text{ cpm}}{1984 \text{ cpm}} \times 0.00440 \text{ g} \text{ Mn}=0.00564 \text{ g} \text{ Mn} \nonumber \]Because the original mass of steel is 1.000 g, the %w/w Mn is 0.564%.Among the advantages of neutron activation are its applicability to almost all elements in the periodic table and that it is nondestructive to the sample. Consequently, NAA is an important technique for analyzing archeological and forensic samples, as well as works of art.This page titled 32.3: Neutron Activation Methods is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

32.4: Isotope Dilution Methods

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Another important radiochemical method for the analysis of nonradioactive analytes is isotope dilution. An external source of analyte is prepared in a radioactive form with a known activity, \(A_T\), for its radioactive decay—we call this form of the analyte a tracer. To prepare a sample for analysis we add a known mass of the tracer, wT, to a portion of sample that contains an unknown mass, wx , of analyte. After homogenizing the sample and tracer, we isolate wA grams of analyte by using a series of appropriate chemical and physical treatments. Because these chemical and physical treatments cannot distinguish between radioactive and nonradioactive forms of the analyte, the isolated material contains both. Finally, we measure the activity of the isolated sample, AA. If we recover all the analyte—both the radioactive tracer and the nonradioactive analyte—then AA and \(A_T\) are equal and wx = wA – wT. Normally, we fail to recover all the analyte. In this case \(A_A\) is less than \(A_T\), and\[A_{A}=A_{T} \times \frac{w_{A}}{w_{x}+w_{T}} \label{13.8} \]The ratio of weights in Equation \ref{13.8} accounts for any loss of activity that results from our failure to recover all the analyte. Solving Equation \ref{13.8} for wx gives\[w_{x}=\frac{A_{T}}{A_{A}} w_{A}-w_{T} \label{13.9} \]How we process the sample depends on the analyte and the sample’s matrix. We might, for example, digest the sample to bring the analyte into solution. After filtering the sample to remove the residual solids, we might precipitate the analyte, isolate it by filtration, dry it in an oven, and obtain its weight.Given that the goal of an analysis is to determine the amount of nonradioactive analyte in our sample, the realization that we might not recover all the analyte might strike you as unsettling. A single liquid–liquid extraction rarely has an extraction efficiency of 100%. One advantage of isotope dilution is that the extraction efficiency for the nonradioactive analyte and for the tracer are the same. If we recover 50% of the tracer, then we also recover 50% of the nonradioactive analyte. Because we know how much tracer we added to the sample, we can determine how much of the nonradioactive analyte is in the sample.The concentration of insulin in a production vat is determined by isotope dilution. A 1.00-mg sample of insulin labeled with 14C that has an activity of 549 cpm is added to a 10.0-mL sample taken from the production vat. After homogenizing the sample, a portion of the insulin is separated and purified, yielding 18.3 mg of pure insulin. The activity for the isolated insulin is measured at 148 cpm. How many mg of insulin are in the original sample?Substituting known values into Equation \ref{13.8} gives\[w_{x}=\frac{549 \text{ cpm}}{148 \text{ cpm}} \times 18.3 \text{ mg}-1.00 \text{ mg}=66.9 \text{ mg} \text { insulin } \nonumber \]Equation \ref{13.8} and Equation \ref{13.9} are valid only if the tracer’s half-life is considerably longer than the time it takes to conduct the analysis. If this is not the case, then the decrease in activity is due both to the incomplete recovery and the natural decrease in the tracer’s activity. Table 32.4.1 provides a list of several common tracers for isotope dilution.An important feature of isotope dilution is that it is not necessary to recover all the analyte to determine the amount of analyte present in the original sample. Isotope dilution, therefore, is useful for the analysis of samples with complex matrices, where a complete recovery of the analyte is difficult.This page titled 32.4: Isotope Dilution Methods is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

33.1: Overview of Automated Methods of Analysis

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An automated method of analysis is one in which the one or more steps in an analysis are completed without the direct action of the analyst. Instead, the instrument itself completes these actions. Some of these actions are carried out discretely, such as an autosampler that can complete all facets of sample preparation, from collecting discrete samples, adding reagents, and diluting the mixture to a desired volume, prior to the analysts analyzing the samples. Another example of a discrete instrument is an automated titrator (see ) that relieves the analyst from manually operating a buret. Instead, the analyst introduces the sample into the automated titrator and lets the instrument complete the titration.Other automated instruments are continuous in nature, in which samples are injected, either manually or with an autosampler, into a flowing stream of reagents that can serve to transport the samples to a detector and can serve as a source of reagents that convert the analyte into a form suitable for analysis.Both discrete and continuous automated methods of analysis have the advantage of allowing for a high throughput of samples and providing for greater reproducibility in results by relieving the analyst of the tedium associated with completing repetitive tasks. In general, continuous automated methods can handle more samples per unit time than can a discrete method.This page titled 33.1: Overview of Automated Methods of Analysis is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

33.2: Flow-Injection Analysis

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In this section we consider the technique of flow injection analysis—a continuous automated method—in which we inject the sample into a flowing carrier stream that gives rise to a transient signal at the detector. The shape of this transient signal depends on the physical and chemical kinetic processes that take place in the carrier stream during the time between injection and detection.Flow injection analysis (FIA) was developed in the mid-1970s as a highly efficient technique for the automated analyses of samples [see, for example, (a) Ruzicka, J.; Hansen, E. H. Anal. Chim. Acta 1975, 78, 145–157; (b) Stewart, K. K.; Beecher, G. R.; Hare, P. E. Anal. Biochem. 1976, 70, 167–173; (c) Valcárcel, M.; Luque de Castro, M. D. Flow Injection Analysis: Principles and Applications, Ellis Horwood: Chichester, England, 1987]. Unlike the centrifugal analyzer described later in this chapter, in which the number of samples is limited by the transfer disk’s size, FIA allows for the rapid, sequential analysis of an unlimited number of samples. FIA is one example of a continuous-flow analyzer, in which we sequentially introduce samples at regular intervals into a liquid carrier stream that transports them to the detector.A schematic diagram detailing the basic components of a flow injection analyzer is shown in Figure 33.2.1 . The reagent that serves as the carrier is stored in a reservoir, and a propelling unit maintains a constant flow of the carrier through a system of tubing that comprises the transport system. We inject the sample directly into the flowing carrier stream, where it travels through one or more mixing and reaction zones before it reaches the detector’s flow-cell. Figure 33.2.1 is the simplest design for a flow injection analyzer, which consists of a single channel and a single reagent reservoir. Multiple channel instruments that merge together separate channels, each of which introduces a new reagent into the carrier stream, also are possible.When we first inject a sample into the carrier stream it has the rectangular flow profile of width w shown in Figure 33.2.2 a. As the sample moves through the mixing zone and the reaction zone, the width of its flow profile increases as the sample disperses into the carrier stream. Dispersion results from two processes: convection due to the flow of the carrier stream and diffusion due to the concentration gradient between the sample and the carrier stream. Convection occurs by laminar flow. The linear velocity of the sample at the tube’s walls is zero, but the sample at the center of the tube moves with a linear velocity twice that of the carrier stream. The result is the parabolic flow profile shown in Figure 33.2.2 b. Convection is the primary means of dispersion in the first 100 ms following the sample’s injection.The second contribution to the sample’s dispersion is diffusion due to the concentration gradient that exists between the sample and the carrier stream. As shown in Figure 33.2.2 , diffusion occurs parallel (axially) and perpendicular (radially) to the direction in which the carrier stream is moving. Only radial diffusion is important in a flow injection analysis. Radial diffusion decreases the sample’s linear velocity at the center of the tubing, while the sample at the edge of the tubing experiences an increase in its linear velocity. Diffusion helps to maintain the integrity of the sample’s flow profile (Figure 33.2.2 c) and prevents adjacent samples in the carrier stream from dispersing into one another. Both convection and diffusion make significant contributions to dispersion from approximately 3–20 s after the sample’s injection. This is the normal time scale for a flow injection analysis. After approximately 25 s, diffusion is the only significant contributor to dispersion, resulting in a flow profile similar to that shown in Figure 33.2.2 d.An FIA curve, or fiagram, is a plot of the detector’s signal as a function of time. Figure 33.2.4 shows a typical fiagram for conditions in which both convection and diffusion contribute to the sample’s dispersion. Also shown on the figure are several parameters that characterize a sample’s fiagram. Two parameters define the time for a sample to move from the injector to the detector. Travel time, ta, is the time between the sample’s injection and the arrival of its leading edge at the detector. Residence time, T, on the other hand, is the time required to obtain the maximum signal. The difference between the residence time and the travel time is \(t^{\prime}\), which approaches zero when convection is the primary means of dispersion, and increases in value as the contribution from diffusion becomes more important.The time required for the sample to pass through the detector’s flow cell—and for the signal to return to the baseline—is also described by two parameters. The baseline-to-baseline time, \(\Delta t\), is the time between the arrival of the sample’s leading edge to the departure of its trailing edge. The elapsed time between the maximum signal and its return to the baseline is the return time, \(T^{\prime}\). The final characteristic parameter of a fiagram is the sample’s peak height, h.Of the six parameters shown in Figure 33.2.4 , the most important are peak height and the return time. Peak height is important because it is directly or indirectly related to the analyte’s concentration. The sensitivity of an FIA method, therefore, is determined by the peak height. The return time is important because it determines the frequency with which we may inject samples. Figure 33.2.5 shows that if we inject a second sample at a time \(T^{\prime}\) after we inject the first sample, there is little overlap of the two FIA curves. By injecting samples at intervals of \(T^{\prime}\), we obtain the maximum possible sampling rate.Peak heights and return times are influenced by the dispersion of the sample’s flow profile and by the physical and chemical properties of the flow injection system. Physical parameters that affect h and \(T^{\prime}\) include the volume of sample we inject, the flow rate, the length, diameter and geometry of the mixing zone and the reaction zone, and the presence of junctions where separate channels merge together. The kinetics of any chemical reactions between the sample and the reagents in the carrier stream also influence the peak height and return time.Unfortunately, there is no good theory that we can use to consistently predict the peak height and the return time for a given set of physical and chemical parameters. The design of a flow injection analyzer for a particular analytical problem still occurs largely by a process of experimentation. Nevertheless, we can make some general observations about the effects of physical and chemical parameters. In the absence of chemical effects, we can improve sensitivity—that is, obtain larger peak heights—by injecting larger samples, by increasing the flow rate, by decreasing the length and diameter of the tubing in the mixing zone and the reaction zone, and by merging separate channels before the point where the sample is injected. With the exception of sample volume, we can increase the sampling rate—that is, decrease the return time—by using the same combination of physical parameters. Larger sample volumes, however, lead to longer return times and a decrease in sample throughput. The effect of chemical reactivity depends on whether the species we are monitoring is a reactant or a product. For example, if we are monitoring a reactant, we can improve sensitivity by choosing conditions that decrease the residence time, T, or by adjusting the carrier stream’s composition so that the reaction occurs more slowly.The basic components of a flow injection analyzer are shown in Figure 33.2.6 and include a pump to propel the carrier stream and the reagent streams, a means to inject the sample into the carrier stream, and a detector to monitor the composition of the carrier stream. Connecting these units is a transport system that brings together separate channels and provides time for the sample to mix with the carrier stream and to react with the reagent streams. We also can incorporate separation modules into the transport system. Each of these components is considered in greater detail in this section.The propelling unit moves the carrier stream through the flow injection analyzer. Although several different propelling units have been used, the most common is a peristaltic pump, which, as shown in Figure 33.2.7 , consists of a set of rollers attached to the outside of a rotating drum. Tubing from the reagent reservoirs fits between the rollers and a fixed plate. As the drum rotates the rollers squeeze the tubing, forcing the contents of the tubing to move in the direction of the rotation. Peristaltic pumps provide a constant flow rate, which is controlled by the drum’s speed of rotation and the inner diameter of the tubing. Flow rates from 0.0005–40 mL/min are possible, which is more than adequate to meet the needs of FIA where flow rates of 0.5–2.5 mL/min are common. One limitation to a peristaltic pump is that it produces a pulsed flow—particularly at higher flow rates—that may lead to oscillations in the signal.The sample, typically 5–200 μL, is injected into the carrier stream. Although syringe injections through a rubber septum are possible, the more common method—as seen in Figure 33.2.6 —is to use a rotary, or loop injector similar to that used in an HPLC. This type of injector provides for a reproducible sample volume and is easily adaptable to automation, an important feature when high sampling rates are needed.The most common detectors for flow injection analysis are the electrochemical and optical detectors used in HPLC. These detectors are discussed in Chapter 28 and are not considered further in this section. FIA detectors also have been designed around the use of ion selective electrodes and atomic absorption spectroscopy.The heart of a flow injection analyzer is the transport system that brings together the carrier stream, the sample, and any reagents that react with the sample. Each reagent stream is considered a separate channel, and all channels must merge before the carrier stream reaches the detector. The complete transport system is called a manifold.The simplest manifold has a single channel, the basic outline of which is shown in Figure 33.2.8 . This type of manifold is used for direct analysis of analyte that does not require a chemical reaction. In this case the carrier stream serves only as a means for rapidly and reproducibly transporting the sample to the detector. For example, this manifold design has been used for sample introduction in atomic absorption spectroscopy, achieving sampling rates as high as 700 samples/h. A single-channel manifold also is used for determining a sample’s pH or determining the concentration of metal ions using an ion selective electrode.We can also use the single-channel manifold in Figure 33.2.8 for an analysis in which we monitor the product of a chemical reaction between the sample and a reactant. In this case the carrier stream both transports the sample to the detector and reacts with the sample. Because the sample must mix with the carrier stream, a lower flow rate is used. One example is the determination of chloride in water, which is based on the following sequence of reactions.\[\mathrm{Hg}(\mathrm{SCN})_{2}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \: \mathrm{HgCl}_{2}(a q)+2 \mathrm{SCN}^{-}(a q) \nonumber \]\[\mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{SCN})^{2+}(a q) \nonumber \]The carrier stream consists of an acidic solution of Hg(SCN)2 and Fe3+. Injecting a sample that contains chloride into the carrier stream displaces thiocyanate from Hg(SCN)2. The displaced thiocyanate then reacts with Fe3+ to form the red-colored Fe(SCN)2+ complex, the absorbance of which is monitored at a wavelength of 480 nm. Sampling rates of approximately 120 samples per hour have been achieved with this system [Hansen, E. H.; Ruzicka, J. J. Chem. Educ. 1979, 56, 677–680].Most flow injection analyses that include a chemical reaction use a manifold with two or more channels. Including additional channels provides more control over the mixing of reagents and the interaction between the reagents and the sample. Two configurations are possible for a dual-channel system. A dual-channel manifold, such as the one shown in Figure 33.2.9 a, is used when the reagents cannot be premixed because of their reactivity. For example, in acidic solutions phosphate reacts with molybdate to form the heteropoly acid H3P(Mo12O40). In the presence of ascorbic acid the molybdenum in the heteropoly acid is reduced from Mo(VI) to Mo(V), forming a blue-colored complex that is monitored spectrophotometrically at 660 nm [Hansen, E. H.; Ruzicka, J. J. Chem. Educ. 1979, 56, 677–680]. Because ascorbic acid reduces molybdate, the two reagents are placed in separate channels that merge just before the loop injector.A dual-channel manifold also is used to add a second reagent after injecting the sample into a carrier stream, as shown in Figure 33.2.9 b. This style of manifold is used for the quantitative analysis of many analytes, including the determination of a wastewater’s chemical oxygen demand (COD) [Korenaga, T.; Ikatsu, H. Anal. Chim. Acta 1982, 141, 301–309]. Chemical oxygen demand is a measure of the amount organic matter in the wastewater sample. In the conventional method of analysis, COD is determined by refluxing the sample for 2 h in the presence of acid and a strong oxidizing agent, such as K2Cr2O7 or KMnO4. When refluxing is complete, the amount of oxidant consumed in the reaction is determined by a redox titration. In the flow injection version of this analysis, the sample is injected into a carrier stream of aqueous H2SO4, which merges with a solution of the oxidant from a secondary channel. The oxidation reaction is kinetically slow and, as a result, the mixing coil and the reaction coil are very long—typically 40 m—and submerged in a thermostated bath. The sampling rate is lower than that for most flow injection analyses, but at 10–30 samples/h it is substantially greater than the redox titrimetric method.More complex manifolds involving three or more channels are common, but the possible combination of designs is too numerous to discuss. One example of a four-channel manifold is shown in Figure 33.2.10 .By incorporating a separation module into the flow injection manifold we can include a separation—dialysis, gaseous diffusion and liquid-liquid extractions are examples—in a flow injection analysis. Although these separations are never complete, they are reproducible if we carefully control the experimental conditions.Dialysis and gaseous diffusion are accomplished by placing a semipermeable membrane between the carrier stream containing the sample and an acceptor stream, as shown in Figure 33.2.11 . As the sample stream passes through the separation module, a portion of those species that can cross the semipermeable membrane do so, entering the acceptor stream. This type of separation module is common for the analysis of clinical samples, such as serum and urine, where a dialysis membrane separates the analyte from its complex matrix. Semipermeable gaseous diffusion membranes are used for the determination of ammonia and carbon dioxide in blood. For example, ammonia is determined by injecting the sample into a carrier stream of aqueous NaOH. Ammonia diffuses across the semipermeable membrane into an acceptor stream that contains an acid–base indicator. The resulting acid–base reaction between ammonia and the indicator is monitored spectrophotometrically.Liquid–liquid extractions are accomplished by merging together two immiscible fluids, each carried in a separate channel. The result is a segmented flow through the separation module, consisting of alternating portions of the two phases. At the outlet of the separation module the two fluids are separated by taking advantage of the difference in their densities. Figure 33.2.12 shows a typical configuration for a separation module in which the sample is injected into an aqueous phase and extracted into a less dense organic phase that passes through the detector.In a quantitative flow injection method a calibration curve is determined by injecting a series of external standards that contain known concentrations of analyte. The calibration curve’s format—examples include plots of absorbance versus concentration and of potential versus concentration—depends on the method of detection.Flow injection analysis has been used to analyze a wide variety of samples, including environmental, clinical, agricultural, industrial, and pharmaceutical samples. The majority of analyses involve environmental and clinical samples, which is the focus of this section.Quantitative flow injection methods have been developed for cationic, anionic, and molecular pollutants in wastewater, freshwaters, groundwaters, and marine waters, three examples of which were described in the previous section. Table 33.2.1 provides a partial listing of other analytes that have been determined using FIA, many of which are modifications of standard spectrophotometric and potentiometric methods. An additional advantage of FIA for environmental analysis is the ability to provide for the continuous, in situ monitoring of pollutants in the field [Andrew, K. N.; Blundell, N. J.; Price, D.; Worsfold, P. J. Anal. Chem. 1994, 66, 916A–922A].Several standard methods for the analysis of water involve an acid–base, complexation, or redox titration. It is easy to adapt these titrations to FIA using a single-channel manifold similar to that shown in Figure 33.2.8 [Ramsing, A. U.; Ruzicka, J.; Hansen, E. H. Anal. Chim. Acta 1981, 129, 1–17]. The titrant—whose concentration must be stoichiometrically less than that of the analyte—and a visual indicator are placed in the reagent reservoir and pumped continuously through the manifold. When we inject the sample it mixes thoroughly with the titrant in the carrier stream. The reaction between the analyte, which is in excess, and the titrant produces a relatively broad rectangular flow profile for the sample. As the sample moves toward the detector, additional mixing occurs and the width of the sample’s flow profile decreases. When the sample passes through the detector, we determine the width of its flow profile, \(\Delta T\), by monitoring the indicator’s absorbance. A calibration curve of \(\Delta T\) versus log[analyte] is prepared using standard solutions of analyte.Flow injection analysis has also found numerous applications in the analysis of clinical samples, using both enzymatic and nonenzymatic methods. Table 33.2.2 summarizes several examples.The majority of flow injection analysis applications are modifications of conventional titrimetric, spectrophotometric, and electrochemical methods of analysis; thus, it is appropriate to compare FIA methods to these conventional methods. The scale of operations for FIA allows for the routine analysis of minor and trace analytes, and for macro, meso, and micro samples. The ability to work with microliter injection volumes is useful when the sample is scarce. Conventional methods of analysis usually have smaller detection limits.The accuracy and precision of FIA methods are comparable to conventional methods of analysis; however, the precision of FIA is influenced by several variables that do not affect conventional methods, including the stability of the flow rate and the reproducibility of the sample’s injection. In addition, results from FIA are more susceptible to temperature variations.In general, the sensitivity of FIA is less than that for conventional methods of analysis for at least two reasons. First, as with chemical kinetic methods, measurements in FIA are made under nonequilibrium conditions when the signal has yet to reach its maximum value. Second, dispersion dilutes the sample as it moves through the manifold. Because the variables that affect sensitivity are known, we can design the FIA manifold to optimize the method’s sensitivity.Selectivity for an FIA method often is better than that for the corresponding conventional method of analysis. In many cases this is due to the kinetic nature of the measurement process, in which potential interferents may react more slowly than the analyte. Contamination from external sources also is less of a problem because reagents are stored in closed reservoirs and are pumped through a system of transport tubing that is closed to the environment.Finally, FIA is an attractive technique when considering time, cost, and equipment. When using an autosampler, a flow injection method can achieve very high sampling rates. A sampling rate of 20–120 samples/h is not unusual and sampling rates as high as 1700 samples/h are possible. Because the volume of the flow injection manifold is small, typically less than 2 mL, the consumption of reagents is substantially smaller than that for a conventional method. This can lead to a significant decrease in the cost per analysis. Flow injection analysis does require the need for additional equipment—a pump, a loop injector, and a manifold—which adds to the cost of an analysis.For a review of the importance of flow injection analysis, see Hansen, E. H.; Miró, M. “How Flow-Injection Analysis (FIA) Over the Past 25 Years has Changed Our Way of Performing Chemical Analyses,” TRAC, Trends Anal. Chem. 2007, 26, 18–26.This page titled 33.2: Flow-Injection Analysis is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

33.3: Other Automated Methods of Analysis

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In the last two sections we introduced two examples of automated methods of analysis: a brief mention of automated titrators and a more extensive coverage of flow-injection analysis. In this section we consider three additional examples of automated methods of analysis: the stopped-flow analyzer, the centrifugal analyzer, and disposable single-test analyzers based on thin films, screen-printing, and paper.A variety of instruments have been developed to automate the kinetic analysis of fast reactions. One example, which is shown in Figure 33.3.1 , is the stopped-flow analyzer. The sample and the reagents are loaded into separate syringes and precisely measured volumes are dispensed into a mixing chamber by the action of a syringe drive. The continued action of the syringe drive pushes the mixture through an observation cell and into a stopping syringe. The back pressure generated when the stopping syringe hits the stopping block completes the mixing, after which the reaction’s progress is monitored spectrophotometrically. With a stopped-flow analyzer it is possible to complete the mixing of sample and reagent, and initiate the kinetic measurements in approximately 0.5 ms. By attaching an autosampler to the sample syringe it is possible to analyze up to several hundred samples per hour.Another instrument for kinetic measurements is the centrifugal analyzer, a partial cross section of which is shown in Figure 33.3.2 . The sample and the reagents are placed in separate wells, which are oriented radially around a circular transfer disk. As the centrifuge spins, the centrifugal force pulls the sample and the reagents into the cuvette where mixing occurs. A single optical source and detector, located below and above the transfer disk’s outer edge, measures the absorbance each time the cuvette passes through the optical beam. When using a transfer disk with 30 cuvettes and rotating at 600 rpm, we can collect 10 data points per second for each sample.The ability to collect lots of data and to collect it quickly requires appropriate hardware and software. Not surprisingly, automated kinetic analyzers developed in parallel with advances in analog and digital circuitry—the hardware—and computer software for smoothing, integrating, and differentiating the analytical signal. For an early discussion of the importance of hardware and software, see Malmstadt, H. V.; Delaney, C. J.; Cordos, E. A. “Instruments for Rate Determinations,” Anal. Chem. 1972, 44, 79A–89A.In comparison to other techniques, potentiometry provides a rapid, relatively low-cost means for analyzing samples. The limiting factor when analyzing a large number of samples is the need to rinse the electrode between samples. The use of inexpensive, disposable ion-selective electrodes can increase a lab’s sample throughput. Figure 33.3.3 shows one example of a disposable ISE for Ag+ [Tymecki, L.; Zwierkowska, E.; Głąb, S.; Koncki, R. Sens. Actuators B 2003, 96, 482–488]. Commercial instruments for measuring pH or potential are available in a variety of price ranges, and includes portable models for use in the field.This page titled 33.3: Other Automated Methods of Analysis is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

34.1: Overview

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Particles come in many forms. Some are very small, such as nanoparticles with dimensions of 1-100 nm and that might consist of just a few hundred atoms, and some are much larger, as in the beads of ion-exchange resin shown in , which range in size from approximately 300 µm to 850 µm. Or, consider soils, which generally are subdivided into four types of particles: clay, which has particles with diameters smaller than 2 µm, silt, which has particle with diameters that range from 2 µm to 50 µm, sand, which has particle with diameters from 50 µm to 2000 µm, and gravel, which has particles with diameters larger than 2000 µm in size.We often hold as an image that particles are spherical in shape, which means we can characterize them by reporting a single number: the particle's diameter. Many particulate materials, however, are not uniform in shape. Although many of the resin beads in appear spherical—the largest bead in the small cluster at the left certainly looks spherical—others of the resin beads are distorted in shape, often appearing somewhat flattened. Still, it is not unusual to treat particles as if they are spheres. There are a number of reasons for this. If the method we use to determine size is not based on a static image (as is the case in ), but on a suspension of particles that are rotating rapidly on the timescale of our measurement, then the particle's shape averages out to a sphere even if the particle itself is not a sphere. The size we report, in this case, is called an equivalent spherical diameter (ESD), which may vary from method-to-method.Suppose we use a method to determine the size of 10000 particles. A simple way to display the data is to use a histogram that reports the frequency of particles in different size ranges, as we see in . We can characterize this distribution by reporting one or more measures of its central tendency and a measure of its spread.Typical measures of central tendency are the mode, which is the most common result, the median, which is the result that falls exactly in the middle of all recorded values, and the mean, which is the numerical average. For the data in , the mode is 0.255 µm (the center of the bin that begins at 0.200 µm and ends at 0.250 µm), the median is 0.265 µm, and the mean is 0.287 µm. If the distribution was symmetrical, then the mode, median, and mean would be identical; here, the distribution has a long tail to the right, which increases the mean relative to the median, and increases the mean and the median relative to the mode.A common way to report the spread is to use the width of the distribution at a frequency that is half of the maximum frequency; this is called the full-width-at-half-maximum (FWHM). For the data in , the maximum frequency is 1230 counts. The FWHM is at a frequency of 615 and runs from a diameter of 0.050 µm to 0.450 µm, or FWHM = 0.450 – 0.050 = 0.400 µm.Another way to characterize the distribution of particle sizes is to plot the cumulative frequency (as a percent) as a function of the diameter of the particles. shows this for the data in using both the binned data for the histogram (shown as the circular black points), and using all of the underlying data (shown as the dashed blue line). The red, purple, and green lines show the particle diameters that include 10% (D10), 50% (D50), and 90% (D90) of all particles. The value of 0.264 µm indicates that half of the particles have diameters less than 0.264 µm and that half have diameters greater than 0.264 µm. One measure of the distribution's relative width is the span, which is defined as\[\text{span} = \frac{\text{D90} - \text{D10}}{\text{D50}} = \frac{0.511 - 0.093}{0.264} = 1.59 \nonumber \]There are a variety of methods that we can use to determine the distribution in the sizes of a particulate material, more than we can cover in a single chapter. Instead, we will consider four common methods: sieving, sedimentation, imaging, and light scattering. When choosing a method, the size and form of the particles are important factors. Sieving, for example, is a practical choice when working with solid particulates that have diameters as small as 20 µm and as large as 125 mm (Note the change from µm to mm!). Sedimentation is useful for particles with diameters of 1 µm, which we can extend to diameters as small as 1 nm by using a centrifuge. Image analysis is useful for particles between 0.5 µm and 1500 µm. Finally, light scattering is useful for particles as small as 0.8 nm.This page titled 34.1: Overview is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

34.2: Measuring Particle Size Using Sieves

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The particulates in a solid matrix are separated by size using one or more sieves ). Sieves are available in a variety of mesh sizes, ranging from approximately 25 mm to 40 μm. By stacking together sieves of different mesh size—with the largest mesh at the top and the smallest mesh at the bottom—we can isolate particulates into several narrow size ranges. Using the sieves in , for example, we can separate a solid into particles with diameters >1700 μm, with diameters between 1700 μm and 500 μm, with diameters between 500 μm and 250 μm, and those with a diameter <250 μm. The sample is place in the uppermost sieve and mechanical shaking used to effect the separation.Because we cannot use more than a limited number of sieves in a single stack, the methods for analyzing the particle size data presented in Chapter 34.1 will be discrete in nature instead of continuous; thus, histograms will have a relatively small number of bins and a cumulative distribution will consist of a discrete number of points. One limitation to a sieve is that irregularly shaped particles are sized based on their two smallest dimensions.This page titled 34.2: Measuring Particle Size Using Sieves is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

34.3: Measuring Particle Size by Sedimentation

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When a particle that is larger than 5 µm is placed in suspension it will slowly settle toward the bottom of its container due to the force of gravity, a process called sedimentation. The time it takes for a particle to move a fixed distance is inversely proportional to the difference in the density of the particle and the density of the fluid in which the particles are suspended, and inversely proportional to the square of the particle's diameter. Larger (and denser) particles, therefore settle out more quickly than do smaller particles, as we see in .To follow the process of sedimentation, a light source is passed through a narrow portion of the sample and the amount of light passing through the sample monitored as a function of time. Once the largest particles pass through the sampling zone, the transmittance of light increases. Standards with well characterized particle sizes are used to calibrate the instrument.For smaller particles, which may remain suspended due to Brownian motion, sedimentation can be carried out using a centrifuge, a technique known as differential centrifigual separation (DCS). As shown in , the sample is introduced in the center of a disk that contains the fluid through which the particles will move. As the disk spins, larger particles move more quickly, eventually reaching the detector located at the outer edge of the disk.This page titled 34.3: Measuring Particle Size by Sedimentation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

34.4: Measuring Particle Size Using Image Analysis

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The chapter overview includes a photograph of an ion-exchange resin's beads. The photograph includes a scale and, in principle, we could use the photograph and scale to estimate the size of the resin's beads. Although the estimates likely are pretty crude, this still serves as an example of image analysis in which we equip an optical microscope or electron microscope with a digital camera that can capture images of the microscope's field of view. Software is used to differentiate the particles from the background, to establish the particle's boundaries, and to determine the particle's size. As shown in , the sample is dispersed on an optical platform and light is passed through the optical platform where it is magnified and focused before capturing the image using a camera. The optical platform can be manually or automatically moved in the xy-plane to capture more images, as in . The software then sorts the particles into groups based on size and reports a count of particles in each group, as in . Because the particles remained immobile, this is called a static image analysis.One limitation to static imaging analysis is that it generally samples a small number of particles as they must be sufficiently dispersed on the optical platform to allow the individual particles to be imaged, analyzed, and counted. In dynamic imaging analysis, the sample is placed in a flow cell set perpendicular to the camera and the light source. Images are collected by using a high-speed flash and shutter speed to capture a sequence of images that are analyzed. By essentially creating an infinite optical platform, dynamic imaging analysis can achieve analysis rates of 10000 particles per minute.This page titled 34.4: Measuring Particle Size Using Image Analysis is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

34.5: Measuring Particle Size Using Light Scattering

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The blue color of the sky during the day and the red color of the sun at sunset are the result of light scattered by small particles of dust, molecules of water, and other gases in the atmosphere. The efficiency of a photon’s scattering depends on its wavelength. We see the sky as blue during the day because violet and blue light scatter to a greater extent than other, longer wavelengths of light. For the same reason, the sun appears red at sunset because red light is less efficiently scattered and is more likely to pass through the atmosphere than other wavelengths of light. The scattering of radiation has been studied since the late 1800s, with applications beginning soon thereafter. The earliest quantitative applications of scattering, which date from the early 1900s, used the elastic scattering of light by colloidal suspensions to determine the concentration of colloidal particles.If we send a focused, monochromatic beam of radiation with a wavelength \(\lambda\) through a medium of particles with dimensions \(< 1.5 \lambda\), the radiation scatters in all directions. For example, visible radiation of 500 nm is scattered by particles as large as 750 nm in the longest dimension. Two general categories of scattering are recognized. In elastic scattering, radiation is first absorbed by the particles and then emitted without undergoing a change in the radiation’s energy. When the radiation emerges with a change in energy, the scattering is inelastic. Only elastic scattering is considered in this chapter.Elastic scattering is divided into two types: Rayleigh, or small-particle scattering, and large-particle scattering. Rayleigh scattering occurs when the scattering particle’s largest dimension is less than 5% of the radiation’s wavelength. The intensity of the scattered radiation is proportional to its frequency to the fourth power, \(\nu^4\)—which accounts for the greater scattering of blue light than red light—and is distributed symmetrically (Figure 34.5.1 a). For larger particles, scattering increases in the forward direction and decreases in the backward direction as the result of constructive and destructive interferences (Figure 34.5.1 b).Small particle, or Rayleigh scattering, measured at an angle of \(\theta\) is the ratio of the intensity of the scattered light, \(I\), to the intensity of the light source, \(I_o\), and is expressed as\[R_{\theta} = \frac{I}{I_o} = K r^6 \nonumber \]where \(r_0\) is the radius of the particle and \(K\) is a constant that is a function of the angle of scattering, the wavelength of light used, the refractive index of the particle, and the distance to the particle, \(R\).In dynamic light scattering (DLS), we use a laser as a light source (see for an illustration). When the light from the source reaches the sample, which is in a sample cell, it scatters in all directions, as shown in .A detector is placed at a fixed angle to collect the light that scatters at that angle. The resulting intensity of scattered light is measure as a function of time. Because the particles in the sample are moving due to Brownian motion, the intensity of light varies with time yielding a noisy signal. Smaller particles diffuse more rapidly than larger particles, which means that fluctuations in intensity with a small particle occurs more rapidly than for a large particle, as seen in .To process the data in DLS, we examine the correlation of the signal with itself over small increments of time. This is accomplished by shifting the signal by a small amount (we call this the delay time, \(\tau\)) and computing the correlation between the original signal and the delayed signal. For short delay times, the correlation in intensities is close to 1 because the particles have not had time to move, and for longer delay times the correlation in intensities is close to 0 because the particles have moved significantly; in between these limits, the correlation undergoes an exponential decay. shows examples of the resulting correlograms for large particles and for small particles. The correlation function, \(G(\tau)\), is defined as\[G(\tau) = A[1 + Be^{-2 \Gamma \tau}] \label{gtau} \]The terms \(A\) and \(B\) are, respectively, the baseline and the intercept of the correlation function, and \(\Gamma = Dq^2\), where \(D\) is the translational diffusion coefficient and where \(q\) is equivalent to \((4 \pi n/ \lambda_0) sin(\theta/2)\) where \(n\) is the refractive index, \(\lambda_0\) is the wavelength of the laser, and \(\theta\) is the angle at which scattered light is collected. The relationship between the size of the particles and the translational diffusion coefficient is give by the Stokes-Einstein equation\[d = \frac{kT}{3 \pi \eta D} \nonumber \]where \(k\) is Boltzmann's constant, \(T\) is the absolute temperature, and \(\eta\) is the viscosity. Fitting one or more equations for \(G(\tau)\) to the correlogram yields the distribution of particle sizes.In dynamic light scattering we are interested in how the intensity of scattering changes with time; in static light scattering, we are interested in how the average intensity of scattered light varies with the concentration of particles, \(c\), and the angle, \(\theta\), at which scattering is measured. The extent of scattering, \(R_{\theta}\), for each combination of \(c\) and \(\theta\) is plotted as \(K c / R_{\theta}\), where \(K\) is a constant that is a function of the solvent's refractive index, the change in refracative index with concentration, and Avogadro's number as a function of the angle; the value of \(S\) for the x-axis is chosen to maintain a separation between the data. A typical plot, which is known as a Zimm plot, is shown in .Each of the solid brown points gives the value of \(K c / R_{\theta}\) for a combination of concentration and angle. For each angle, the change in \(K c / R_{\theta}\) is extrapolated back to a concentration of zero (the dashed green lines) and for each concentration, the change in \(K c / R_{\theta}\) is extrapolated back to an angle of zero (the dashed blue lines). The resulting extrapolation of \(c_0\) to \(\theta = 0\) gives an intercept that is the inverse of the particles molecular weight, \(M\), and the slope of the extrapolations of \(\theta_0\) to a concentration of zero gives \(R_g\), which is the particle's radius of gyration (it is moving, after all), which is its effective particle size.34.5: Measuring Particle Size Using Light Scattering is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

35.1: Evaluation of Analytical Data

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Instrumental_Analysis_(LibreTexts)/35%3A_Appendicies/35.01%3A_Evaluation_of_Analytical_Data

The material in this appendix is adapted from the textbook Chemometrics Using R, which is available through LibreTexts using this link. In addition to the material here, the textbook contains instructions on how to use the statistical programming language R to carry out the calculations.At the heart of any analysis is data. Sometimes our data describes a category and sometimes it is numerical; sometimes our data conveys order and sometimes it does not; sometimes our data has an absolute reference and sometimes it has an arbitrary reference; and sometimes our data takes on discrete values and sometimes it takes on continuous values. Whatever its form, when we gather data our intent is to extract from it information that can help us solve a problem.If we are to consider how to describe data, then we need some data with which we can work. Ideally, we want data that is easy to gather and easy to understand. It also is helpful if you can gather similar data on your own so you can repeat what we cover here. A simple system that meets these criteria is to analyze the contents of bags of M&Ms. Although this system may seem trivial, keep in mind that reporting the percentage of yellow M&Ms in a bag is analogous to reporting the concentration of Cu2+ in a sample of an ore or water: both express the amount of an analyte present in a unit of its matrix.At the beginning of this chapter we identified four contrasting ways to describe data: categorical vs. numerical, ordered vs. unordered, absolute reference vs. arbitrary reference, and discrete vs. continuous. To give meaning to these descriptive terms, let’s consider the data in Table \(\PageIndex{1}\), which includes the year the bag was purchased and analyzed, the weight listed on the package, the type of M&Ms, the number of yellow M&Ms in the bag, the percentage of the M&Ms that were red, the total number of M&Ms in the bag and their corresponding ranks.The entries in Table \(\PageIndex{1}\) are organized by column and by row. The first row—sometimes called the header row—identifies the variables that make up the data. Each additional row is the record for one sample and each entry in a sample’s record provides information about one of its variables; thus, the data in the table lists the result for each variable and for each sample.Of the variables included in Table \(\PageIndex{1}\), some are categorical and some are numerical. A categorical variable provides qualitative information that we can use to describe the samples relative to each other, or that we can use to organize the samples into groups (or categories). For the data in Table \(\PageIndex{1}\), bag id, type, and rank are categorical variables.A numerical variable provides quantitative information that we can use in a meaningful calculation; for example, we can use the number of yellow M&Ms and the total number of M&Ms to calculate a new variable that reports the percentage of M&Ms that are yellow. For the data in Table \(\PageIndex{1}\), year, weight (oz), number yellow, % red M&Ms, and total M&Ms are numerical variables.We can also use a numerical variable to assign samples to groups. For example, we can divide the plain M&Ms in Table \(\PageIndex{1}\) into two groups based on the sample’s weight. What makes a numerical variable more interesting, however, is that we can use it to make quantitative comparisons between samples; thus, we can report that there are \(14.4 \times\) as many plain M&Ms in a 10-oz. bag as there are in a 0.8-oz. bag.\[\frac{333 + 331}{24 + 22} = \frac{664}{46} = 14.4 \nonumber \]Although we could classify year as a categorical variable—not an unreasonable choice as it could serve as a useful way to group samples—we list it here as a numerical variable because it can serve as a useful predictive variable in a regression analysis. On the other hand rank is not a numerical variable—even if we rewrite the ranks as numerals—as there are no meaningful calculations we can complete using this variable.Categorical variables are described as nominal or ordinal. A nominal categorical variable does not imply a particular order; an ordinal categorical variable, on the other hand, coveys a meaningful sense of order. For the categorical variables in Table \(\PageIndex{1}\), bag id and type are nominal variables, and rank is an ordinal variable.A numerical variable is described as either ratio or interval depending on whether it has (ratio) or does not have (interval) an absolute reference. Although we can complete meaningful calculations using any numerical variable, the type of calculation we can perform depends on whether or not the variable’s values have an absolute reference.A numerical variable has an absolute reference if it has a meaningful zero—that is, a zero that means a measured quantity of none—against which we reference all other measurements of that variable. For the numerical variables in Table \(\PageIndex{1}\), weight (oz), number yellow, % red, and total M&Ms are ratio variables because each has a meaningful zero; year is an interval variable because its scale is referenced to an arbitrary point in time, 1 BCE, and not to the beginning of time.For a ratio variable, we can make meaningful absolute and relative comparisons between two results, but only meaningful absolute comparisons for an interval variable. For example, consider sample e, which was collected in 1994 and has 331 M&Ms, and sample d, which was collected in 2000 and has 24 M&Ms. We can report a meaningful absolute comparison for both variables: sample e is six years older than sample d and sample e has 307 more M&Ms than sample d. We also can report a meaningful relative comparison for the total number of M&Ms—there are\[\frac{331}{24} = 13.8 \times \nonumber \]as many M&Ms in sample e as in sample d—but we cannot report a meaningful relative comparison for year because a sample collected in 2000 is not\[\frac{2000}{1994} = 1.003 \times \nonumber \]older than a sample collected in 1994.Finally, the granularity of a numerical variable provides one more way to describe our data. For example, we can describe a numerical variable as discrete or continuous. A numerical variable is discrete if it can take on only specific values—typically, but not always, an integer value—between its limits; a continuous variable can take on any possible value within its limits. For the numerical data in Table \(\PageIndex{1}\), year, number yellow, and total M&Ms are discrete in that each is limited to integer values. The numerical variables weight (oz) and % red, on the other hand, are continuous variables. Note that weight is a continuous variable even if the device we use to measure weight yields discrete values.The old saying that "a picture is worth a 1000 words" may not be universally true, but it true when it comes to the analysis of data. A good visualization of data, for example, allows us to see patterns and relationships that are less evident when we look at data arranged in a table, and it provides a powerful way to tell our data's story. Suppose we want to study the composition of 1.69-oz (47.9-g) packages of plain M&Ms. We obtain 30 bags of M&Ms (ten from each of three stores) and remove the M&Ms from each bag one-by-one, recording the number of blue, brown, green, orange, red, and yellow M&Ms. We also record the number of yellow M&Ms in the first five candies drawn from each bag, and record the actual net weight of the M&Ms in each bag. Table \(\PageIndex{2}\) summarizes the data collected on these samples. The bag id identifies the order in which the bags were opened and analyzed.Having collected our data, we next examine it for possible problems, such as missing values (Did we forget to record the number of brown M&Ms in any of our samples?), for errors introduced when we recorded the data (Is the decimal point recorded incorrectly for any of the net weights?), or for unusual results (Is it really the case that this bag has only yellow M&M?). We also examine our data to identify interesting observations that we may wish to explore (It appears that most net weights are greater than the net weight listed on the individual packages. Why might this be? Is the difference significant?) When our data set is small we usually can identify possible problems and interesting observations without much difficulty; however, for a large data set, this becomes a challenge. Instead of trying to examine individual values, we can look at our results visually. While it may be difficult to find a single, odd data point when we have to individually review 1000 samples, it often jumps out when we look at the data using one or more of the approaches we will explore in this chapter.A dot plot displays data for one variable, with each sample’s value plotted on the x-axis. The individual points are organized along the y-axis with the first sample at the bottom and the last sample at the top. shows a dot plot for the number of brown M&Ms in the 30 bags of M&Ms from Table \(\PageIndex{2}\). The distribution of points appears random as there is no correlation between the sample id and the number of brown M&Ms. We would be surprised if we discovered that the points were arranged from the lower-left to the upper-right as this implies that the order in which we open the bags determines whether they have many or a few brown M&Ms.A dot plot provides a quick way to give us confidence that our data are free from unusual patterns, but at the cost of space because we use the y-axis to include the sample id as a variable. A stripchart uses the same x-axis as a dot plot, but does not use the y-axis to distinguish between samples. Because all samples with the same number of brown M&Ms will appear in the same place—making it impossible to distinguish them from each other—we stack the points vertically to spread them out, as shown in .Both the dot plot in and the stripchart in suggest that there is a smaller density of points at the lower limit and the upper limit of our results. We see, for example, that there is just one bag each with 8, 16, 18, 19, 20, and 21 brown M&Ms, but there are six bags each with 13 and 17 brown M&Ms.Because a stripchart does not use the y-axis to provide meaningful categorical information, we can easily display several stripcharts at once. shows this for the data in Table \(\PageIndex{2}\). Instead of stacking the individual points, we jitter them by applying a small, random offset to each point. Among the things we learn from this stripchart are that only brown and yellow M&Ms have counts of greater than 20 and that only blue and green M&Ms have counts of three or fewer M&Ms.The stripchart in is easy for us to examine because the number of samples, 30 bags, and the number of M&Ms per bag is sufficiently small that we can see the individual points. As the density of points becomes greater, a stripchart becomes less useful. A box and whisker plot provides a similar view but focuses on the data in terms of the range of values that encompass the middle 50% of the data. shows the box and whisker plot for brown M&Ms using the data in Table \(\PageIndex{2}\). The 30 individual samples are superimposed as a stripchart. The central box divides the x-axis into three regions: bags with fewer than 13 brown M&Ms (seven samples), bags with between 13 and 17 brown M&Ms (19 samples), and bags with more than 17 brown M&Ms (four samples). The box's limits are set so that it includes at least the middle 50% of our data. In this case, the box contains 19 of the 30 samples (63%) of the bags, because moving either end of the box toward the middle results in a box that includes less than 50% of the samples. The difference between the box's upper limit and its lower limit is called the interquartile range (IQR). The thick line in the box is the median, or middle value (more on this and the IQR in the next chapter). The dashed lines at either end of the box are called whiskers, and they extend to the largest or the smallest result that is within \(\pm 1.5 \times \text{IQR}\) of the box's right or left edge, respectively.Because a box and whisker plot does not use the y-axis to provide meaningful categorical information, we can easily display several plots in the same frame. shows this for the data in Table \(\PageIndex{2}\). Note that when a value falls outside of a whisker, as is the case here for yellow M&Ms, it is flagged by displaying it as an open circle.One use of a box and whisker plot is to examine the distribution of the individual samples, particularly with respect to symmetry. With the exception of the single sample that falls outside of the whiskers, the distribution of yellow M&Ms appears symmetrical: the median is near the center of the box and the whiskers extend equally in both directions. The distribution of the orange M&Ms is asymmetrical: half of the samples have 4–7 M&Ms (just four possible outcomes) and half have 7–15 M&Ms (nine possible outcomes), suggesting that the distribution is skewed toward higher numbers of orange M&Ms (see Chapter 5 for more information about the distribution of samples). shows box-and-whisker plots for yellow M&Ms grouped according to the store where the bags of M&Ms were purchased. Although the box and whisker plots are quite different in terms of the relative sizes of the boxes and the relative length of the whiskers, the dot plots suggest that the distribution of the underlying data is relatively similar in that most bags contain 12–18 yellow M&Ms and just a few bags deviate from these limits. These observations are reassuring because we do not expect the choice of store to affect the composition of bags of M&Ms. If we saw evidence that the choice of store affected our results, then we would look more closely at the bags themselves for evidence of a poorly controlled variable, such as type (Did we accidentally purchase bags of peanut butter M&Ms from one store?) or the product’s lot number (Did the manufacturer change the composition of colors between lots?).Although a dot plot, a stripchart and a box-and-whisker plot provide some qualitative evidence of how a variable’s values are distributed—we will have more to say about the distribution of data in Chapter 5—they are less useful when we need a more quantitative picture of the distribution. For this we can use a bar plot that displays a count of each discrete outcome. shows bar plots for orange and for yellow M&Ms using the data in Table \(\PageIndex{2}\).Here we see that the most common number of orange M&Ms per bag is four, which is also the smallest number of orange M&Ms per bag, and that there is a general decrease in the number of bags as the number of orange M&M per bag increases. For the yellow M&Ms, the most common number of M&Ms per bag is 16, which falls near the middle of the range of yellow M&Ms.A bar plot is a useful way to look at the distribution of discrete results, such as the counts of orange or yellow M&Ms, but it is not useful for continuous data where each result is unique. A histogram, in which we display the number of results that fall within a sequence of equally spaced bins, provides a view that is similar to that of a bar plot but that works with continuous data. , for example, shows a histogram for the net weights of the 30 bags of M&Ms in Table \(\PageIndex{2}\). Individual values are shown by the vertical hash marks at the bottom of the histogram.In the last section we used data collected from 30 bags of M&Ms to explore different ways to visualize data. In this section we consider several ways to summarize data using the net weights of the same bags of M&Ms. Here is the raw data.Without completing any calculations, what conclusions can we make by just looking at this data? Here are a few:Both visualizations provide a good qualitative picture of the data, suggesting that the individual results are scattered around some central value with more results closer to that central value that at distance from it. Neither visualization, however, describes the data quantitatively. What we need is a convenient way to summarize the data by reporting where the data is centered and how varied the individual results are around that center.There are two common ways to report the center of a data set: the mean and the median.The mean, \(\overline{Y}\), is the numerical average obtained by adding together the results for all n observations and dividing by the number of observations\[\overline{Y} = \frac{ \sum_{i = 1}^n Y_{i} } {n} = \frac{49.287 + 48.870 + \cdots + 48.317} {30} = 48.980 \text{ g} \nonumber \]The median, \(\widetilde{Y}\), is the middle value after we order our observations from smallest-to-largest, as we show here for our data.If we have an odd number of samples, then the median is simply the middle value, or\[\widetilde{Y} = Y_{\frac{n + 1}{2}} \nonumber \]where n is the number of samples. If, as is the case here, n is even, then\[\widetilde{Y} = \frac {Y_{\frac{n}{2}} + Y_{\frac{n}{2}+1}} {2} = \frac {48.692 + 48.777}{2} = 48.734 \text{ g} \nonumber \]When our data has a symmetrical distribution, as we believe is the case here, then the mean and the median will have similar values.There are five common measures of the variation of data about its center: the variance, the standard deviation, the range, the interquartile range, and the median average difference.The variance, s2, is an average squared deviation of the individual observations relative to the mean\[s^{2} = \frac { \sum_{i = 1}^n \big(Y_{i} - \overline{Y} \big)^{2} } {n - 1} = \frac { \big(49.287 - 48.980\big)^{2} + \cdots + \big(48.317 - 48.980\big)^{2} } {30 - 1} = 2.052 \nonumber \]and the standard deviation, s, is the square root of the variance, which gives it the same units as the mean.\[s = \sqrt{\frac { \sum_{i = 1}^n \big(Y_{i} - \overline{Y} \big)^{2} } {n - 1}} = \sqrt{\frac { \big(49.287 - 48.980\big)^{2} + \cdots + \big(48.317 - 48.980\big)^{2} } {30 - 1}} = 1.432 \nonumber \]The range, w, is the difference between the largest and the smallest value in our data set.\[w = 51.730 \text{ g} - 46.405 \text{ g} = 5.325 \text{ g} \nonumber \]The interquartile range, IQR, is the difference between the median of the bottom 25% of observations and the median of the top 25% of observations; that is, it provides a measure of the range of values that spans the middle 50% of observations. There is no single, standard formula for calculating the IQR, and different algorithms yield slightly different results. We will adopt the algorithm described here:1. Divide the sorted data set in half; if there is an odd number of values, then remove the median for the complete data set. For our data, the lower half isand the upper half is2. Find FL, the median for the lower half of the data, which for our data is 48.196 g.3. Find FU , the median for the upper half of the data, which for our data is 50.037 g.4. The IQR is the difference between FU and FL.\[F_{U} - F_{L} = 50.037 \text{ g} - 48.196 \text{ g} = 1.841 \text{ g} \nonumber \]The median absolute deviation, MAD, is the median of the absolute deviations of each observation from the median of all observations. To find the MAD for our set of 30 net weights, we first subtract the median from each sample in Table \(\PageIndex{3}\).Next we take the absolute value of each difference and sort them from smallest-to-largest.Finally, we report the median for these sorted values as\[\frac{0.7425 + 0.8935}{2} = 0.818 \nonumber \]A good question to ask is why we might desire more than one way to report the center of our data and the variation in our data about the center. Suppose that the result for the last of our 30 samples was reported as 483.17 instead of 48.317. Whether this is an accidental shifting of the decimal point or a true result is not relevant to us here; what matters is its effect on what we report. Here is a summary of the effect of this one value on each of our ways of summarizing our data.Note that the mean, the variance, the standard deviation, and the range are very sensitive to the change in the last result, but the median, the IQR, and the MAD are not. The median, the IQR, and the MAD are considered robust statistics because they are less sensitive to an unusual result; the others are, of course, non-robust statistics. Both types of statistics have value to us, a point we will return to from time-to-time.When we measure something, such as the percentage of yellow M&Ms in a bag of M&Ms, we expect two things:Visualizations of data—such as dot plots, stripcharts, boxplot-and-whisker plots, bar plots, histograms, and scatterplots—often suggest there is an underlying structure to our data. For example, we have seen that the distribution of yellow M&Ms in bags of M&Ms is more or less symmetrical around its median, while the distribution of orange M&Ms was skewed toward higher values. This underlying structure, or distribution, of our data as it effects how we choose to analyze our data. In this chapter we will take a closer look at several ways in which data are distributed.Before we consider different types of distributions, let's define some key terms. You may wish, as well, to review the discussion of different types of data in Chapter 2.A population includes every possible measurement we could make on a system, while a sample is the subset of a population on which we actually make measurements. These definitions are fluid. A single bag of M&Ms is a population if we are interested only in that specific bag, but it is but one sample from a box that contains a gross of individual bags. That box, itself, can be a population, or it can be one sample from a much larger production lot. And so on.In a discrete distribution the possible results take on a limited set of specific values that are independent of how we make our measurements. When we determine the number of yellow M&Ms in a bag, the results are limited to integer values. We may find 13 yellow M&Ms or 24 yellow M&Ms, but we cannot obtain a result of 15.43 yellow M&Ms.For a continuous distribution the result of a measurement can take on any possible value between a lower limit and an upper limit, even though our measuring device has a limited precision; thus, when we weigh a bag of M&Ms on a three-digit balance and obtain a result of 49.287 g we know that its true mass is greater than 49.2865... g and less than 49.2875... g.There are four important types of distributions that we will consider in this chapter: the uniform distribution, the binomial distribution, the Poisson distribution, and the normal, or Gaussian, distribution. In the previous sections we used the analysis of bags of M&Ms to explore ways to visualize data and to summarize data. Here we will use the same data set to explore the distribution of data.In a uniform distribution, all outcomes are equally probable. Suppose the population of M&Ms has a uniform distribution. If this is the case, then, with six colors, we expect each color to appear with a probability of 1/6 or 16.7%. shows a comparison of the theoretical results if we draw 1699 M&Ms—the total number of M&Ms in our sample of 30 bags—from a population with a uniform distribution (on the left) to the actual distribution of the 1699 M&Ms in our sample (on the right). It seems unlikely that the population of M&Ms has a uniform distribution of colors!A binomial distribution shows the probability of obtaining a particular result in a fixed number of trials, where the odds of that result happening in a single trial are known. Mathematically, a binomial distribution is defined by the equation\[P(X, N) = \frac {N!} {X! (N - X)!} \times p^{X} \times (1 - p)^{N - X} \nonumber \]where P(X,N) is the probability that the event happens X times in N trials, and where p is the probability that the event happens in a single trial. The binomial distribution has a theoretical mean, \(\mu\), and a theoretical variance, \(\sigma^2\), of\[\mu = Np \quad \quad \quad \sigma^2 = Np(1 - p) \nonumber \] compares the expected binomial distribution for drawing 0, 1, 2, 3, 4, or 5 yellow M&Ms in the first five M&Ms—assuming that the probability of drawing a yellow M&M is 435/1699, the ratio of the number of yellow M&Ms and the total number of M&Ms—to the actual distribution of results. The similarity between the theoretical and the actual results seems evident; in a later section we will consider ways to test this claim.The binomial distribution is useful if we wish to model the probability of finding a fixed number of yellow M&Ms in a sample of M&Ms of fixed size—such as the first five M&Ms that we draw from a bag—but not the probability of finding a fixed number of yellow M&Ms in a single bag because there is some variability in the total number of M&Ms per bag.A Poisson distribution gives the probability that a given number of events will occur in a fixed interval in time or space if the event has a known average rate and if each new event is independent of the preceding event. Mathematically a Poisson distribution is defined by the equation\[P(X, \lambda) = \frac {e^{-\lambda} \lambda^X} {X !} \nonumber \]where \(P(X, \lambda)\) is the probability that an event happens X times given the event’s average rate, \(\lambda\). The Poisson distribution has a theoretical mean, \(\mu\), and a theoretical variance, \(\sigma^2\), that are each equal to \(\lambda\).The bar plot in shows the actual distribution of green M&Ms in 35 small bags of M&Ms (as reported by M. A. Xu-Friedman “Illustrating concepts of quantal analysis with an intuitive classroom model,” Adv. Physiol. Educ. 2013, 37, 112–116). Superimposed on the bar plot is the theoretical Poisson distribution based on their reported average rate of 3.4 green M&Ms per bag. The similarity between the theoretical and the actual results seems evident; in Chapter 6 we will consider ways to test this claim.A uniform distribution, a binomial distribution, and a Poisson distribution predict the probability of a discrete event, such as the probability of finding exactly two green M&Ms in the next bag of M&Ms that we open. Not all of the data we collect is discrete. The net weights of bags of M&Ms is an example of continuous data as the mass of an individual bag is not restricted to a discrete set of allowed values. In many cases we can model continuous data using a normal (or Gaussian) distribution, which gives the probability of obtaining a particular outcome, P(x), from a population with a known mean, \(\mu\), and a known variance, \(\sigma^2\). Mathematically a normal distribution is defined by the equation\[P(x) = \frac {1} {\sqrt{2 \pi \sigma^2}} e^{-(x - \mu)^2/(2 \sigma^2)} \nonumber \] shows the expected normal distribution for the net weights of our sample of 30 bags of M&Ms if we assume that their mean, \(\overline{X}\), of 48.98 g and standard deviation, s, of 1.433 g are good predictors of the population’s mean, \(\mu\), and standard deviation, \(\sigma\). Given the small sample of 30 bags, the agreement between the model and the data seems reasonable.Suppose we have a population for which one of its properties has a uniform distribution where every result between 0 and 1 is equally probable. If we analyze 10,000 samples we should not be surprised to find that the distribution of these 10000 results looks uniform, as shown by the histogram on the left side of . If we collect 1000 pooled samples—each of which consists of 10 individual samples for a total of 10,000 individual samples—and report the average results for these 1000 pooled samples, we see something interesting as their distribution, as shown by the histogram on the right, looks remarkably like a normal distribution. When we draw single samples from a uniform distribution, each possible outcome is equally likely, which is why we see the distribution on the left. When we draw a pooled sample that consists of 10 individual samples, however, the average values are more likely to be near the middle of the distribution’s range, as we see on the right, because the pooled sample likely includes values drawn from both the lower half and the upper half of the uniform distribution.This tendency for a normal distribution to emerge when we pool samples is known as the central limit theorem. As shown in , we see a similar effect with populations that follow a binomial distribution or a Poisson distribution.You might reasonably ask whether the central limit theorem is important as it is unlikely that we will complete 1000 analyses, each of which is the average of 10 individual trials. This is deceiving. When we acquire a sample of soil, for example, it consists of many individual particles each of which is an individual sample of the soil. Our analysis of this sample, therefore, is the mean for a large number of individual soil particles. Because of this, the central limit theorem is relevant.In the last section we examined four ways in which the individual samples we collect and analyze are distributed about a central value: a uniform distribution, a binomial distribution, a Poisson distribution, and a normal distribution. We also learned that regardless of how individual samples are distributed, the distribution of averages for multiple samples often follows a normal distribution. This tendency for a normal distribution to emerge when we report averages for multiple samples is known as the central limit theorem. In this chapter we look more closely at the normal distribution—examining some of its properties—and consider how we can use these properties to say something more meaningful about our data than simply reporting a mean and a standard deviation.Mathematically a normal distribution is defined by the equation\[P(x) = \frac {1} {\sqrt{2 \pi \sigma^2}} e^{-(x - \mu)^2/(2 \sigma^2)} \nonumber \]where \(P(x)\) is the probability of obtaining a result, \(x\), from a population with a known mean, \(\mu\), and a known standard deviation, \(\sigma\). shows the normal distribution curves for \(\mu = 0\) with standard deviations of 5, 10, and 20.Because the equation for a normal distribution depends solely on the population’s mean, \(\mu\), and its standard deviation, \(\sigma\), the probability that a sample drawn from a population has a value between any two arbitrary limits is the same for all populations. For example, shows that 68.26% of all samples drawn from a normally distributed population have values within the range \(\mu \pm 1\sigma\), and only 0.14% have values greater than \(\mu + 3\sigma\).This feature of a normal distribution—that the area under the curve is the same for all values of \(\sigma\)—allows us to create a probability table (see Appendix 2) based on the relative deviation, \(z\), between a limit, x, and the mean, \(\mu\).\[z = \frac {x - \mu} {\sigma} \nonumber \]The value of \(z\) gives the area under the curve between that limit and the distribution’s closest tail, as shown in .Suppose we know that \(\mu\) is 5.5833 ppb Pb and that \(\sigma\) is 0.0558 ppb Pb for a particular standard reference material (SRM). What is the probability that we will obtain a result that is greater than 5.650 ppb if we analyze a single, random sample drawn from the SRM? shows the normal distribution curve given values of 5.5833 ppb Pb for \(\mu\) and of 0.0558 ppb Pb \(\sigma\). The shaded area in the figures is the probability of obtaining a sample with a concentration of Pb greater than 5.650 ppm. To determine the probability, we first calculate \(z\)\[z = \frac {x - \mu} {\sigma} = \frac {5.650 - 5.5833} {0.0558} = 1.195 \nonumber \]Next, we look up the probability in Appendix 2 for this value of \(z\), which is the average of 0.1170 (for \(z = 1.19\)) and 0.1151 (for \(z = 1.20\)), or a probability of 0.1160; thus, we expect that 11.60% of samples will provide a result greater than 5.650 ppb Pb.Example \(\PageIndex{1}\) considers a single limit—the probability that a result exceeds a single value. But what if we want to determine the probability that a sample has between 5.580 g Pb and 5.625 g Pb?In this case we are interested in the shaded area shown in . First, we calculate \(z\) for the upper limit\[z = \frac {5.625 - 5.5833} {0.0558} = 0.747 \nonumber \]and then we calculate \(z\) for the lower limit\[z = \frac {5.580 - 5.5833} {0.0558} = -0.059 \nonumber \]Then, we look up the probability in Appendix 2 that a result will exceed our upper limit of 5.625, which is 0.2275, or 22.75%, and the probability that a result will be less than our lower limit of 5.580, which is 0.4765, or 47.65%. The total unshaded area is 71.4% of the total area, so the shaded area corresponds to a probability of\[100.00 - 22.75 - 47.65 = 100.00 - 71.40 = 29.6 \% \nonumber \]In the previous section, we learned how to predict the probability of obtaining a particular outcome if our data are normally distributed with a known \(\mu\) and a known \(\sigma\). For example, we estimated that 11.60% of samples drawn at random from a standard reference material will have a concentration of Pb greater than 5.650 ppb given a \(\mu\) of 5.5833 ppb and a \(\sigma\) of 0.0558 ppb. In essence, we determined how many standard deviations 5.650 is from \(\mu\) and used this to define the probability given the standard area under a normal distribution curve.We can look at this in a different way by asking the following question: If we collect a single sample at random from a population with a known \(\mu\) and a known \(\sigma\), within what range of values might we reasonably expect to find the sample’s result 95% of the time? Rearranging the equation\[z = \frac {x - \mu} {\sigma} \nonumber \]and solving for \(x\) gives\[x = \mu \pm z \sigma = 5.5833 \pm (1.96)(0.0558) = 5.5833 \pm 0.1094 \nonumber \]where a \(z\) of 1.96 corresponds to 95% of the area under the curve; we call this a 95% confidence interval for a single sample.It generally is a poor idea to draw a conclusion from the result of a single experiment; instead, we usually collect several samples and ask the question this way: If we collect \(n\) random samples from a population with a known \(\mu\) and a known \(\sigma\), within what range of values might we reasonably expect to find the mean of these samples 95% of the time?We might reasonably expect that the standard deviation for the mean of several samples is smaller than the standard deviation for a set of individual samples; indeed it is and it is given as\[\sigma_{\bar{x}} = \frac {\sigma} {\sqrt{n}} \nonumber \]where \(\frac {\sigma} {\sqrt{n}}\) is called the standard error of the mean. For example, if we collect three samples from the standard reference material described above, then we expect that the mean for these three samples will fall within a range\[\bar{x} = \mu \pm z \sigma_{\bar{X}} = \mu \pm \frac {z \sigma} {\sqrt{n}} = 5.5833 \pm \frac{(1.96)(0.0558)} {\sqrt{3}} = 5.5833 \pm 0.0631 \nonumber \]that is \(\pm 0.0631\) ppb around \(\mu\), a range that is smaller than that of \(\pm 0.1094\) ppb when we analyze individual samples. Note that the relative value to us of increasing the sample’s size diminishes as \(n\) increases because of the square root term, as shown in .Our treatment thus far assumes we know \(\mu\) and \(\sigma\) for the parent population, but we rarely know these values; instead, we examine samples drawn from the parent population and ask the following question: Given the sample’s mean, \(\bar{x}\), and its standard deviation, \(s\), what is our best estimate of the population’s mean, \(\mu\), and its standard deviation, \(\sigma\).To make this estimate, we replace the population’s standard deviation, \(\sigma\), with the standard deviation, \(s\), for our samples, replace the population’s mean, \(\mu\), with the mean, \(\bar{x}\), for our samples, replace \(z\) with \(t\), where the value of \(t\) depends on the number of samples, \(n\)\[\bar{x} = \mu \pm \frac{ts}{\sqrt{n}} \nonumber \]and then rearrange the equation to solve for \(\mu\).\[\mu = \bar{x} \pm \frac {ts} {\sqrt{n}} \nonumber \]We call this a confidence interval. Values for \(t\) are available in tables (see Appendix 3) and depend on the probability level, \(\alpha\), where \((1 − \alpha) \times 100\) is the confidence level, and the degrees of freedom, \(n − 1\); note that for any probability level, \(t \longrightarrow z\) as \(n \longrightarrow \infty\).We need to give special attention to what this confidence interval means and to what it does not mean:A confidence interval is a useful way to report the result of an analysis because it sets limits on the expected result. In the absence of determinate error, or bias, a confidence interval based on a sample’s mean indicates the range of values in which we expect to find the population’s mean. When we report a 95% confidence interval for the mass of a penny as 3.117 g ± 0.047 g, for example, we are stating that there is only a 5% probability that the penny’s expected mass is less than 3.070 g or more than 3.164 g.Because a confidence interval is a statement of probability, it allows us to consider comparative questions, such as these:“Are the results for a newly developed method to determine cholesterol in blood significantly different from those obtained using a standard method?”“Is there a significant variation in the composition of rainwater collected at different sites downwind from a coal-burning utility plant?”In this chapter we introduce a general approach that uses experimental data to ask and answer such questions, an approach we call significance testing.The reliability of significance testing recently has received much attention—see Nuzzo, R. “Scientific Method: Statistical Errors,” Nature, 2014, 506, 150–152 for a general discussion of the issues—so it is appropriate to begin this chapter by noting the need to ensure that our data and our research question are compatible so that we do not read more into a statistical analysis than our data allows; see Leek, J. T.; Peng, R. D. “What is the Question? Science, 2015, 347, 1314-1315 for a useful discussion of six common research questions.In the context of analytical chemistry, significance testing often accompanies an exploratory data analysis"Is there a reason to suspect that there is a difference between these two analytical methods when applied to a common sample?"or an inferential data analysis."Is there a reason to suspect that there is a relationship between these two independent measurements?"A statistically significant result for these types of analytical research questions generally leads to the design of additional experiments that are better suited to making predictions or to explaining an underlying causal relationship. A significance test is the first step toward building a greater understanding of an analytical problem, not the final answer to that problem!Let’s consider the following problem. To determine if a medication is effective in lowering blood glucose concentrations, we collect two sets of blood samples from a patient. We collect one set of samples immediately before we administer the medication, and we collect the second set of samples several hours later. After we analyze the samples, we report their respective means and variances. How do we decide if the medication was successful in lowering the patient’s concentration of blood glucose?One way to answer this question is to construct a normal distribution curve for each sample, and to compare the two curves to each other. Three possible outcomes are shown in . In , there is a complete separation of the two normal distribution curves, which suggests the two samples are significantly different from each other. In , the normal distribution curves for the two samples almost completely overlap each other, which suggests the difference between the samples is insignificant. , however, presents us with a dilemma. Although the means for the two samples seem different, the overlap of their normal distribution curves suggests that a significant number of possible outcomes could belong to either distribution. In this case the best we can do is to make a statement about the probability that the samples are significantly different from each other.The process by which we determine the probability that there is a significant difference between two samples is called significance testing or hypothesis testing. Before we discuss specific examples let's first establish a general approach to conducting and interpreting a significance test.The purpose of a significance test is to determine whether the difference between two or more results is sufficiently large that we are comfortable stating that the difference cannot be explained by indeterminate errors. The first step in constructing a significance test is to state the problem as a yes or no question, such as“Is this medication effective at lowering a patient’s blood glucose levels?”A null hypothesis and an alternative hypothesis define the two possible answers to our yes or no question. The null hypothesis, H0, is that indeterminate errors are sufficient to explain any differences between our results. The alternative hypothesis, HA, is that the differences in our results are too great to be explained by random error and that they must be determinate in nature. We test the null hypothesis, which we either retain or reject. If we reject the null hypothesis, then we must accept the alternative hypothesis and conclude that the difference is significant.Failing to reject a null hypothesis is not the same as accepting it. We retain a null hypothesis because we have insufficient evidence to prove it incorrect. It is impossible to prove that a null hypothesis is true. This is an important point and one that is easy to forget. To appreciate this point let’s use this data for the mass of 100 circulating United States pennies.After looking at the data we might propose the following null and alternative hypotheses.H0: The mass of a circulating U.S. penny is between 2.900 g and 3.200 gHA: The mass of a circulating U.S. penny may be less than 2.900 g or more than 3.200 gTo test the null hypothesis we find a penny and determine its mass. If the penny’s mass is 2.512 g then we can reject the null hypothesis and accept the alternative hypothesis. Suppose that the penny’s mass is 3.162 g. Although this result increases our confidence in the null hypothesis, it does not prove that the null hypothesis is correct because the next penny we sample might weigh less than 2.900 g or more than 3.200 g.After we state the null and the alternative hypotheses, the second step is to choose a confidence level for the analysis. The confidence level defines the probability that we will incorrectly reject the null hypothesis when it is, in fact, true. We can express this as our confidence that we are correct in rejecting the null hypothesis (e.g. 95%), or as the probability that we are incorrect in rejecting the null hypothesis. For the latter, the confidence level is given as \(\alpha\), where\[\alpha = 1 - \frac {\text{confidence interval (%)}} {100} \nonumber \]For a 95% confidence level, \(\alpha\) is 0.05.The third step is to calculate an appropriate test statistic and to compare it to a critical value. The test statistic’s critical value defines a breakpoint between values that lead us to reject or to retain the null hypothesis, which is the fourth, and final, step of a significance test. As we will see in the sections that follow, how we calculate the test statistic depends on what we are comparing.The four steps for a statistical analysis of data using a significance test:Suppose we want to evaluate the accuracy of a new analytical method. We might use the method to analyze a Standard Reference Material that contains a known concentration of analyte, \(\mu\). We analyze the standard several times, obtaining a mean value, \(\overline{X}\), for the analyte’s concentration. Our null hypothesis is that there is no difference between \(\overline{X}\) and \(\mu\)\[H_0 \text{: } \overline{X} = \mu \nonumber \]If we conduct the significance test at \(\alpha = 0.05\), then we retain the null hypothesis if a 95% confidence interval around \(\overline{X}\) contains \(\mu\). If the alternative hypothesis is\[H_\text{A} \text{: } \overline{X} \neq \mu \nonumber \]then we reject the null hypothesis and accept the alternative hypothesis if \(\mu\) lies in the shaded areas at either end of the sample’s probability distribution curve ). Each of the shaded areas accounts for 2.5% of the area under the probability distribution curve, for a total of 5%. This is a two-tailed significance test because we reject the null hypothesis for values of \(\mu\) at either extreme of the sample’s probability distribution curve.We can write the alternative hypothesis in two additional ways\[H_\text{A} \text{: } \overline{X} > \mu \nonumber \]\[H_\text{A} \text{: } \overline{X} < \mu \nonumber \]rejecting the null hypothesis if \(\mu\) falls within the shaded areas shown in or , respectively. In each case the shaded area represents 5% of the area under the probability distribution curve. These are examples of a one-tailed significance test.For a fixed confidence level, a two-tailed significance test is the more conservative test because rejecting the null hypothesis requires a larger difference between the results we are comparing. In most situations we have no particular reason to expect that one result must be larger (or must be smaller) than the other result. This is the case, for example, when we evaluate the accuracy of a new analytical method. A two-tailed significance test, therefore, usually is the appropriate choice.We reserve a one-tailed significance test for a situation where we specifically are interested in whether one result is larger (or smaller) than the other result. For example, a one-tailed significance test is appropriate if we are evaluating a medication’s ability to lower blood glucose levels. In this case we are interested only in whether the glucose levels after we administer the medication are less than the glucose levels before we initiated treatment. If a patient’s blood glucose level is greater after we administer the medication, then we know the answer—the medication did not work—and we do not need to conduct a statistical analysis.Because a significance test relies on probability, its interpretation is subject to error. In a significance test, \(\alpha\) defines the probability of rejecting a null hypothesis that is true. When we conduct a significance test at \(\alpha = 0.05\), there is a 5% probability that we will incorrectly reject the null hypothesis. This is known as a type 1 error, and its risk is always equivalent to \(\alpha\). A type 1 error in a two-tailed or a one-tailed significance tests corresponds to the shaded areas under the probability distribution curves in .A second type of error occurs when we retain a null hypothesis even though it is false. This is a type 2 error, and the probability of its occurrence is \(\beta\). Unfortunately, in most cases we cannot calculate or estimate the value for \(\beta\). The probability of a type 2 error, however, is inversely proportional to the probability of a type 1 error.Minimizing a type 1 error by decreasing \(\alpha\) increases the likelihood of a type 2 error. When we choose a value for \(\alpha\) we must compromise between these two types of error. Most of the examples in this text use a 95% confidence level (\(\alpha = 0.05\)) because this usually is a reasonable compromise between type 1 and type 2 errors for analytical work. It is not unusual, however, to use a more stringent (e.g. \(\alpha = 0.01\)) or a more lenient (e.g. \(\alpha = 0.10\)) confidence level when the situation calls for it.A normal distribution is the most common distribution for the data we collect. Because the area between any two limits of a normal distribution curve is well defined, it is straightforward to construct and evaluate significance tests.One way to validate a new analytical method is to analyze a sample that contains a known amount of analyte, \(\mu\). To judge the method’s accuracy we analyze several portions of the sample, determine the average amount of analyte in the sample, \(\overline{X}\), and use a significance test to compare \(\overline{X}\) to \(\mu\). The null hypothesis is that the difference between \(\overline{X}\) and \(\mu\) is explained by indeterminate errors that affect our determination of \(\overline{X}\). The alternative hypothesis is that the difference between \(\overline{X}\) and \(\mu\) is too large to be explained by indeterminate error.\[H_0 \text{: } \overline{X} = \mu \nonumber \]\[H_A \text{: } \overline{X} \neq \mu \nonumber \]The test statistic is texp, which we substitute into the confidence interval for \(\mu\)\[\mu = \overline{X} \pm \frac {t_\text{exp} s} {\sqrt{n}} \nonumber \]Rearranging this equation and solving for \(t_\text{exp}\)\[t_\text{exp} = \frac {|\mu - \overline{X}| \sqrt{n}} {s} \nonumber \]gives the value for \(t_\text{exp}\) when \(\mu\) is at either the right edge or the left edge of the sample's confidence interval ).To determine if we should retain or reject the null hypothesis, we compare the value of texp to a critical value, \(t(\alpha, \nu)\), where \(\alpha\) is the confidence level and \(\nu\) is the degrees of freedom for the sample. The critical value \(t(\alpha, \nu)\) defines the largest confidence interval explained by indeterminate error. If \(t_\text{exp} > t(\alpha, \nu)\), then our sample’s confidence interval is greater than that explained by indeterminate errors b). In this case, we reject the null hypothesis and accept the alternative hypothesis. If \(t_\text{exp} \leq t(\alpha, \nu)\), then our sample’s confidence interval is smaller than that explained by indeterminate error, and we retain the null hypothesis c). Example \(\PageIndex{24}\) provides a typical application of this significance test, which is known as a t-test of \(\overline{X}\) to \(\mu\). You will find values for \(t(\alpha, \nu)\) in Appendix 3.Before determining the amount of Na2CO3 in a sample, you decide to check your procedure by analyzing a standard sample that is 98.76% w/w Na2CO3. Five replicate determinations of the %w/w Na2CO3 in the standard gives the following results\(98.71 \% \quad 98.59 \% \quad 98.62 \% \quad 98.44 \% \quad 98.58 \%\)Using \(\alpha = 0.05\), is there any evidence that the analysis is giving inaccurate results?The mean and standard deviation for the five trials are\[\overline{X} = 98.59 \quad \quad \quad s = 0.0973 \nonumber \]Because there is no reason to believe that the results for the standard must be larger or smaller than \(\mu\), a two-tailed t-test is appropriate. The null hypothesis and alternative hypothesis are\[H_0 \text{: } \overline{X} = \mu \quad \quad \quad H_\text{A} \text{: } \overline{X} \neq \mu \nonumber \]The test statistic, texp, is\[t_\text{exp} = \frac {|\mu - \overline{X}|\sqrt{n}} {2} = \frac {|98.76 - 98.59| \sqrt{5}} {0.0973} = 3.91 \nonumber \]The critical value for t(0.05, 4) from Appendix 3 is 2.78. Since texp is greater than t(0.05, 4), we reject the null hypothesis and accept the alternative hypothesis. At the 95% confidence level the difference between \(\overline{X}\) and \(\mu\) is too large to be explained by indeterminate sources of error, which suggests there is a determinate source of error that affects the analysis.There is another way to interpret the result of this t-test. Knowing that texp is 3.91 and that there are 4 degrees of freedom, we use Appendix 3 to estimate the value of \(\alpha\) that corresponds to a t(\(\alpha\), 4) of 3.91. From Appendix 3, t(0.02, 4) is 3.75 and t(0.01, 4) is 4.60. Although we can reject the null hypothesis at the 98% confidence level, we cannot reject it at the 99% confidence level. For a discussion of the advantages of this approach, see J. A. C. Sterne and G. D. Smith “Sifting the evidence—what’s wrong with significance tests?” BMJ 2001, 322, 226–231.Earlier we made the point that we must exercise caution when we interpret the result of a statistical analysis. We will keep returning to this point because it is an important one. Having determined that a result is inaccurate, as we did in Example \(\PageIndex{3}\), the next step is to identify and to correct the error. Before we expend time and money on this, however, we first should critically examine our data. For example, the smaller the value of s, the larger the value of texp. If the standard deviation for our analysis is unrealistically small, then the probability of a type 2 error increases. Including a few additional replicate analyses of the standard and reevaluating the t-test may strengthen our evidence for a determinate error, or it may show us that there is no evidence for a determinate error.If we analyze regularly a particular sample, we may be able to establish an expected variance, \(\sigma^2\), for the analysis. This often is the case, for example, in a clinical lab that analyzes hundreds of blood samples each day. A few replicate analyses of a single sample gives a sample variance, s2, whose value may or may not differ significantly from \(\sigma^2\).We can use an F-test to evaluate whether a difference between s2 and \(\sigma^2\) is significant. The null hypothesis is \(H_0 \text{: } s^2 = \sigma^2\) and the alternative hypothesis is \(H_\text{A} \text{: } s^2 \neq \sigma^2\). The test statistic for evaluating the null hypothesis is Fexp, which is given as either\[F_\text{exp} = \frac {s^2} {\sigma^2} \text{ if } s^2 > \sigma^2 \text{ or } F_\text{exp} = \frac {\sigma^2} {s^2} \text{ if } \sigma^2 > s^2 \nonumber \]depending on whether s2 is larger or smaller than \(\sigma^2\). This way of defining Fexp ensures that its value is always greater than or equal to one.If the null hypothesis is true, then Fexp should equal one; however, because of indeterminate errors, Fexp, usually is greater than one. A critical value, \(F(\alpha, \nu_\text{num}, \nu_\text{den})\), is the largest value of Fexp that we can attribute to indeterminate error given the specified significance level, \(\alpha\), and the degrees of freedom for the variance in the numerator, \(\nu_\text{num}\), and the variance in the denominator, \(\nu_\text{den}\). The degrees of freedom for s2 is n – 1, where n is the number of replicates used to determine the sample’s variance, and the degrees of freedom for \(\sigma^2\) is defined as infinity, \(\infty\). Critical values of F for \(\alpha = 0.05\) are listed in Appendix 4 for both one-tailed and two-tailed F-tests.A manufacturer’s process for analyzing aspirin tablets has a known variance of 25. A sample of 10 aspirin tablets is selected and analyzed for the amount of aspirin, yielding the following results in mg aspirin/tablet.\(254 \quad 249 \quad 252 \quad 252 \quad 249 \quad 249 \quad 250 \quad 247 \quad 251 \quad 252\)Determine whether there is evidence of a significant difference between the sample’s variance and the expected variance at \(\alpha = 0.05\).The variance for the sample of 10 tablets is 4.3. The null hypothesis and alternative hypotheses are\[H_0 \text{: } s^2 = \sigma^2 \quad \quad \quad H_\text{A} \text{: } s^2 \neq \sigma^2 \nonumber \]and the value for Fexp is\[F_\text{exp} = \frac {\sigma^2} {s^2} = \frac {25} {4.3} = 5.8 \nonumber \]The critical value for F(0.05, \(\infty\), 9) from Appendix 4 is 3.333. Since Fexp is greater than F(0.05, \(\infty\), 9), we reject the null hypothesis and accept the alternative hypothesis that there is a significant difference between the sample’s variance and the expected variance. One explanation for the difference might be that the aspirin tablets were not selected randomly.We can extend the F-test to compare the variances for two samples, A and B, by rewriting our equation for Fexp as\[F_\text{exp} = \frac {s_A^2} {s_B^2} \nonumber \]defining A and B so that the value of Fexp is greater than or equal to 1.The table below shows results for two experiments to determine the mass of a circulating U.S. penny. Determine whether there is a difference in the variances of these analyses at \(\alpha = 0.05\).The standard deviations for the two experiments are 0.051 for the first experiment (A) and 0.037 for the second experiment (B). The null and alternative hypotheses are\[H_0 \text{: } s_A^2 = s_B^2 \quad \quad \quad H_\text{A} \text{: } s_A^2 \neq s_B^2 \nonumber \]and the value of Fexp is\[F_\text{exp} = \frac {s_A^2} {s_B^2} = \frac {(0.051)^2} {(0.037)^2} = \frac {0.00260} {0.00137} = 1.90 \nonumber \]From Appendix 4 the critical value for F(0.05, 6, 4) is 9.197. Because Fexp < F(0.05, 6, 4), we retain the null hypothesis. There is no evidence at \(\alpha = 0.05\) to suggest that the difference in variances is significant.Three factors influence the result of an analysis: the method, the sample, and the analyst. We can study the influence of these factors by conducting experiments in which we change one factor while holding constant the other factors. For example, to compare two analytical methods we can have the same analyst apply each method to the same sample and then examine the resulting means. In a similar fashion, we can design experiments to compare two analysts or to compare two samples.Before we consider the significance tests for comparing the means of two samples, we need to understand the difference between unpaired data and paired data. This is a critical distinction and learning to distinguish between these two types of data is important. Here are two simple examples that highlight the difference between unpaired data and paired data. In each example the goal is to compare two balances by weighing pennies.In both examples the samples of 10 pennies were drawn from the same population; the difference is how we sampled that population. We will learn why this distinction is important when we review the significance test for paired data; first, however, we present the significance test for unpaired data.One simple test for determining whether data are paired or unpaired is to look at the size of each sample. If the samples are of different size, then the data must be unpaired. The converse is not true. If two samples are of equal size, they may be paired or unpaired.Consider two analyses, A and B, with means of \(\overline{X}_A\) and \(\overline{X}_B\), and standard deviations of sA and sB. The confidence intervals for \(\mu_A\) and for \(\mu_B\) are\[\mu_A = \overline{X}_A \pm \frac {t s_A} {\sqrt{n_A}} \nonumber \]\[\mu_B = \overline{X}_B \pm \frac {t s_B} {\sqrt{n_B}} \nonumber \]where nA and nB are the sample sizes for A and for B. Our null hypothesis, \(H_0 \text{: } \mu_A = \mu_B\), is that any difference between \(\mu_A\) and \(\mu_B\) is the result of indeterminate errors that affect the analyses. The alternative hypothesis, \(H_A \text{: } \mu_A \neq \mu_B\), is that the difference between \(\mu_A\)and \(\mu_B\) is too large to be explained by indeterminate error.To derive an equation for texp, we assume that \(\mu_A\) equals \(\mu_B\), and combine the equations for the two confidence intervals\[\overline{X}_A \pm \frac {t_\text{exp} s_A} {\sqrt{n_A}} = \overline{X}_B \pm \frac {t_\text{exp} s_B} {\sqrt{n_B}} \nonumber \]Solving for \(|\overline{X}_A - \overline{X}_B|\) and using a propagation of uncertainty, gives\[|\overline{X}_A - \overline{X}_B| = t_\text{exp} \times \sqrt{\frac {s_A^2} {n_A} + \frac {s_B^2} {n_B}} \nonumber \]Finally, we solve for texp\[t_\text{exp} = \frac {|\overline{X}_A - \overline{X}_B|} {\sqrt{\frac {s_A^2} {n_A} + \frac {s_B^2} {n_B}}} \nonumber \]and compare it to a critical value, \(t(\alpha, \nu)\), where \(\alpha\) is the probability of a type 1 error, and \(\nu\) is the degrees of freedom.Thus far our development of this t-test is similar to that for comparing \(\overline{X}\) to \(\mu\), and yet we do not have enough information to evaluate the t-test. Do you see the problem? With two independent sets of data it is unclear how many degrees of freedom we have.Suppose that the variances \(s_A^2\) and \(s_B^2\) provide estimates of the same \(\sigma^2\). In this case we can replace \(s_A^2\) and \(s_B^2\) with a pooled variance, \(s_\text{pool}^2\), that is a better estimate for the variance. Thus, our equation for \(t_\text{exp}\) becomes\[t_\text{exp} = \frac {|\overline{X}_A - \overline{X}_B|} {s_\text{pool} \times \sqrt{\frac {1} {n_A} + \frac {1} {n_B}}} = \frac {|\overline{X}_A - \overline{X}_B|} {s_\text{pool}} \times \sqrt{\frac {n_A n_B} {n_A + n_B}} \nonumber \]where spool, the pooled standard deviation, is\[s_\text{pool} = \sqrt{\frac {(n_A - 1) s_A^2 + (n_B - 1)s_B^2} {n_A + n_B - 2}} \nonumber \]The denominator of this equation shows us that the degrees of freedom for a pooled standard deviation is \(n_A + n_B - 2\), which also is the degrees of freedom for the t-test. Note that we lose two degrees of freedom because the calculations for \(s_A^2\) and \(s_B^2\) require the prior calculation of \(\overline{X}_A\) amd \(\overline{X}_B\).So how do you determine if it is okay to pool the variances? Use an F-test.If \(s_A^2\) and \(s_B^2\) are significantly different, then we calculate texp using the following equation. In this case, we find the degrees of freedom using the following imposing equation.\[\nu = \frac {\left( \frac {s_A^2} {n_A} + \frac {s_B^2} {n_B} \right)^2} {\frac {\left( \frac {s_A^2} {n_A} \right)^2} {n_A + 1} + \frac {\left( \frac {s_B^2} {n_B} \right)^2} {n_B + 1}} - 2 \nonumber \]Because the degrees of freedom must be an integer, we round to the nearest integer the value of \(\nu\) obtained from this equation.The equation above for the degrees of freedom is from Miller, J.C.; Miller, J.N. Statistics for Analytical Chemistry, 2nd Ed., Ellis-Horward: Chichester, UK, 1988. In the 6th Edition, the authors note that several different equations have been suggested for the number of degrees of freedom for t when sA and sB differ, reflecting the fact that the determination of degrees of freedom an approximation. An alternative equation—which is used by statistical software packages, such as R, Minitab, Excel—is\[\nu = \frac {\left( \frac {s_A^2} {n_A} + \frac {s_B^2} {n_B} \right)^2} {\frac {\left( \frac {s_A^2} {n_A} \right)^2} {n_A - 1} + \frac {\left( \frac {s_B^2} {n_B} \right)^2} {n_B - 1}} = \frac {\left( \frac {s_A^2} {n_A} + \frac {s_B^2} {n_B} \right)^2} {\frac {s_A^4} {n_A^2(n_A - 1)} + \frac {s_B^4} {n_B^2(n_B - 1)}} \nonumber \]For typical problems in analytical chemistry, the calculated degrees of freedom is reasonably insensitive to the choice of equation.Regardless of whether how we calculate texp, we reject the null hypothesis if texp is greater than \(t(\alpha, \nu)\) and retain the null hypothesis if texp is less than or equal to \(t(\alpha, \nu)\).Example \(\PageIndex{3}\) provides results for two experiments to determine the mass of a circulating U.S. penny. Determine whether there is a difference in the means of these analyses at \(\alpha = 0.05\).First we use an F-test to determine whether we can pool the variances. We completed this analysis in Example \(\PageIndex{5}\), finding no evidence of a significant difference, which means we can pool the standard deviations, obtaining\[s_\text{pool} = \sqrt{\frac {(7 - 1)(0.051)^2 + (5 - 1)(0.037)^2} {7 + 5 - 2}} = 0.0459 \nonumber \]with 10 degrees of freedom. To compare the means we use the following null hypothesis and alternative hypotheses\[H_0 \text{: } \mu_A = \mu_B \quad \quad \quad H_A \text{: } \mu_A \neq \mu_B \nonumber \]Because we are using the pooled standard deviation, we calculate texp as\[t_\text{exp} = \frac {|3.117 - 3.081|} {0.0459} \times \sqrt{\frac {7 \times 5} {7 + 5}} = 1.34 \nonumber \]The critical value for t(0.05, 10), from Appendix 3, is 2.23. Because texp is less than t(0.05, 10) we retain the null hypothesis. For \(\alpha = 0.05\) we do not have evidence that the two sets of pennies are significantly different.One method for determining the %w/w Na2CO3 in soda ash is to use an acid–base titration. When two analysts analyze the same sample of soda ash they obtain the results shown here.Analyst A: \(86.82 \% \quad 87.04 \% \quad 86.93 \% \quad 87.01 \% \quad 86.20 \% \quad 87.00 \%\)Analyst B: \(81.01 \% \quad 86.15 \% \quad 81.73 \% \quad 83.19 \% \quad 80.27 \% \quad 83.93 \% \quad\)Determine whether the difference in the mean values is significant at \(\alpha = 0.05\).We begin by reporting the mean and standard deviation for each analyst.\[\overline{X}_A = 86.83\% \quad \quad s_A = 0.32\% \nonumber \]\[\overline{X}_B = 82.71\% \quad \quad s_B = 2.16\% \nonumber \]To determine whether we can use a pooled standard deviation, we first complete an F-test using the following null and alternative hypotheses.\[H_0 \text{: } s_A^2 = s_B^2 \quad \quad \quad H_A \text{: } s_A^2 \neq s_B^2 \nonumber \]Calculating Fexp, we obtain a value of\[F_\text{exp} = \frac {(2.16)^2} {(0.32)^2} = 45.6 \nonumber \]Because Fexp is larger than the critical value of 7.15 for F(0.05, 5, 5) from Appendix 4, we reject the null hypothesis and accept the alternative hypothesis that there is a significant difference between the variances; thus, we cannot calculate a pooled standard deviation.To compare the means for the two analysts we use the following null and alternative hypotheses.\[H_0 \text{: } \overline{X}_A = \overline{X}_B \quad \quad \quad H_A \text{: } \overline{X}_A \neq \overline{X}_B \nonumber \]Because we cannot pool the standard deviations, we calculate texp as\[t_\text{exp} = \frac {|86.83 - 82.71|} {\sqrt{\frac {(0.32)^2} {6} + \frac {(2.16)^2} {6}}} = 4.62 \nonumber \]and calculate the degrees of freedom as\[\nu = \frac {\left( \frac {(0.32)^2} {6} + \frac {(2.16)^2} {6} \right)^2} {\frac {\left( \frac {(0.32)^2} {6} \right)^2} {6 + 1} + \frac {\left( \frac {(2.16)^2} {6} \right)^2} {6 + 1}} - 2 = 5.3 \approx 5 \nonumber \]From Appendix 3, the critical value for t(0.05, 5) is 2.57. Because texp is greater than t(0.05, 5) we reject the null hypothesis and accept the alternative hypothesis that the means for the two analysts are significantly different at \(\alpha = 0.05\).Suppose we are evaluating a new method for monitoring blood glucose concentrations in patients. An important part of evaluating a new method is to compare it to an established method. What is the best way to gather data for this study? Because the variation in the blood glucose levels amongst patients is large we may be unable to detect a small, but significant difference between the methods if we use different patients to gather data for each method. Using paired data, in which the we analyze each patient’s blood using both methods, prevents a large variance within a population from adversely affecting a t-test of means.Typical blood glucose levels for most non-diabetic individuals ranges between 80–120 mg/dL (4.4–6.7 mM), rising to as high as 140 mg/dL (7.8 mM) shortly after eating. Higher levels are common for individuals who are pre-diabetic or diabetic.When we use paired data we first calculate the individual differences, di, between each sample's paired resykts. Using these individual differences, we then calculate the average difference, \(\overline{d}\), and the standard deviation of the differences, sd. The null hypothesis, \(H_0 \text{: } d = 0\), is that there is no difference between the two samples, and the alternative hypothesis, \(H_A \text{: } d \neq 0\), is that the difference between the two samples is significant.The test statistic, texp, is derived from a confidence interval around \(\overline{d}\)\[t_\text{exp} = \frac {|\overline{d}| \sqrt{n}} {s_d} \nonumber \]where n is the number of paired samples. As is true for other forms of the t-test, we compare texp to \(t(\alpha, \nu)\), where the degrees of freedom, \(\nu\), is n – 1. If texp is greater than \(t(\alpha, \nu)\), then we reject the null hypothesis and accept the alternative hypothesis. We retain the null hypothesis if texp is less than or equal to t(a, o). This is known as a paired t-test.Marecek et. al. developed a new electrochemical method for the rapid determination of the concentration of the antibiotic monensin in fermentation vats [Marecek, V.; Janchenova, H.; Brezina, M.; Betti, M. Anal. Chim. Acta 1991, 244, 15–19]. The standard method for the analysis is a test for microbiological activity, which is both difficult to complete and time-consuming. Samples were collected from the fermentation vats at various times during production and analyzed for the concentration of monensin using both methods. The results, in parts per thousand (ppt), are reported in the following table.Is there a significant difference between the methods at \(\alpha = 0.05\)?Acquiring samples over an extended period of time introduces a substantial time-dependent change in the concentration of monensin. Because the variation in concentration between samples is so large, we use a paired t-test with the following null and alternative hypotheses.\[H_0 \text{: } \overline{d} = 0 \quad \quad \quad H_A \text{: } \overline{d} \neq 0 \nonumber \]Defining the difference between the methods as\[d_i = (X_\text{elect})_i - (X_\text{micro})_i \nonumber \]we calculate the difference for each sample.The mean and the standard deviation for the differences are, respectively, 2.25 ppt and 5.63 ppt. The value of texp is\[t_\text{exp} = \frac {|2.25| \sqrt{11}} {5.63} = 1.33 \nonumber \]which is smaller than the critical value of 2.23 for t(0.05, 10) from Appendix 3. We retain the null hypothesis and find no evidence for a significant difference in the methods at \(\alpha = 0.05\).One important requirement for a paired t-test is that the determinate and the indeterminate errors that affect the analysis must be independent of the analyte’s concentration. If this is not the case, then a sample with an unusually high concentration of analyte will have an unusually large di. Including this sample in the calculation of \(\overline{d}\) and sd gives a biased estimate for the expected mean and standard deviation. This rarely is a problem for samples that span a limited range of analyte concentrations, such as those in Example \(\PageIndex{6}\) or Exercise \(\PageIndex{8}\). When paired data span a wide range of concentrations, however, the magnitude of the determinate and indeterminate sources of error may not be independent of the analyte’s concentration; when true, a paired t-test may give misleading results because the paired data with the largest absolute determinate and indeterminate errors will dominate \(\overline{d}\). In this situation a regression analysis, which is the subject of the next chapter, is more appropriate method for comparing the data.The importance of distinguishing between paired and unpaired data is worth examining more closely. The following is data from some work I completed with a colleague in which we were looking at concentration of Zn in Lake Erie at the air-water interface and the sediment-water interface.The mean and the standard deviation for the ppm Zn at the air-water interface are 0.5178 ppm and 0.01732 ppm, and the mean and the standard deviation for the ppm Zn at the sediment-water interface are 0.4445 ppm and 0.1418 ppm. We can use these values to draw normal distributions for both by letting the means and the standard deviations for the samples, \(\overline{X}\) and \(s\), serve as estimates for the means and the standard deviations for the population, \(\mu\) and \(\sigma\). As we see in the following figurethe two distributions overlap strongly, suggesting that a t-test of their means is not likely to find evidence of a difference. And yet, we also see that for each site, the concentration of Zn at the sediment-water interface is less than that at the air-water interface. In this case, the difference between the concentration of Zn at individual sites is sufficiently large that it masks our ability to see the difference between the two interfaces.If we take the differences between the air-water and sediment-water interfaces, we have values of 0.015, 0.028, 0.067, 0.121, 0.102, and 0.107 ppm Zn, with a mean of 0.07333 ppm Zn and a standard deviation of 0.04410 ppm Zn. Superimposing all three normal distributionsshows clearly that most of the normal distribution for the differences lies above zero, suggesting that a t-test might show evidence that the difference is significant.Table \(\PageIndex{11}\) provides one more data set giving the masses for a sample of pennies. Do you notice anything unusual in this data? Of the 100 pennies included in our earlier table, no penny has a mass of less than 3 g. In this table, however, the mass of one penny is less than 3 g. We might ask whether this penny’s mass is so different from the other pennies that it is in error.A measurement that is not consistent with other measurements is called an outlier. An outlier might exist for many reasons: the outlier might belong to a different populationIs this a Canadian penny?or the outlier might be a contaminated or an otherwise altered sampleIs the penny damaged or unusually dirty?or the outlier may result from an error in the analysisDid we forget to tare the balance?Regardless of its source, the presence of an outlier compromises any meaningful analysis of our data. There are many significance tests that we can use to identify a potential outlier, three of which we present here.One of the most common significance tests for identifying an outlier is Dixon’s Q-test. The null hypothesis is that there are no outliers, and the alternative hypothesis is that there is an outlier. The Q-test compares the gap between the suspected outlier and its nearest numerical neighbor to the range of the entire data set ).The test statistic, Qexp, is\[Q_\text{exp} = \frac {\text{gap}} {\text{range}} = \frac {|\text{outlier's value} - \text{nearest value}|} {\text{largest value} - \text{smallest value}} \nonumber \]This equation is appropriate for evaluating a single outlier. Other forms of Dixon’s Q-test allow its extension to detecting multiple outliers [Rorabacher, D. B. Anal. Chem. 1991, 63, 139–146].The value of Qexp is compared to a critical value, \(Q(\alpha, n)\), where \(\alpha\) is the probability that we will reject a valid data point (a type 1 error) and n is the total number of data points. To protect against rejecting a valid data point, usually we apply the more conservative two-tailed Q-test, even though the possible outlier is the smallest or the largest value in the data set. If Qexp is greater than \(Q(\alpha, n)\), then we reject the null hypothesis and may exclude the outlier. We retain the possible outlier when Qexp is less than or equal to \(Q(\alpha, n)\). Table \(\PageIndex{12}\) provides values for \(Q(\alpha, n)\) for a data set that has 3–10 values. A more extensive table is in Appendix 5. Values for \(Q(\alpha, n)\) assume an underlying normal distribution.Although Dixon’s Q-test is a common method for evaluating outliers, it is no longer favored by the International Standards Organization (ISO), which recommends the Grubb’s test. There are several versions of Grubb’s test depending on the number of potential outliers. Here we will consider the case where there is a single suspected outlier.For details on this recommendation, see International Standards ISO Guide 5752-2 “Accuracy (trueness and precision) of measurement methods and results–Part 2: basic methods for the determination of repeatability and reproducibility of a standard measurement method,” 1994.The test statistic for Grubb’s test, Gexp, is the distance between the sample’s mean, \(\overline{X}\), and the potential outlier, \(X_\text{out}\), in terms of the sample’s standard deviation, s.\[G_\text{exp} = \frac {|X_\text{out} - \overline{X}|} {s} \nonumber \]We compare the value of Gexp to a critical value \(G(\alpha, n)\), where \(\alpha\) is the probability that we will reject a valid data point and n is the number of data points in the sample. If Gexp is greater than \(G(\alpha, n)\), then we may reject the data point as an outlier, otherwise we retain the data point as part of the sample. Table \(\PageIndex{13}\) provides values for G(0.05, n) for a sample containing 3–10 values. A more extensive table is in Appendix 6. Values for \(G(\alpha, n)\) assume an underlying normal distribution.Our final method for identifying an outlier is Chauvenet’s criterion. Unlike Dixon’s Q-Test and Grubb’s test, you can apply this method to any distribution as long as you know how to calculate the probability for a particular outcome. Chauvenet’s criterion states that we can reject a data point if the probability of obtaining the data point’s value is less than \((2n^{-1})\), where n is the size of the sample. For example, if n = 10, a result with a probability of less than \((2 \times 10)^{-1}\), or 0.05, is considered an outlier.To calculate a potential outlier’s probability we first calculate its standardized deviation, z \[z = \frac {|X_\text{out} - \overline{X}|} {s} \nonumber \]where \(X_\text{out}\) is the potential outlier, \(\overline{X}\) is the sample’s mean and s is the sample’s standard deviation. Note that this equation is identical to the equation for Gexp in the Grubb’s test. For a normal distribution, we can find the probability of obtaining a value of z using the probability table in Appendix 2.Table \(\PageIndex{11}\) contains the masses for nine circulating United States pennies. One entry, 2.514 g, appears to be an outlier. Determine if this penny is an outlier using a Q-test, Grubb’s test, and Chauvenet’s criterion. For the Q-test and Grubb’s test, let \(\alpha = 0.05\).For the Q-test the value for \(Q_\text{exp}\) is\[Q_\text{exp} = \frac {|2.514 - 3.039|} {3.109 - 2.514} = 0.882 \nonumber \]From Table \(\PageIndex{12}\), the critical value for Q(0.05, 9) is 0.493. Because Qexp is greater than Q(0.05, 9), we can assume the penny with a mass of 2.514 g likely is an outlier.For Grubb’s test we first need the mean and the standard deviation, which are 3.011 g and 0.188 g, respectively. The value for Gexp is\[G_\text{exp} = \frac {|2.514 - 3.011|} {0.188} = 2.64 \nonumber \]Using Table \(\PageIndex{13}\), we find that the critical value for G(0.05, 9) is 2.215. Because Gexp is greater than G(0.05, 9), we can assume that the penny with a mass of 2.514 g likely is an outlier.For Chauvenet’s criterion, the critical probability is \((2 \times 9)^{-1}\), or 0.0556. The value of z is the same as Gexp, or 2.64. Using Appendix 1, the probability for z = 2.64 is 0.00415. Because the probability of obtaining a mass of 0.2514 g is less than the critical probability, we can assume the penny with a mass of 2.514 g likely is an outlier.You should exercise caution when using a significance test for outliers because there is a chance you will reject a valid result. In addition, you should avoid rejecting an outlier if it leads to a precision that is much better than expected based on a propagation of uncertainty. Given these concerns it is not surprising that some statisticians caution against the removal of outliers [Deming, W. E. Statistical Analysis of Data; Wiley: New York, 1943 (republished by Dover: New York, 1961); p. 171].You also can adopt a more stringent requirement for rejecting data. When using the Grubb’s test, for example, the ISO 5752 guidelines suggest retaining a value if the probability for rejecting it is greater than \(\alpha = 0.05\), and flagging a value as a “straggler” if the probability for rejecting it is between \(\alpha = 0.05\) and \(\alpha = 0.01\). A “straggler” is retained unless there is compelling reason for its rejection. The guidelines recommend using \(\alpha = 0.01\) as the minimum criterion for rejecting a possible outlier.On the other hand, testing for outliers can provide useful information if we try to understand the source of the suspected outlier. For example, the outlier in Table \(\PageIndex{11}\) represents a significant change in the mass of a penny (an approximately 17% decrease in mass), which is the result of a change in the composition of the U.S. penny. In 1982 the composition of a U.S. penny changed from a brass alloy that was 95% w/w Cu and 5% w/w Zn (with a nominal mass of 3.1 g), to a pure zinc core covered with copper (with a nominal mass of 2.5 g) [Richardson, T. H. J. Chem. Educ. 1991, 68, 310–311]. The pennies in Table \(\PageIndex{11}\), therefore, were drawn from different populations.A calibration curve is one of the most important tools in analytical chemistry as it allows us to determine the concentration of an analyte in a sample by measuring the signal it generates when placed in an instrument, such as a spectrophotometer. To determine the analyte's concentration we must know the relationship between the signal we measure , \(S\), and the analyte's concentration, \(C_A\), which we can write as\[S = k_A C_A + S_{blank} \nonumber \]where \(k_A\) is the calibration curve's sensitivity and \(S_{blank}\) is the signal in the absence of analyte.How do we find the best estimate for this relationship between the signal and the concentration of analyte? When a calibration curve is a straight-line, we represent it using the following mathematical model\[y = \beta_0 + \beta_1 x \nonumber \]where y is the analyte’s measured signal, S, and x is the analyte’s known concentration, \(C_A\), in a series of standard solutions. The constants \(\beta_0\) and \(\beta_1\) are, respectively, the calibration curve’s expected y-intercept and its expected slope. Because of uncertainty in our measurements, the best we can do is to estimate values for \(\beta_0\) and \(\beta_1\), which we represent as b0 and b1. The goal of a linear regression analysis is to determine the best estimates for b0 and b1.The most common method for completing a linear regression makes three assumptions:Because we assume that the indeterminate errors are the same for all standards, each standard contributes equally in our estimate of the slope and the y-intercept. For this reason the result is considered an unweighted linear regression.The second assumption generally is true because of the central limit theorem, which we considered earlier. The validity of the two remaining assumptions is less obvious and you should evaluate them before you accept the results of a linear regression. In particular the first assumption is always suspect because there certainly is some indeterminate error in the measurement of x. When we prepare a calibration curve, however, it is not unusual to find that the uncertainty in the signal, S, is significantly greater than the uncertainty in the analyte’s concentration, \(C_A\). In such circumstances the first assumption usually is reasonable.To understand the logic of a linear regression consider the example in , which shows three data points and two possible straight-lines that might reasonably explain the data. How do we decide how well these straight-lines fit the data, and how do we determine which, if either, is the best straight-line?Let’s focus on the solid line in . The equation for this line is\[\hat{y} = b_0 + b_1 x \nonumber \]where b0 and b1 are estimates for the y-intercept and the slope, and \(\hat{y}\) is the predicted value of y for any value of x. Because we assume that all uncertainty is the result of indeterminate errors in y, the difference between y and \(\hat{y}\) for each value of x is the residual error, r, in our mathematical model.\[r_i = (y_i - \hat{y}_i) \nonumber \] shows the residual errors for the three data points. The smaller the total residual error, R, which we define as\[R = \sum_{i = 1}^{n} (y_i - \hat{y}_i)^2 \nonumber \]the better the fit between the straight-line and the data. In a linear regression analysis, we seek values of b0 and b1 that give the smallest total residual error.The reason for squaring the individual residual errors is to prevent a positive residual error from canceling out a negative residual error. You have seen this before in the equations for the sample and population standard deviations introduced in Chapter 4. You also can see from this equation why a linear regression is sometimes called the method of least squares.Although we will not formally develop the mathematical equations for a linear regression analysis, you can find the derivations in many standard statistical texts [ See, for example, Draper, N. R.; Smith, H. Applied Regression Analysis, 3rd ed.; Wiley: New York, 1998]. The resulting equation for the slope, b1, is\[b_1 = \frac {n \sum_{i = 1}^{n} x_i y_i - \sum_{i = 1}^{n} x_i \sum_{i = 1}^{n} y_i} {n \sum_{i = 1}^{n} x_i^2 - \left( \sum_{i = 1}^{n} x_i \right)^2} \nonumber \]and the equation for the y-intercept, b0, is\[b_0 = \frac {\sum_{i = 1}^{n} y_i - b_1 \sum_{i = 1}^{n} x_i} {n} \nonumber \]Although these equations appear formidable, it is necessary only to evaluate the following four summations\[\sum_{i = 1}^{n} x_i \quad \sum_{i = 1}^{n} y_i \quad \sum_{i = 1}^{n} x_i y_i \quad \sum_{i = 1}^{n} x_i^2 \nonumber \]Many calculators, spreadsheets, and other statistical software packages are capable of performing a linear regression analysis based on this model; see Section 8.5 for details on completing a linear regression analysis using R. For illustrative purposes the necessary calculations are shown in detail in the following example.Using the calibration data in the following table, determine the relationship between the signal, \(y_i\), and the analyte's concentration, \(x_i\), using an unweighted linear regression.We begin by setting up a table to help us organize the calculation.Adding the values in each column gives\[\sum_{i = 1}^{n} x_i = 1.500 \quad \sum_{i = 1}^{n} y_i = 182.31 \quad \sum_{i = 1}^{n} x_i y_i = 66.701 \quad \sum_{i = 1}^{n} x_i^2 = 0.550 \nonumber \]Substituting these values into the equations for the slope and the y-intercept gives\[b_1 = \frac {(6 \times 66.701) - (1.500 \times 182.31)} {(6 \times 0.550) - (1.500)^2} = 120.706 \approx 120.71 \nonumber \]\[b_0 = \frac {182.31 - (120.706 \times 1.500)} {6} = 0.209 \approx 0.21 \nonumber \]The relationship between the signal, \(S\), and the analyte's concentration, \(C_A\), therefore, is\[S = 120.71 \times C_A + 0.21 \nonumber \]For now we keep two decimal places to match the number of decimal places in the signal. The resulting calibration curve is shown in .As we see in , because of indeterminate errors in the signal, the regression line does not pass through the exact center of each data point. The cumulative deviation of our data from the regression line—the total residual error—is proportional to the uncertainty in the regression. We call this uncertainty the standard deviation about the regression, sr, which is equal to\[s_r = \sqrt{\frac {\sum_{i = 1}^{n} \left( y_i - \hat{y}_i \right)^2} {n - 2}} \nonumber \]where yi is the ith experimental value, and \(\hat{y}_i\) is the corresponding value predicted by the regression equation \(\hat{y} = b_0 + b_1 x\). Note that the denominator indicates that our regression analysis has n – 2 degrees of freedom—we lose two degree of freedom because we use two parameters, the slope and the y-intercept, to calculate \(\hat{y}_i\).A more useful representation of the uncertainty in our regression analysis is to consider the effect of indeterminate errors on the slope, b1, and the y-intercept, b0, which we express as standard deviations.\[s_{b_1} = \sqrt{\frac {n s_r^2} {n \sum_{i = 1}^{n} x_i^2 - \left( \sum_{i = 1}^{n} x_i \right)^2}} = \sqrt{\frac {s_r^2} {\sum_{i = 1}^{n} \left( x_i - \overline{x} \right)^2}} \nonumber \]\[s_{b_0} = \sqrt{\frac {s_r^2 \sum_{i = 1}^{n} x_i^2} {n \sum_{i = 1}^{n} x_i^2 - \left( \sum_{i = 1}^{n} x_i \right)^2}} = \sqrt{\frac {s_r^2 \sum_{i = 1}^{n} x_i^2} {n \sum_{i = 1}^{n} \left( x_i - \overline{x} \right)^2}} \nonumber \]We use these standard deviations to establish confidence intervals for the expected slope, \(\beta_1\), and the expected y-intercept, \(\beta_0\)\[\beta_1 = b_1 \pm t s_{b_1} \nonumber \]\[\beta_0 = b_0 \pm t s_{b_0} \nonumber \]where we select t for a significance level of \(\alpha\) and for n – 2 degrees of freedom. Note that these equations do not contain the factor of \((\sqrt{n})^{-1}\) seen in the confidence intervals for \(\mu\) because the confidence interval here is based on a single regression line.Calculate the 95% confidence intervals for the slope and y-intercept from Example \(\PageIndex{10}\).We begin by calculating the standard deviation about the regression. To do this we must calculate the predicted signals, \(\hat{y}_i\) , using the slope and the y-intercept from Example \(\PageIndex{10}\), and the squares of the residual error, \((y_i - \hat{y}_i)^2\). Using the last standard as an example, we find that the predicted signal is\[\hat{y}_6 = b_0 + b_1 x_6 = 0.209 + (120.706 \times 0.500) = 60.562 \nonumber \]and that the square of the residual error is\[(y_i - \hat{y}_i)^2 = (60.42 - 60.562)^2 = 0.2016 \approx 0.202 \nonumber \]The following table displays the results for all six solutions.Adding together the data in the last column gives the numerator in the equation for the standard deviation about the regression; thus\[s_r = \sqrt{\frac {0.6512} {6 - 2}} = 0.4035 \nonumber \]Next we calculate the standard deviations for the slope and the y-intercept. The values for the summation terms are from Example \(\PageIndex{10}\).\[s_{b_1} = \sqrt{\frac {6 \times (0.4035)^2} {(6 \times 0.550) - (1.500)^2}} = 0.965 \nonumber \]\[s_{b_0} = \sqrt{\frac {(0.4035)^2 \times 0.550} {(6 \times 0.550) - (1.500)^2}} = 0.292 \nonumber \]Finally, the 95% confidence intervals (\(\alpha = 0.05\), 4 degrees of freedom) for the slope and y-intercept are\[\beta_1 = b_1 \pm ts_{b_1} = 120.706 \pm (2.78 \times 0.965) = 120.7 \pm 2.7 \nonumber \]\[\beta_0 = b_0 \pm ts_{b_0} = 0.209 \pm (2.78 \times 0.292) = 0.2 \pm 0.80 \nonumber \]where t(0.05, 4) from Appendix 3 is 2.78. The standard deviation about the regression, sr, suggests that the signal, Sstd, is precise to one decimal place. For this reason we report the slope and the y-intercept to a single decimal place.Once we have our regression equation, it is easy to determine the concentration of analyte in a sample. When we use a normal calibration curve, for example, we measure the signal for our sample, Ssamp, and calculate the analyte’s concentration, CA, using the regression equation.\[C_A = \frac {S_{samp} - b_0} {b_1} \nonumber \]What is less obvious is how to report a confidence interval for CA that expresses the uncertainty in our analysis. To calculate a confidence interval we need to know the standard deviation in the analyte’s concentration, \(s_{C_A}\), which is given by the following equation\[s_{C_A} = \frac {s_r} {b_1} \sqrt{\frac {1} {m} + \frac {1} {n} + \frac {\left( \overline{S}_{samp} - \overline{S}_{std} \right)^2} {(b_1)^2 \sum_{i = 1}^{n} \left( C_{std_i} - \overline{C}_{std} \right)^2}} \nonumber \]where m is the number of replicates we use to establish the sample’s average signal, Ssamp, n is the number of calibration standards, Sstd is the average signal for the calibration standards, and \(C_{std_i}\) and \(\overline{C}_{std}\) are the individual and the mean concentrations for the calibration standards. Knowing the value of \(s_{C_A}\), the confidence interval for the analyte’s concentration is\[\mu_{C_A} = C_A \pm t s_{C_A} \nonumber \]where \(\mu_{C_A}\) is the expected value of CA in the absence of determinate errors, and with the value of t is based on the desired level of confidence and n – 2 degrees of freedom.A close examination of these equations should convince you that we can decrease the uncertainty in the predicted concentration of analyte, \(C_A\) if we increase the number of standards, \(n\), increase the number of replicate samples that we analyze, \(m\), and if the sample’s average signal, \(\overline{S}_{samp}\), is equal to the average signal for the standards, \(\overline{S}_{std}\). When practical, you should plan your calibration curve so that Ssamp falls in the middle of the calibration curve. For more information about these regression equations see (a) Miller, J. N. Analyst 1991, 116, 3–14; (b) Sharaf, M. A.; Illman, D. L.; Kowalski, B. R. Chemometrics, Wiley-Interscience: New York, 1986, pp. 126-127; (c) Analytical Methods Committee “Uncertainties in concentrations estimated from calibration experiments,” AMC Technical Brief, March 2006.The equation for the standard deviation in the analyte's concentration is written in terms of a calibration experiment. A more general form of the equation, written in terms of x and y, is given here.\[s_{x} = \frac {s_r} {b_1} \sqrt{\frac {1} {m} + \frac {1} {n} + \frac {\left( \overline{Y} - \overline{y} \right)^2} {(b_1)^2 \sum_{i = 1}^{n} \left( x_i - \overline{x} \right)^2}} \nonumber \]Three replicate analyses for a sample that contains an unknown concentration of analyte, yields values for Ssamp of 29.32, 29.16 and 29.51 (arbitrary units). Using the results from Example \(\PageIndex{10}\) and Example \(\PageIndex{11}\), determine the analyte’s concentration, CA, and its 95% confidence interval.The average signal, \(\overline{S}_{samp}\), is 29.33, which, using the slope and the y-intercept from Example \(\PageIndex{10}\), gives the analyte’s concentration as\[C_A = \frac {\overline{S}_{samp} - b_0} {b_1} = \frac {29.33 - 0.209} {120.706} = 0.241 \nonumber \]To calculate the standard deviation for the analyte’s concentration we must determine the values for \(\overline{S}_{std}\) and for \(\sum_{i = 1}^{2} (C_{std_i} - \overline{C}_{std})^2\). The former is just the average signal for the calibration standards, which, using the data in Table \(\PageIndex{10}\), is 30.385. Calculating \(\sum_{i = 1}^{2} (C_{std_i} - \overline{C}_{std})^2\) looks formidable, but we can simplify its calculation by recognizing that this sum-of-squares is the numerator in a standard deviation equation; thus,\[\sum_{i = 1}^{n} (C_{std_i} - \overline{C}_{std})^2 = (s_{C_{std}})^2 \times (n - 1) \nonumber \]where \(s_{C_{std}}\) is the standard deviation for the concentration of analyte in the calibration standards. Using the data in Table \(\PageIndex{10}\) we find that \(s_{C_{std}}\) is 0.1871 and\[\sum_{i = 1}^{n} (C_{std_i} - \overline{C}_{std})^2 = (0.1872)^2 \times (6 - 1) = 0.175 \nonumber \]Substituting known values into the equation for \(s_{C_A}\) gives\[s_{C_A} = \frac {0.4035} {120.706} \sqrt{\frac {1} {3} + \frac {1} {6} + \frac {(29.33 - 30.385)^2} {(120.706)^2 \times 0.175}} = 0.0024 \nonumber \]Finally, the 95% confidence interval for 4 degrees of freedom is\[\mu_{C_A} = C_A \pm ts_{C_A} = 0.241 \pm (2.78 \times 0.0024) = 0.241 \pm 0.007 \nonumber \] shows the calibration curve with curves showing the 95% confidence interval for CA.You should never accept the result of a linear regression analysis without evaluating the validity of the model. Perhaps the simplest way to evaluate a regression analysis is to examine the residual errors. As we saw earlier, the residual error for a single calibration standard, ri, is\[r_i = (y_i - \hat{y}_i) \nonumber \]If the regression model is valid, then the residual errors should be distributed randomly about an average residual error of zero, with no apparent trend toward either smaller or larger residual errors ). Trends such as those in and provide evidence that at least one of the model’s assumptions is incorrect. For example, a trend toward larger residual errors at higher concentrations, , suggests that the indeterminate errors affecting the signal are not independent of the analyte’s concentration. In , the residual errors are not random, which suggests we cannot model the data using a straight-line relationship. Regression methods for the latter two cases are discussed in the following sections.Use your results from Exercise \(\PageIndex{10}\) to construct a residual plot and explain its significance.To create a residual plot, we need to calculate the residual error for each standard. The following table contains the relevant information.The figure below shows a plot of the resulting residual errors. The residual errors appear random, although they do alternate in sign, and they do not show any significant dependence on the analyte’s concentration. Taken together, these observations suggest that our regression model is appropriate.Our treatment of linear regression to this point assumes that any indeterminate errors that affect y are independent of the value of x. If this assumption is false, then we must include the variance for each value of y in our determination of the y-intercept, b0, and the slope, b1; thus\[b_0 = \frac {\sum_{i = 1}^{n} w_i y_i - b_1 \sum_{i = 1}^{n} w_i x_i} {n} \nonumber \]\[b_1 = \frac {n \sum_{i = 1}^{n} w_i x_i y_i - \sum_{i = 1}^{n} w_i x_i \sum_{i = 1}^{n} w_i y_i} {n \sum_{i =1}^{n} w_i x_i^2 - \left( \sum_{i = 1}^{n} w_i x_i \right)^2} \nonumber \]where wi is a weighting factor that accounts for the variance in yi \[w_i = \frac {n (s_{y_i})^{-2}} {\sum_{i = 1}^{n} (s_{y_i})^{-2}} \nonumber \]and \(s_{y_i}\) is the standard deviation for yi. In a weighted linear regression, each xy-pair’s contribution to the regression line is inversely proportional to the precision of yi; that is, the more precise the value of y, the greater its contribution to the regression.Shown here are data for an external standardization in which sstd is the standard deviation for three replicate determination of the signal. Determine the calibration curve’s equation using a weighted linear regression. As you work through this example, remember that x corresponds to Cstd, and that y corresponds to Sstd.We begin by setting up a table to aid in calculating the weighting factors.Adding together the values in the fourth column gives\[\sum_{i = 1}^{n} (s_{y_i})^{-2} \nonumber \]which we use to calculate the individual weights in the last column. As a check on your calculations, the sum of the individual weights must equal the number of calibration standards, n. The sum of the entries in the last column is 6.0000, so all is well. After we calculate the individual weights, we use a second table to aid in calculating the four summation terms in the equations for the slope, \(b_1\), and the y-intercept, \(b_0\).Adding the values in the last four columns gives\[\sum_{i = 1}^{n} w_i x_i = 0.3644 \quad \sum_{i = 1}^{n} w_i y_i = 44.9499 \quad \sum_{i = 1}^{n} w_i x_i^2 = 0.0499 \quad \sum_{i = 1}^{n} w_i x_i y_i = 6.1451 \nonumber \]which gives the estimated slope and the estimated y-intercept as\[b_1 = \frac {(6 \times 6.1451) - (0.3644 \times 44.9499)} {(6 \times 0.0499) - (0.3644)^2} = 122.985 \nonumber \]\[b_0 = \frac{44.9499 - (122.985 \times 0.3644)} {6} = 0.0224 \nonumber \]The calibration equation is\[S_{std} = 122.98 \times C_{std} + 0.2 \nonumber \] shows the calibration curve for the weighted regression determined here and the calibration curve for the unweighted regression. Although the two calibration curves are very similar, there are slight differences in the slope and in the y-intercept. Most notably, the y-intercept for the weighted linear regression is closer to the expected value of zero. Because the standard deviation for the signal, Sstd, is smaller for smaller concentrations of analyte, Cstd, a weighted linear regression gives more emphasis to these standards, allowing for a better estimate of the y-intercept.Equations for calculating confidence intervals for the slope, the y-intercept, and the concentration of analyte when using a weighted linear regression are not as easy to define as for an unweighted linear regression [Bonate, P. J. Anal. Chem. 1993, 65, 1367–1372]. The confidence interval for the analyte’s concentration, however, is at its optimum value when the analyte’s signal is near the weighted centroid, yc , of the calibration curve.\[y_c = \frac {1} {n} \sum_{i = 1}^{n} w_i x_i \nonumber \]If we remove our assumption that indeterminate errors affecting a calibration curve are present only in the signal (y), then we also must factor into the regression model the indeterminate errors that affect the analyte’s concentration in the calibration standards (x). The solution for the resulting regression line is computationally more involved than that for either the unweighted or weighted regression lines. Although we will not consider the details in this textbook, you should be aware that neglecting the presence of indeterminate errors in x can bias the results of a linear regression.See, for example, Analytical Methods Committee, “Fitting a linear functional relationship to data with error on both variable,” AMC Technical Brief, March, 2002), as well as this chapter’s Additional Resources.A straight-line regression model, despite its apparent complexity, is the simplest functional relationship between two variables. What do we do if our calibration curve is curvilinear—that is, if it is a curved-line instead of a straight-line? One approach is to try transforming the data into a straight-line. Logarithms, exponentials, reciprocals, square roots, and trigonometric functions have been used in this way. A plot of log(y) versus x is a typical example. Such transformations are not without complications, of which the most obvious is that data with a uniform variance in y will not maintain that uniform variance after it is transformed.It is worth noting here that the term “linear” does not mean a straight-line. A linear function may contain more than one additive term, but each such term has one and only one adjustable multiplicative parameter. The function\[y = ax + bx^2 \nonumber \]is an example of a linear function because the terms x and x2 each include a single multiplicative parameter, a and b, respectively. The function\[y = x^b \nonumber \]is nonlinear because b is not a multiplicative parameter; it is, instead, a power. This is why you can use linear regression to fit a polynomial equation to your data.Sometimes it is possible to transform a nonlinear function into a linear function. For example, taking the log of both sides of the nonlinear function above gives a linear function.\[\log(y) = b \log(x) \nonumber \]Another approach to developing a linear regression model is to fit a polynomial equation to the data, such as \(y = a + b x + c x^2\). You can use linear regression to calculate the parameters a, b, and c, although the equations are different than those for the linear regression of a straight-line. If you cannot fit your data using a single polynomial equation, it may be possible to fit separate polynomial equations to short segments of the calibration curve. The result is a single continuous calibration curve known as a spline function. The use of R for curvilinear regression is included in Chapter 8.5.For details about curvilinear regression, see (a) Sharaf, M. A.; Illman, D. L.; Kowalski, B. R. Chemometrics, Wiley-Interscience: New York, 1986; (b) Deming, S. N.; Morgan, S. L. Experimental Design: A Chemometric Approach, Elsevier: Amsterdam, 1987.The regression models in this chapter apply only to functions that contain a single dependent variable and a single independent variable. One example is the simplest form of Beer's law in which the absorbance, \(A\), of a sample at a single wavelength, \(\lambda\), depends upon the concentration of a single analyte, \(C_A\)\[A_{\lambda} = \epsilon_{\lambda, A} b C_A \nonumber \]where \(\epsilon_{\lambda, A}\) is the analyte's molar absorptivity at the selected wavelength and \(b\) is the pathlength through the sample. In the presence of an interferent, \(I\), however, the signal may depend on the concentrations of both the analyte and the interferent\[A_{\lambda} = \epsilon_{\lambda, A} b C_A + \epsilon_{\lambda, I} b C_I \nonumber \]where \(\epsilon_{\lambda, I}\) is the interferent’s molar absorptivity and CI is the interferent’s concentration. This is an example of multivariable regression, which is covered in more detail in Chapter 9 when we consider the optimization of experiments where there is a single dependent variable and two or more independent variables.In multivariate regression we have both multiple dependent variables, such as the absorbance of samples at two or more wavelengths, and multiple independent variables, such as the concentrations of two or more analytes in the samples. As discussed in Chapter 0.2, we can represent this using matrix notation\[\begin{bmatrix} \cdots & \cdots & \cdots \\ \vdots & A & \vdots \\ \cdots & \cdots & \cdots \end{bmatrix}_{r \times c} = \begin{bmatrix} \cdots & \cdots & \cdots \\ \vdots & \epsilon b & \vdots \\ \cdots & \cdots & \cdots \end{bmatrix}_{r \times n} \times \begin{bmatrix} \cdots & \cdots & \cdots \\ \vdots & C & \vdots \\ \cdots & \cdots & \cdots \end{bmatrix}_{n \times c} \nonumber \]where there are \(r\) wavelengths, \(c\) samples, and \(n\) analytes. Each column in the \(\epsilon b\) matrix, for example, holds the \(\epsilon b\) value for a different analyte at one of \(r\) wavelengths, and each row in the \(C\) matrix is the concentration of one of the \(n\) analytes in one of the \(c\) samples. We will consider this approach in more detail in Chapter 11.For a nice discussion of the difference between multivariable regression and multivariate regression, see Hidalgo, B.; Goodman, M. "Multivariate or Multivariable Regression," Am. J. Public Health, 2013, 103, 39-40.This page titled 35.1: Evaluation of Analytical Data is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

35.2: Single-Sided Normal Distribution

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Table \(\PageIndex{1}\), at the bottom of this appendix, gives the proportion, P, of the area under a normal distribution curve that lies to the right of a deviation, z \[z = \frac {X -\mu} {\sigma} \nonumber \]where X is the value for which the deviation is defined, \(\mu\) is the distribution’s mean value and \(\sigma\) is the distribution’s standard deviation. For example, the proportion of the area under a normal distribution to the right of a deviation of 0.04 is 0.4840 (see entry in red in the table), or 48.40% of the total area (see the area shaded blue in ). The proportion of the area to the left of the deviation is 1 – P. For a deviation of 0.04, this is 1 – 0.4840, or 51.60%. . Normal distribution curve showing the area under a curve greater than a deviation of +0.04 (blue) and with a deviation less than –0.04 (green).When the deviation is negative—that is, when X is smaller than \(\mu\)—the value of z is negative. In this case, the values in the table give the area to the left of z. For example, if z is –0.04, then 48.40% of the area lies to the left of the deviation (see area shaded green in .To use the single-sided normal distribution table, sketch the normal distribution curve for your problem and shade the area that corresponds to your answer (for example, see , which is for Example 4.4.2).This divides the normal distribution curve into three regions: the area that corresponds to our answer (shown in blue), the area to the right of this, and the area to the left of this. Calculate the values of z for the limits of the area that corresponds to your answer. Use the table to find the areas to the right and to the left of these deviations. Subtract these values from 100% and, voilà, you have your answer.This page titled 35.2: Single-Sided Normal Distribution is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

35.3: Critical Values for t-Test

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Assuming we have calculated texp, there are two approaches to interpreting a t-test. In the first approach we choose a value of \(\alpha\) for rejecting the null hypothesis and read the value of \(t(\alpha,\nu)\) from the table below. If \(t_\text{exp} > t(\alpha,\nu)\), we reject the null hypothesis and accept the alternative hypothesis. In the second approach, we find the row in the table below that corresponds to the available degrees of freedom and move across the row to find (or estimate) the a that corresponds to \(t_\text{exp} = t(\alpha,\nu)\); this establishes largest value of \(\alpha\) for which we can retain the null hypothesis. Finding, for example, that \(\alpha\) is 0.10 means that we retain the null hypothesis at the 90% confidence level, but reject it at the 89% confidence level. The examples in this textbook use the first approach.The values in this table are for a two-tailed t-test. For a one-tailed test, divide the \(\alpha\) values by 2. For example, the last column has an \(\alpha\) value of 0.005 and a confidence interval of 99.5% when conducting a one-tailed t-test.This page titled 35.3: Critical Values for t-Test is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

35.4: Critical Values for F-Test

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The following tables provide values for \(F(0.05, \nu_\text{num}, \nu_\text{denom})\) for one-tailed and for two-tailed F-tests. To use these tables, we first decide whether the situation calls for a one-tailed or a two-tailed analysis and calculate Fexp\[F_\text{exp} = \frac {s_A^2} {s_B^2} \nonumber \]where \(S_A^2\) is greater than \(s_B^2\). Next, we compare Fexp to \(F(0.05, \nu_\text{num}, \nu_\text{denom})\) and reject the null hypothesis if \(F_\text{exp} > F(0.05, \nu_\text{num}, \nu_\text{denom})\). You may replace s with \(\sigma\) if you know the population’s standard deviation.This page titled 35.4: Critical Values for F-Test is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

35.5: Critical Values for Dixon's Q-Test

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The following table provides critical values for \(Q(\alpha, n)\), where \(\alpha\) is the probability of incorrectly rejecting the suspected outlier and \(n\) is the number of samples in the data set. There are several versions of Dixon’s Q-Test, each of which calculates a value for Qij where i is the number of suspected outliers on one end of the data set and j is the number of suspected outliers on the opposite end of the data set. The critical values for Q here are for a single outlier, Q10, where\[Q_\text{exp} = Q_{10} = \frac {|\text{outlier's value} - \text{nearest value}|} {\text{largest value} - \text{smallest value}} \nonumber \]The suspected outlier is rejected if Qexp is greater than \(Q(\alpha, n)\). For additional information consult Rorabacher, D. B. “Statistical Treatment for Rejection of Deviant Values: Critical Values of Dixon’s ‘Q’ Parameter and Related Subrange Ratios at the 95% confidence Level,” Anal. Chem. 1991, 63, 139–146.This page titled 35.5: Critical Values for Dixon's Q-Test is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

35.6: Critical Values for Grubb's Test

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The following table provides critical values for \(G(\alpha, n)\), where \(\alpha\) is the probability of incorrectly rejecting the suspected outlier and n is the number of samples in the data set. There are several versions of Grubb’s Test, each of which calculates a value for Gij where i is the number of suspected outliers on one end of the data set and j is the number of suspected outliers on the opposite end of the data set. The critical values for G given here are for a single outlier, G10, where\[G_\text{exp} = G_{10} = \frac {|X_{out} - \overline{X}|} {s} \nonumber \]The suspected outlier is rejected if Gexp is greater than \(G(\alpha, n)\).This page titled 35.6: Critical Values for Grubb's Test is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

35.7: Activity Coefficients

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Careful measurements on the metal–ligand complex Fe(SCN)2+ suggest its stability decreases in the presence of inert ions [Lister, M. W.; Rivington, D. E. Can. J. Chem. 1995, 33, 1572–1590]. We can demonstrate this by adding an inert salt to an equilibrium mixture of Fe3+ and SCN–. Figure 35.7.1 a shows the result of mixing together equal volumes of 1.0 mM FeCl3 and 1.5 mM KSCN, both of which are colorless. The solution’s reddish–orange color is due to the formation of Fe(SCN)2+.\[\mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{SCN})^{2+}(a q) \label{6.1} \]Adding 10 g of KNO3 to the solution and stirring to dissolve the solid, produces the result shown in Figure 35.7.1 b. The solution’s lighter color suggests that adding KNO3 shifts reaction \ref{6.1} to the left, decreasing the concentration of Fe(SCN)2+ and increasing the concentrations of Fe3+ and SCN–. The result is a decrease in the complex’s formation constant, K1.\[K_{1}=\frac{\left[\mathrm{Fe}(\mathrm{SCN})^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]} \label{6.2} \]Why should adding an inert electrolyte affect a reaction’s equilibrium position? We can explain the effect of KNO3 on the formation of Fe(SCN)2+ if we consider the reaction on a microscopic scale. The solution in Figure 35.7.1 b contains a variety of cations and anions: Fe3+, SCN–, K+, \(\text{NO}_3^-\), H3O+, and OH–. Although the solution is homogeneous, on average, there are slightly more anions in regions near the Fe3+ ions, and slightly more cations in regions near the SCN– ions. As shown in Figure 35.7.2 , each Fe3+ ion and each SCN– ion is surrounded by an ionic atmosphere of opposite charge (\(\delta^–\) and \(\delta^+\)) that partially screen the ions from each other. Because each ion’s apparent charge at the edge of its ionic atmosphere is less than its actual charge, the force of attraction between the two ions is smaller. As a result, the formation of Fe(SCN)2+ is slightly less favorable and the formation constant in Equation \ref{6.2} is slightly smaller. Higher concentrations of KNO3 increase \(\delta^–\) and \(\delta^+\), resulting in even smaller values for the formation constant.To factor the concentration of ions into the formation constant for Fe(SCN)2+, we need a way to express that concentration in a meaningful way. Because both an ion’s concentration and its charge are important, we define the solution’s ionic strength, \(\mu\) as\[\mu=\frac{1}{2} \sum_{i=1}^{n} c_{i} z_{i}^{2} \nonumber \]where ci and zi are the concentration and charge of the ith ion.Calculate the ionic strength of a solution of 0.10 M NaCl. Repeat the calculation for a solution of 0.10 M Na2SO4.The ionic strength for 0.10 M NaCl is\[\begin{array}{c}{\mu=\frac{1}{2}\left\{\left[\mathrm{Na}^{+}\right] \times(+1)^{2}+\left[\mathrm{Cl}^{-}\right] \times(-1)^{2}\right\}} \\ {\mu=\frac{1}{2}\left\{(0.10) \times(+1)^{2}+(0.10) \times(-1)^{2}\right\}=0.10 \ \mathrm{M}}\end{array} \nonumber \]For 0.10 M Na2SO4 the ionic strength is\[\begin{array}{c}{\mu=\frac{1}{2}\left\{\left[\mathrm{Na}^{+}\right] \times(+1)^{2}+\left[\mathrm{SO}_{4}^{2-}\right] \times(-2)^{2}\right\}} \\ {\mu=\frac{1}{2}\left\{(0.20) \times(+1)^{2}+(0.10) \times(-2)^{2}\right\}=0.30 \ \mathrm{M}}\end{array} \nonumber \]In calculating the ionic strengths of these solutions we are ignoring the presence of H3O+ and OH–, and, in the case of Na2SO4, the presence of \(\text{HSO}_4^-\) from the base dissociation reaction of \(\text{SO}_4^{2-}\). In the case of 0.10 M NaCl, the concentrations of H3O+ and OH– are \(1.0 \times 10^{-7}\), which is significantly smaller than the concentrations of Na+ and Cl–. Because \(\text{SO}_4^{2-}\) is a very weak base (Kb = \(1.0 \times 10^{-12}\)), the solution is only slightly basic (pH = 7.5), and the concentrations of H3O+, OH–, and \(\text{HSO}_4^-\) are negligible. Although we can ignore the presence of H3O+, OH–, and \(\text{HSO}_4^-\) when we calculate the ionic strength of these two solutions, be aware that an equilibrium reaction can generate ions that might affect the solution’s ionic strength.Note that the unit for ionic strength is molarity, but that a salt’s ionic strength need not match its molar concentration. For a 1:1 salt, such as NaCl, ionic strength and molar concentration are identical. The ionic strength of a 2:1 electrolyte, such as Na2SO4, is three times larger than the electrolyte’s molar concentration.Figure 35.7.1 shows that adding KNO3 to a mixture of Fe3+ and SCN– decreases the formation constant for Fe(SCN)2+. This creates a contradiction. Earlier in this chapter we showed that there is a relationship between a reaction’s standard‐state free energy, ∆Go, and its equilibrium constant, K.\[\triangle G^{\circ}=-R T \ln K \nonumber \]Because a reaction has only one standard‐state, its equilibrium constant must be independent of solution conditions. Although ionic strength affects the apparent formation constant for Fe(SCN)2+, reaction \ref{6.1} must have an underlying thermodynamic formation constant that is independent of ionic strength.The apparent formation constant for Fe(SCN)2+, as shown in Equation \ref{6.2}, is a function of concentrations. In place of concentrations, we define the true thermodynamic equilibrium constant using activities. The activity of species A, aA, is the product of its concentration, [A], and a solution‐dependent activity coefficient, \(\gamma_A\)\[a_{A}=[A] \gamma_{A} \nonumber \]The true thermodynamic formation constant for Fe(SCN)2+, therefore, is\[K_{1}=\frac{a_{\mathrm{Fe}(S \mathrm{CN})^{2+}}}{a_{\mathrm{Fe}^{3+}} \times a_{\mathrm{SCN}^-}}=\frac{\left[\mathrm{Fe}(\mathrm{SCN})^{2+}\right] \gamma_{\mathrm{Fe}(\mathrm{SCN})^{2+}}}{\left[\mathrm{Fe}^{3+}\right] \gamma_{\mathrm{Fe}^{3+}}\left[\mathrm{SCN}^{-}\right] \gamma_{\mathrm{SCN}^{-}}} \nonumber \]Unless otherwise specified, the equilibrium constants in the appendices are thermodynamic equilibrium constants.A species’ activity coefficient corrects for any deviation between its physical concentration and its ideal value. For a gas, a pure solid, a pure liquid, or a non‐ionic solute, the activity coefficient is approximately one under most reasonable experimental conditions.For a gas the proper terms are fugacity and fugacity coefficient, instead of activity and activity coefficient.For a reaction that involves only these species, the difference between activity and concentration is negligible. The activity coefficient for an ion, however, depends on the solution’s ionic strength, the ion’s charge, and the ion’s size. It is possible to estimate activity coefficients using the extended Debye‐Hückel equation\[\log \gamma_{A}=\frac{-0.51 \times z_{A}^{2} \times \sqrt{\mu}}{1+3.3 \times \alpha_{A} \times \sqrt{\mu}} \label{6.3} \]where zA is the ion’s charge, \(\alpha_A\) is the hydrated ion’s effective diameter in nanometers (Table 6.2), \(\mu\) is the solution’s ionic strength, and 0.51 and 3.3 are constants appropriate for an aqueous solution at 25oC. A hydrated ion’s effective radius is the radius of the ion plus those water molecules closely bound to the ion. The effective radius is greater for smaller, more highly charged ions than it is for larger, less highly charged ions.Several features of Equation \ref{6.3} deserve our attention. First, as the ionic strength approaches zero an ion’s activity coefficient approaches a value of one. In a solution where \(\mu = 0\), an ion’s activity and its concentration are identical. We can take advantage of this fact to determine a reaction’s thermodynamic equilibrium constant by measuring the apparent equilibrium constant for several increasingly smaller ionic strengths and extrapolating back to an ionic strength of zero. Second, an activity coefficient is smaller, and the effect of activity is more important, for an ion with a higher charge and a smaller effective radius. Finally, the extended Debye‐Hückel equation provides a reasonable estimate of an ion’s activity coefficient when the ionic strength is less than 0.1. Modifications to Equation \ref{6.3} extend the calculation of activity coefficients to higher ionic strengths [Davies, C. W. Ion Association, Butterworth: London, 1962].Earlier in this chapter we calculated the solubility of Pb(IO3)2 in deionized water, obtaining a result of \(4.0 \times 10^{-5}\) mol/L. Because the only significant source of ions is from the solubility reaction, the ionic strength is very low and we can assume that \(\gamma \approx 1\) for both Pb2+ and \(\text{IO}_3^-\). In calculating the solubility of Pb(IO3)2 in deionized water, we do not need to account for ionic strength. But what if we need to know the solubility of Pb(IO3)2 in a solution that contains other, inert ions? In this case we need to include activity coefficients in our calculation.Calculate the solubility of Pb(IO3)2 in a matrix of 0.020 M Mg(NO3)2.We begin by calculating the solution’s ionic strength. Since Pb(IO3)2 is only sparingly soluble, we will assume we can ignore its contribution to the ionic strength; thus\[\mu=\frac{1}{2}\left\{(0.020)(+2)^{2}+(0.040)(-1)^{2}\right\}=0.060 \ \mathrm{M} \nonumber \]Next, we use Equation \ref{6.3} to calculate the activity coefficients for Pb2+ and \(\text{IO}_3^-\).\[\log \gamma_{\mathrm{Pb}^{2+}}=\frac{-0.51 \times(+2)^{2} \times \sqrt{0.060}}{1+3.3 \times 0.45 \times \sqrt{0.060}}=-0.366 \nonumber \]\[\gamma_{\mathrm{Pb}^{2+}}=0.431 \nonumber \]\[\log \gamma_{\mathrm{IO}_{3}^{-}}=\frac{-0.51 \times(-1)^{2} \times \sqrt{0.060}}{1+3.3 \times 0.45 \times \sqrt{0.060}}=-0.0916 \nonumber \]\[\gamma_{\mathrm{IO}_{3}^-}=0.810 \nonumber \]Defining the equilibrium concentrations of Pb2+ and \(\text{IO}_3^-\) in terms of the variable x and substituting into the thermodynamic solubility product for Pb(IO3)2 leaves us with\[K_{\mathrm{sp}}=a_{\mathrm{Pb}^{2+}} \times a_{\mathrm{IO}_{3}^-}^{2}=\gamma_{\mathrm{Pb}^{2+}}\left[\mathrm{Pb}^{2+}\right] \times \gamma_{\mathrm{IO}_3^-}^{2}\left[\mathrm{IO}_{3}^{-}\right]^{2}=2.5 \times 10^{-13} \nonumber \]\[K_{\mathrm{sp}}=(0.431)(x)(0.810)^{2}(2 x)^{2}=2.5 \times 10^{-13} \nonumber \]\[K_{\mathrm{sp}}=1.131 x^{3}=2.5 \times 10^{-13} \nonumber \]Solving for x gives \(6.0 \times 10^{-5}\) and a molar solubility of \(6.0 \times 10^{-5}\) mol/L for Pb(IO3)2. If we ignore activity, as we did in our earlier calculation, we report the molar solubility as \(4.0 \times 10^{-5}\) mol/L. Failing to account for activity in this case underestimates the molar solubility of Pb(IO3)2 by 33%.The solution’s equilibrium composition is\[\begin{array}{c}{\left[\mathrm{Pb}^{2+}\right]=6.0 \times 10^{-5} \ \mathrm{M}} \\ {\left[\mathrm{IO}_{3}^{-}\right]=1.2 \times 10^{-4} \ \mathrm{M}} \\ {\left[\mathrm{Mg}^{2+}\right]=0.020 \ \mathrm{M}} \\ {\left[\mathrm{NO}_{3}^{-}\right]=0.040 \ \mathrm{M}}\end{array} \nonumber \]Because the concentrations of both Pb2+ and \(\text{IO}_3^-\) are much smaller than the concentrations of Mg2+ and \(\text{NO}_3^-\) our decision to ignore the contribution of Pb2+ and \(\text{IO}_3^-\) to the ionic strength is reasonable.How do we handle the calculation if we can not ignore the concentrations of Pb2+ and \(\text{IO}_3^-\) when calculating the ionic strength. One approach is to use the method of successive approximations. First, we recalculate the ionic strength using the concentrations of all ions, including Pb2+ and \(\text{IO}_3^-\). Next, we recalculate the activity coefficients for Pb2+ and \(\text{IO}_3^-\) using this new ionic strength and then recalculate the molar solubility. We continue this cycle until two successive calculations yield the same molar solubility within an acceptable margin of error.Calculate the molar solubility of Hg2Cl2 in 0.10 M NaCl, taking into account the effect of ionic strength. Compare your answer to that from Exercise 6.7.2 in which you ignored the effect of ionic strength.We begin by calculating the solution’s ionic strength. Because NaCl is a 1:1 ionic salt, the ionic strength is the same as the concentration of NaCl; thus \(\mu\) = 0.10 M. This assumes, of course, that we can ignore the contributions of \(\text{Hg}_2^{2+}\) and Cl– from the solubility of Hg2Cl2.Next we use Equation \ref{6.3} to calculate the activity coefficients for \(\text{Hg}_2^{2+}\) and Cl–.\[\log \gamma_{\mathrm{Hg}_{2}^{2+}}=\frac{-0.51 \times(+2)^{2} \times \sqrt{0.10}}{1+3.3 \times 0.40 \times \sqrt{0.10}}=-0.455 \nonumber \]\[\gamma_{\mathrm{H} \mathrm{g}_{2}^{2+}}=0.351 \nonumber \]\[\log \gamma_{\mathrm{Cl}^{-}}=\frac{-0.51 \times(-1)^{2} \times \sqrt{0.10}}{1+3.3 \times 0.3 \times \sqrt{0.10}}=-0.12 \nonumber \]\[\gamma_{\mathrm{Cl}^-}=0.75 \nonumber \]Defining the equilibrium concentrations of \(\text{Hg}_2^{2+}\) and Cl– in terms of the variable xand substituting into the thermodynamic solubility product for Hg2Cl2, leave us with\[K_{\mathrm{sp}}=a_{\mathrm{Hg}_{2}^{2+}}\left(a_{\mathrm{Cl}^-}\right)^{2} = \gamma_{\mathrm{Hg}_{2}^{2+}}\left[\mathrm{Hg}_{2}^{2+}\right]\left(\gamma_{\mathrm{Cl}^{-}}\right)^{2}\left[\mathrm{Cl}^{-}\right]^{2}=1.2 \times 10^{-18} \nonumber \]Because the value of x likely is small, let’s simplify this equation to\[(0.351)(x)(0.75)^{2}(0.1)^{2}=1.2 \times 10^{-18} \nonumber \]Solving for x gives its value as \(6.1 \times 10^{-16}\). Because x is the concentration of \(\text{Hg}_2^{2+}\) and 2x is the concentration of Cl–, our decision to ignore their contributions to the ionic strength is reasonable. The molar solubility of Hg2Cl2 in 0.10 M NaCl is \(6.1 \times 10^{-16}\) mol/L. In Exercise 6.7.2, where we ignored ionic strength, we determined that the molar solubility of Hg2Cl2 is \(1.2 \times 10^{-16}\) mol/L, a result that is \(5 \times\) smaller than the its actual value.As Example 35.7.2 and Exercise 35.7.1 show, failing to correct for the effect of ionic strength can lead to a significant error in an equilibrium calculation. Nevertheless, it is not unusual to ignore activities and to assume that the equilibrium constant is expressed in terms of concentrations. There is a practical reason for this—in an analysis we rarely know the exact composition, much less the ionic strength of aqueous samples or of solid samples brought into solution. Equilibrium calculations are a useful guide when we develop an analytical method; however, it only is when we complete an analysis and evaluate the results that can we judge whether our theory matches reality. In the end, work in the laboratory is the most critical step in developing a reliable analytical method.This is a good place to revisit the meaning of pH. In Chapter 2 we defined pH as\[\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \nonumber \]Now we see that the correct definition is\[\begin{array}{c}{\mathrm{pH}=-\log a_{\mathrm{H}_{3} \mathrm{O}^{+}}} \\ {\mathrm{pH}=-\log \gamma_{\mathrm{H}_{3} \mathrm{O}^{+}}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}\end{array} \nonumber \]Failing to account for the effect of ionic strength can lead to a significant error in the reported concentration of H3O+. For example, if the pH of a solution is 7.00 and the activity coefficient for H3O+ is 0.90, then the concentration of H3O+ is \(1.11 \times 10^{-7}\) M, not \(1.00 \times 10^{-7}\) M, an error of +11%. Fortunately, when we develop and carry out an analytical method, we are more interested in controlling pH than in calculating [H3O+]. As a result, the difference between the two definitions of pH rarely is of significant concern.This page titled 35.7: Activity Coefficients is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

35.8: Standard Reduction Potentials & Polarographic Half-wave Potentials

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Standard/Formal Reduction PotentialsThe following table provides Eo and Eo ́ values for selected reduction reactions. Values are from the following sources (primarily the first two):Solids, gases, and liquids are identified; all other species are aqueous. Reduction reactions in acidic solution are written using H+ in place of H3O+. You may rewrite a reaction by replacing H+ with H3O+ and adding to the opposite side of the reaction one molecule of H2O per H+; thusH3AsO4 + 2H+ +2e– \(\rightleftharpoons\) HAsO2 +2H2ObecomesH3AsO4 + 2H3O+ +2e– \(\rightleftharpoons\) HAsO2 +4H2OConditions for formal potentials (Eo ́) are listed next to the potential.For most of the reduction half-reactions gathered here, there are minor differences in values provided by the references above. In most cases, these differences are small and will not affect calculations. In a few cases the differences are not insignificant and the user may find discrepancies in calculations. For example, Bard, Parsons, and Jordon report an Eo value of –1.285 V for\[\text{Zn(OH)}_4^{2-} + 2e^- \rightleftharpoons \text{Zn}(s) + 4\text{OH}^-\nonumber \]while Milazzo, Caroli, and Sharma report the value as –1.214 V, Swift reports the value as –1.22, Bratsch reports the value as –1.199 V, and Latimer reports the value as –1.216 V.1.70 in 1 M HClO41.44 in 1 M H2SO41.61 in 1 M HNO31.28 in 1 M HCl0.70 in 1 M HCl0.767 in 1 M HClO40.746 in 1 M HNO30.68 in 1 M H2SO40.44 in 0.3 M H3PO41.25 in 1 M HClO40.77 in 1 M HClThe following table provides E1/2 values for selected reduction reactions. Values are from Dean, J. A. Analytical Chemistry Handbook, McGraw-Hill: New York, 1995.0.1 M KCl0.050 M H2SO41 M HNO3–0.35 \((+3 \ce{->} +2)\)–1.70 \((+2 \ce{->} 0)\)1 M NH4Cl plus 1 M NH31 M NH4+/NH3 buffer (pH 8–9)–0.5 \((+3 \ce{->} +2)\)–1.3 \((+2 \ce{->} 0)\)0.04–0.220.1 M KSCN0.1 M NH4ClO41 M Na2SO40.5 M potassium citrate (pH 7.5)–0.17 \((+3 \ce{->} +2)\)–1.52 \((+2 \ce{->} 0)\)–0.405–0.4351 M HNO31 M KCl–0.70–1.091 M KSCN1 M NH4Cl plus 1 M NH3–0.995–1.330.1 M KCl1 M NH4Cl plus 1 M NH3This page titled 35.8: Standard Reduction Potentials & Polarographic Half-wave Potentials is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

35.9: Recommended Primary Standards

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All compounds are of the highest available purity. Metals are cleaned with dilute acid to remove any surface impurities and rinsed with distilled water. Unless otherwise indicated, compounds are dried to a constant weight at 110 oC. Most of these compounds are soluble in dilute acid (1:1 HCl or 1:1 HNO3), with gentle heating if necessary; some of the compounds are water soluble.Sources:This page titled 35.9: Recommended Primary Standards is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

4.1: Analog and Digital Data

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shows an xy-recorder that we can use to provide a permanent record of a cyclic voltammetry experiment. In this particular experiment, we apply a variable potential to an electrochemical cell and measure the current that flows in response to this potential (see Chapter 25 for a discussion of cyclic voltammetry). The potential and the current, which is converted into a voltage for the purpose of recording the cyclic voltammogram, are fed into the recorder using the cables on the right side of the recorder. The Y1 Range and the X Range controls allow us to adjust the scales of the axes.The vertical bar on the xy-recorder moves toward the recorder's left or right based on the applied potential, and a pen attached to the vertical bar moves toward the recorder's top or bottom based on the measured current. The applied potential and the current are continuous variables within the instrument's range; the resulting cyclic voltammogram in is an analog record of the experiment.Although the analog trace in provides a permanent record of an experiment, it is not in form that gives us access to the raw data. We can take the image and use digitizing software (see here for an open-source digitizer) to extract a digital version of the data, or we can design our instruments to collect the data in digital form by sampling the analog signal at preset intervals and then saving the data. Such files often are in a format that includes metadata that explains how to extract the data from the file. For example, xy-coordinate data for a wide variety of spectroscopy experiments is often stored digitally using a format established by the Joint Committee on Atomic and Molecular Physical Data (JCAMP). Such files have the extension .jdx and can be opened using a variety of different software programs. is a screenshot that illustrates how we can work with digitized data using data analysis software, such as R and RStudio. The upper left panel shows some of the contents of a .jdx file that contains the IR spectrum of methanol (in this case, digitized by NIST from an analog hard copy). The lines preceded by double hashtags (##) are metadata that provide information about the x-axis scale (minimum and maximum limits and increments between values), the y-axis scale (minimum and maximum values), and the number of data points. This is followed by multiple lines of digitized data. Each line of data contains one value of x and five values of y. The R package readJDX was used to extract the information from the .jdx file and to store it in a variable given the name methanol (see upper right panel). Code written in R (see lower left panel) was used to plot (see lower right panel) the spectrum.Although the spectrum for methanol in —with its smooth, continuous line—looks like an analog spectrum, this is a result of choosing to plot the data as a sequence of lines that connect individual points without actually displaying the individual points themselves. , in which we plot only the individual data points, shows us that the spectrum actually consists of discrete, digitized data.This page titled 4.1: Analog and Digital Data is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

4.2: Working With Binary Numbers

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In Chapter 7 we will examine several transducers for counting photons. The transducers are made of an array—some use a linear array and some use a two-dimensional array—of individual detecting units. We will worry about the details of how these transducers work in Chapter 7, but if you take a quick look at individual units. What is interesting about these numbers is that each is a power of two: \(1024 = 2^{10}\), \(2048 = 2^{11}\), and \(1,048,576 = 10^{20}\).Humans are comfortable working with numbers expressed using a decimal notation that relies on 10 unique digits (0, 1, 2, 3, 4, 5, 6, 7, 8, and 9), but computers work with information using a binary notation that is limited to just two unique digits (0 and 1). Although we will not complete calculations using binary numbers, you will see examples of instrumental methods, such as FT-NMR, where the data analysis algorithms (the Fourier transform in this case) require that the number of data points be a power of two. It is useful, therefore, to be familiar with how we represent numbers in both decimal and binary form.My university was founded in 1837, which is a decimal expression of the year. Each of these four digits represents a power of 10, a fact that is clear when we read the number out loud: one thousand—eight hundred— thirty—seven, or, when we write it out this way\[(1 \times 1000) + (8 \times 100) + (3 \times 10) + (7 \times 1) = 1837 \nonumber \]or this way\[(1 \times 10^3) + (8 \times 10^2) + (3 \times 10^1) + (7 \times 10^0) = 1837 \nonumber \]We refer to the 7 being in the ones place (\(10^0 = 1\)), the 3 in the tens place (\(10^1 = 10\)), the 8 in the hundreds place (\(10^2 = 100\)), and the 1 in the thousands place (\(10^3 = 1000\)). shows these three ways of representing a number using a decimal notation.The decimal number 1837 is 11100101101 in binary notation. We can see that this is true if we follow the pattern for decimal numbers in reverse. There are eleven binary digits, so we begin by expressing the number as multiples of the powers of two from \(2^{10}\) to \(2^{0}\), beginning with the digit furthest to the left and moving to the right\[(1 \times 2^{10}) + (1 \times 2^{9}) + (1 \times 2^{8}) + (0 \times 2^{7}) + (0 \times 2^{6}) + (1 \times 2^{5}) + (0 \times 2^{4}) + (1 \times 2^{3}) + (1 \times 2^{2}) + (0 \times 2^{1}) + (1 \times 2^{0}) = 1837 \nonumber \]Each power of two has a decimal equivalent—\(2^4\) is the same as \(2 \times 2 \times 2 \times 2 = 16\), for example—which we can express here as\[(1 \times 1024) + (1 \times 512) + (1 \times 256) + (0 \times 128) + (0 \times 64) + (1 \times 32) + (0 \times 16) + (1 \times 8) + (1 \times 4) + (0 \times 2) + (1 \times 1) = 1837 \nonumber \]Each power of two represents a place as well; thus, the second 0 from the right is in the sixteenths place. provides a visual representation of these ways of expressing a binary number.There are lots of on-line calculators that you can use to convert between decimal and binary representations of numbers, such as the one here. Still, it is useful to be comfortable with converting numbers by hand. Converting a binary number into its decimal equivalent is straightforward, as we showed above for the binary representation of the year in which my university was founded\[11100101101 = (1 \times 1024) + (1 \times 512) + (1 \times 256) + (0 \times 128) + (0 \times 64) + (1 \times 32) + (0 \times 16) + (1 \times 8) + (1 \times 4) + (0 \times 2) + (1 \times 1) = 1837 \nonumber \]Converting a decimal number, such as 1837, into its binary equivalent requires a bit more work; Table \(\PageIndex{1}\) will help us organize the conversion. We begin by writing the dividend, which is 1837, in the left-most column and divide it by 2, writing the quotient of 918 in the second column and the remainder of 1 in the third column; note that dividing by 2 gives a remainder of 0 if the dividend is even or a remainder of 1 if the dividend is odd. The remainder is the exponent for the first place in the binary notation. In this case, we have \(2^0 = 1\). The quotient becomes the dividend for the next cycle, with the process continuing until we achieve a quotient of 0. The binary equivalent of the original decimal is given by reading the remainders from bottom-to-top as 11100101101.This page titled 4.2: Working With Binary Numbers is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

4.3: Cleaning Up Signals and Counting Events

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How an instrument handles signals depends on what is being measured, so we cannot develop here a single model that applies to all instruments. Broadly speaking, however, an instrument is likely to include one or more of the following: the ability to clean up the raw signal and convert it into a form that we can analyze; the ability to count events in binary form; the ability to convert binary information into a digital information; and the ability to convert between digital and analog signals. In this section we will cover the first two of these topics.Suppose our instrument is designed to count discrete events, perhaps a Geiger counter that detects the emission of \(\beta\) particles, or a photodiode that detects photons. Even though a time-dependent count of particles is a digital signal, the raw signal (a voltage) likely consists of digital pulses superimposed on a background signal that contains noise, as seen in . The total signal, therefore, is in analog form.To clean up this signal we want to accomplish two things: remove the noise and ensure that each pulse is counted. A simple way to accomplish this is to set a threshold signal and use a voltage follower operational amplifier (see Chapter 3) to set all voltages below the threshold to a logical value of 0 and all voltages above the threshold to a logical value of 1. As seen in , the choice of the threshold voltage must be chosen carefully if we are to resolve closely spaced pulses and discriminate against noise. Note that the peak-shaped pulses become rectangular pulses.To count the pulses in we can send them though a binary pulse counter (BPC). shows how such a counter works. In this case, the BPC has three registers, each of which can be in a logical state of 0 or 1. With three registers, we are limited to counting no more than \(2^3 = 8\) pulses; a more useful BPC would have more registers. We can treat the pulses as entering the BPC from the right. When a pulse enters a register, it flips each register from 1 to 0 or from 0 to 1, stopping after if first flips a register from 0 to 1. For example, the second pulse flips the right-most register from 1 to 0 and the middle register from 0 to 1; because the middle register initially was at 0, the counting of this pulse comes to an end.This page titled 4.3: Cleaning Up Signals and Counting Events is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

5.1: The Signal-to-Noise Ratio

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When we make a measurement it is the sum of two parts, a determinate, or fixed contribution that arises from the analyte and an indeterminate, or random, contribution that arises from uncertainty in the measurement process. We call the first of these the signal and we call the latter the noise. There are two broad categories of noise: that associated with obtaining samples and that associated with making measurements. Our interest here is in the latter.Noise is a random event characterized by a mean and standard deviation. There are many types of noise, but we will limit ourselves for now to noise that is stationary, in that its mean and its standard deviation are independent of time, and that is heteroscedastic, in that its mean and its variance (and its standard deviation) are independent of the signal's magnitude. shows an example of a noisy signal that meets these criteria. The x-axis here is shown as time—perhaps a chromatogram—but other units, such as wavelength (spectroscopy) or potential (electrochemistry), are possible. shows the underlying noise and shows the underlying signal. Note that the noise in appears consistent in its central tendency (mean) and its spread (variance) along the x-axis and is independent of the signal's strength.Although we characterize noise by its mean and its standard deviation, the most important benchmark is the signal-to-noise ratio, \(S/N\), which we define as\[S/N = \frac{S_\text{analyte}}{s_\text{noise}} \nonumber \]where \(S_\text{analyte}\) is the signal's value at particular location on the x-axis and \(s_\text{noise}\) is the standard deviation of the noise using a signal-free portion of the data. As general rules-of-thumb, we can measure the signal with some confidence when \(S/N \ge 3\) and we can detect the signal with some confidence when \(3 \ge S/N \ge 2\). For the data in , and using the information in the figure caption, the signal-to-noise ratios are, from left-to-right, 10, 6, and 3.To measure the signal with confidence implies we can use the signal's value in a calculation, such as constructing a calibration curve. To detect the signal with confidence means we are certain that a signal is present (and that an analyte responsible for the signal is present) even if we cannot measure the signal with sufficient confidence to allow for a meaningful calculation.This page titled 5.1: The Signal-to-Noise Ratio is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

5.2: Sources of Instrumental Noise

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When we make an analytical measurement, we are interested in both the accuracy and the precision of our results. Noise, as we learned in the previous section, is a random fluctuation in the signal that limits our ability to detect the presence of the underlying signal. There are a variety of ways in which noise can enter into our measurements. Some of these sources of noise are related to the process of collecting and processing samples for analysis; these sources of noise, which we might collectively call chemical sources of noise, are important and receive consideration in those sections of this textbook that consider the application of analytical methods. In this chapter, we will limit ourselves to considering sources of noise that arise from the instruments we use to make measurements. We call these sources of instrumental noise.Even when an external voltage is not applied to an electrical circuit, a small current is present due to the random motion of electrons that arises from the temperature of the surroundings; we can this thermal (or, sometimes, Johnson) noise. The magnitude of this noise in any electrical element increases with temperature, of course, but it also is affected by its resistance, and by how quickly it responds to a change in the signal. Mathematically, we express this as the root-mean-square voltage, \(\nu_{\text{rms}}\), which is given as\[\nu_{\text{rms}} = \sqrt{4 k T R \Delta f} \label{thermal} \]where \(k\) is Boltzmann's constant, \(T\) is the temperature in Kelvin, \(R\) is the resistance in ohms, and \(\Delta f\) is the bandwidth. The latter term is a measure of how quickly the electrical element responds to a change in its input by changing its output from 10% to 90% of its final value, which is called the rise time, \(t_r\), where\[\Delta f = \frac{1}{3 t_r} \nonumber \]For example, if a change in the input increases the output by 1, then the rise time is how long it takes the output to increase from 0.1 to 0.9.A close look at Equation \ref{thermal} shows that we can reduce thermal noise by decreasing the temperature, by decreasing the resistance of the electrical circuit, and by decreasing the bandwidth; the latter, of course, comes at the cost of an increase in the response time, which means the instrument responds more slowly to a change in the signal. Of these, it is often easiest to reduce the temperature by cooling, for example, the instrument's detector.As its name implies, shot noise is a discrete event that happens in response to an event, such as the movement of an electron through the space between two surfaces of opposite charge. These events are random and quantized, and generate random furcations in the current that have a root-mean-square value, \(i_{\text{rms}}\), which is given by\[i_{\text{rms}} = \sqrt{2 I e \Delta f} \label{shot} \]where \(I\) is the average current, \(e\) is the charge on the electron in Coulombs, and \(\Delta f\) is the bandwidth. Of these terms, the only one under our control is the bandwidth; again, decreasing the bandwidth comes at the cost of an instrument that responds more slowly to a change in the signal.Unlike thermal noise or shot noise, flicker noise is related to the frequency of the signal being measured, \(f\), instead of the signal's bandwidth. The sources of flicker noise are not well understood, but it is known that it is inversely proportional to the signal's frequency; thus, flicker noise is sometimes called \(1/f\) noise. Because of the inverse relationship, flicker noise is more important at low frequencies, where it appears as a long-term drift in the signal. It is less important at higher frequencies where thermal noise and shot noise are more important.Our instruments normally do not operate in an environment free from external signals, each of which has a frequency that can be picked up by the instrument. Television signals, cell-phone signals, radio signals, power lines are obvious examples of high-to-moderate frequencies that can serve as noise. Less obvious are lower frequency sources of noise, such as the change in temperature during the day or through the year.This page titled 5.2: Sources of Instrumental Noise is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

5.3: Signal-to-Noise Enhancement

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There are two broad approaches we can use to improve the signal-to-noise ratio: hardware and software. Hardware approaches are built into the instrument and include decisions on how the instrument is set-up for making measurements (for example, the choice of a scan rate or a slit width), and how the signal is processed by the instrument (for example, using electronic filters). A few approaches are briefly considered here; others are included with the discussion of individual instruments. Software solutions are computational approaches in which we manipulate the data either while we are collecting it or after data acquisition is complete.One way to reduce noise is to focus on the circuitry, or hardware, used to measure the signal.One way to reduce environmental noise is to prevent it from entering into the instrument's electronic circuitry. One approach is to use a Faraday cage in which the instrument sits within a room or space covered with a conductive material. Electromagnetic radiation from the environment is absorbed by the conductive material and then shunted away to the ground. Rather than encasing the entire instrument in a Faraday cage, particularly sensitive portions of the circuitry can be shielded.A difference amplifier (see Chapter 3) is an electrical circuit used to determine the difference between two input voltages or currents and to return that difference as a larger voltage or current. As the magnitude of the noise in the two input signals is generally similar in value—that is, it is in phase—while the signal of interest is not, much of the noise's contribution to the signal is subtracted out.When the frequency of the noise is quite different from the frequency of the signal, a simple electrical circuit can be used to remove the high frequency noise and pass the low frequency signal; this is called a low-pass filter. See Chapter 2 for details on low-pass filters.When the signal of interest has a low frequency, the effect of flicker noise becomes significant because a technique that removes low frequency noise will remove the signal as well. Modulation is a process of increasing the frequency of the signal. When complete, a high-pass filter is used to remove the noise. Reversing the modulation returns the original signal, but with much of the noise removed.In this section we will consider three common computational tools for improving the signal-to-noise ratio: signal averaging, digital smoothing, and Fourier filtering.The most important difference between the signal and the noise is that a signal is determinate (fixed in value) and the noise is indeterminate (random in value). If we measure a pure signal several times, we expect its value to be the same each time; thus, if we add together n scans, we expect that the net signal, \(S_n\), is defined as\[S_n = n S \nonumber \]where \(S\) is the signal for a single scan. Because noise is random, its value varies from one run to the next, sometimes with a value that is larger and sometimes with a value that is smaller, and sometimes with a value that is positive and sometimes with a value that is negative. On average, the standard deviation of the noise increases as we make more scans, but it does so at a slower rate than for the signal\[s_n = \sqrt{n} s \nonumber \]where \(s\) is the standard deviation for a single scan and \(s_n\) is the standard deviation after n scans. Combining these two equations, shows us that the signal-to-noise ratio, \(S/N\), after n scans increases as\[(S/N)_n = \frac{S_n}{s_n} = \frac{nS}{\sqrt{n}s} = \sqrt{n}(S/N)_{n = 1} \nonumber \]where \((S/N)_{n = 1}\) is the signal-to-noise ratio for the initial scan. Thus, when \(n = 4\) the signal-to-noise ratio improves by a factor of 2, and when \(n = 16\) the signal-to-noise ratio increases by a factor of 4. shows the improvement in the signal-to-noise ratio for 1, 2, 4, and 8 scans.Signal averaging works well when the time it takes to collect a single scan is short and when the analyte's signal is stable with respect to time both because the sample is stable and the instrument is stable; when this is not the case, then we risk a time-dependent change in \(S_\text{analyte}\) and/or \(s_\text{noise}\) Because the equation for \((S/N)_n\) is proportional to the \(\sqrt{n}\), the relative improvement in the signal-to-noise ratio decreases as \(n\) increases; for example, 16 scans gives a \(4 \times\) improvement in the signal-to-noise ratio, but it takes an additional 48 scans (for a total of 64 scans) to achieve a \(8 \times\) improvement in the signal-to-noise ratio.One characteristic of noise is that its magnitude fluctuates rapidly in contrast to the underlying signal. We see this, for example, in where the underlying signal either remains constant or steadily increases or decreases while the noise fluctuates chaotically. Digital smoothing filters take advantage of this by using a mathematical function to average the data for a small range of consecutive data points, replacing the range's middle value with the average signal over that range.For a moving average filter, also called a boxcar filter, we replace each point by the average signal for that point and an equal number of points on either side; thus, a moving average filter has a width, \(w\), of 3, 5, 7, ... points. For example, suppose the first five points in a sequence arethen a three-point moving average (\(w = 3)\) returns values ofwhere, for example, 0.63 is the average of 0.80, 0.30, and 0.80. Note that we lose \((w - 1)/2 = (3 - 1)/2 = 1\) points at each end of the data set because we do not have a sufficient number of data points to complete a calculation for the first and the last point. shows the improvement in the \(S/N\) ratio when using moving average filters with widths of 5, 9, and 13.One limitation to a moving average filter is that it distorts the original data by removing points from both ends, although this is not a serious concern if the points in question are just noise. Of greater concern is the distortion in a signal's height if we use a range that is too wide; for example, , shows how a 23-point moving average filter (shown in blue) applied to the noisy signal in the upper left quadrant of , reduces the height of the original signal (shown in black). Because the filter's width—shown by the red bar—is similar to the peak's width, as the filter passes through the peak it systematically reduces the signal by averaging together values that are mostly smaller than the maximum signal.A moving average filter weights all points equally; that is, points near the edges of the filter contribute to the average as a level equal to points near the filter's center. A Savitzky-Golay filter uses a polynomial model that weights each point differently, placing more weight on points near the center of the filter and less weight on points at the edge of the filter. Specific values depend on the size of the window and the polynomial model; for example, a five-point filter using a second-order polynomial has weights of\[-3/35 \quad \quad 12/35 \quad \quad 17/35 \quad \quad 12/35 \quad \quad -3/35 \nonumber \]For example, suppose the first five points in a sequence arethen this Savitzky-Golay filter returns values ofwhere, for example, the value for the middle point is\[0.80 \times \frac{-3}{35} + 0.30 \times \frac{12}{35} + 0.80 \times \frac{17}{35} + 0.20 \times \frac{12}{35} + 1.00 \times \frac{-3}{35} = 0.406 \approx 0.41 \nonumber \]Note that we lose \((w - 1)/2 = (5 - 1)/2 = 2\) points at each end of the data set, where w is the filter's range, because we do not have a sufficient number of data points to complete the calculations. For other Savitzky-Golay smoothing filters, see Savitzky, A.; Golay, M. J. E. Anal Chem, 1964, 36, 1627-1639. shows the improvement in the \(S/N\) ratio when using Savitzky-Golay filters using a second-order polynomial with 5, 9, and 13 points.Because a Savitzky-Golay filter weights points differently than does a moving average smoothing filter, a Savitzky-Golay filter introduces less distortion to the signal, as we see in the following figure.This approach to improving the signal-to-noise ratio takes advantage of a mathematical technique called a Fourier transform (FT). The basis of a Fourier transform is that we can express a signal in two separate domains. In the first domain the signal is characterized by one or more peaks, each defined by its position, its width, and its area; this is called the frequency domain. In the second domain, which is called the time domain, the signal consists of a set of oscillations, each defined by its frequency, its amplitude, and its decay rate. The Fourier transform—and the inverse Fourier transform—allow us to move between these two domains.The mathematical details behind the Fourier transform are beyond the level of this textbook; for a more in-depth treatment, consult this series of articles from the Journal of Chemical Education: shows a single peak in the frequency domain and shows its equivalent time domain signal. There are correlations between the two domains:We can use a Fourier transform to improve the signal-to-noise ratio because the signal is a single broad peak and the noise appears as a multitude of very narrow peaks. As noted above, a broad peak in the frequency domain has a fast decaying signal in the time domain, which means that while the beginning of the time domain signal includes contributions from the signal and the noise, the latter part of the time domain signal includes contributions from noise only. The figure below shows how we can take advantage of this to reduce the noise and improve the signal-to-noise ratio for the noisy signal in , which has 256 points along the x-axis and has a signal-to-noise ratio of 5.1. First, we use the Fourier transform to convert its original domain into the new domain, the first 128 points of which are shown in (note: the first half of the data contains the same information as the second half of the data, so we only need to look at the first half of the data). The points at the beginning are dominated by the signal, which is why there is a systematic decrease in the intensity of the oscillations; the remaining points are dominated by noise, which is why the variation in intensity is random. To filter out the noise we retain the first 24 points as they are and set the intensities of the remaining points to zero (the choice of how many points to retain may require some adjustment). As shown in , we repeat this for the remaining 128 points, retaining the last 24 points as they are. Finally, we use an inverse Fourier transform to return to our original domain, with the result in , with the signal-to-noise ratio improving from 5. 1 for the original noisy signal to 11.2 for the filtered signal.This page titled 5.3: Signal-to-Noise Enhancement is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

6.1: General Properties of Electromagnetic Radiation

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Electromagnetic radiation—light—is a form of energy whose behavior is described by the properties of both waves and particles. Some properties of electromagnetic radiation, such as its refraction when it passes from one medium to another (\), are explained best when we describe light as a wave. Other properties, such as absorption and emission, are better described by treating light as a particle. The exact nature of electromagnetic radiation remains unclear, as it has since the development of quantum mechanics in the first quarter of the 20th century [Home, D.; Gribbin, J. New Scientist 1991, 2 Nov. 30–33]. Nevertheless, this dual model of wave and particle behavior provides a useful description for electromagnetic radiation.The frequency and the wavelength of electromagnetic radiation vary over many orders of magnitude. For convenience, we divide electromagnetic radiation into different regions—the electromagnetic spectrum—based on the type of atomic or molecular transitions that gives rise to the absorption or emission of photons ). The boundaries between the regions of the electromagnetic spectrum are not rigid and overlap between spectral regions is possible.This page titled 6.1: General Properties of Electromagnetic Radiation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

6.2: Wave Properties of Electromagnetic Radiation

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Electromagnetic radiation consists of oscillating electric and magnetic fields that propagate through space along a linear path and with a constant velocity. The oscillations in the electric field and the magnetic field are perpendicular to each other and to the direction of the wave’s propagation. shows an example of plane-polarized electromagnetic radiation, which consists of a single oscillating electric field and a single oscillating magnetic field.An electromagnetic wave is characterized by several fundamental properties, including its velocity, amplitude, frequency, phase angle, polarization, and direction of propagation [Ball, D. W. Spectroscopy 1994, 9, 24–25]. Focusing on the oscillations in the electric field, amplitude is the maximum displacement of the electrical field. The wave's frequency, \(\nu\), is the number of oscillations in the electric field per unit time. Wavelength, \(\lambda\) is defined as the distance between successive maxima . shows the initial amplitude as 0; the phase angle \(\Phi\) accounts for the fact that the initial amplitude need not be zero, which we can accomplish by shifting the wave along the direction of propagation.There is a relationship between wavelength and frequency, which is\[\lambda = \frac {c} {\nu} \nonumber \]where \(c\) is the speed of light in a vacuum. Another unit useful unit is the wavenumber, \(\overline{\nu}\), which is the reciprocal of the wavelength\[\overline{\nu} = \frac {1} {\lambda} \nonumber \]Wavenumbers frequently are used to characterize infrared radiation, with the units given in cm–1. Power, \(P\), and intensity, \(I\), are two additional properties of light, both related to the square of the amplitude; power is the energy transferred per second and intensity is the power transferred to a given area.In a vacuum, electromagnetic radiation travels at the speed of light, c, which is \(2.99792 \times 10^8\) m/s. When electromagnetic radiation moves through a medium other than a vacuum, its velocity, v, is less than the speed of light in a vacuum. The difference between v and c is sufficiently small (<0.1%) that the speed of light to three significant figures, \(3.00 \times 10^8\) m/s, is accurate enough for most purposes.When electromagnetic radiation moves between different media—for example, when it moves from air into water—its frequency, \(\nu\), remains constant. Because its velocity depends upon the medium in which it is traveling, the electromagnetic radiation’s wavelength, \(\lambda\), changes. If we replace the speed of light in a vacuum, c, with its speed in the medium, \(v\), then the wavelength is\[\lambda = \frac {v} {\nu} \nonumber \]This change in wavelength as light passes between two media explains the refraction of electromagnetic radiation seen in the photograph of light passing through rain drop, what was included in the previous section. This is discussed in more detail later in this section.In 1817, Josef Fraunhofer studied the spectrum of solar radiation, observing a continuous spectrum with numerous dark lines. Fraunhofer labeled the most prominent of the dark lines with letters. In 1859, Gustav Kirchhoff showed that the D line in the sun’s spectrum was due to the absorption of solar radiation by sodium atoms. The wavelength of the sodium D line is 589 nm. What are the frequency and the wavenumber for this line?The frequency and wavenumber of the sodium D line are\[\nu=\frac{c}{\lambda}=\frac{3.00 \times 10^{8} \ \mathrm{m} / \mathrm{s}}{589 \times 10^{-9} \ \mathrm{m}}=5.09 \times 10^{14} \ \mathrm{s}^{-1} \nonumber \]\[\overline{\nu}=\frac{1}{\lambda}=\frac{1}{589 \times 10^{-9} \ \mathrm{m}} \times \frac{1 \ \mathrm{m}}{100 \ \mathrm{cm}}=1.70 \times 10^{4} \ \mathrm{cm}^{-1} \nonumber \]Another historically important series of spectral lines is the Balmer series of emission lines from hydrogen. One of its lines has a wavelength of 656.3 nm. What are the frequency and the wavenumber for this line?The frequency and wavenumber for the line are\[\nu=\frac{c}{\lambda}=\frac{3.00 \times 10^{8} \ \mathrm{m} / \mathrm{s}}{656.3 \times 10^{-9} \ \mathrm{m}}=4.57 \times 10^{14} \ \mathrm{s}^{-1} \nonumber \]\[\overline{\nu}=\frac{1}{\lambda}=\frac{1}{656.3 \times 10^{-9} \ \mathrm{m}} \times \frac{1 \ \mathrm{m}}{100 \ \mathrm{cm}}=1.524 \times 10^{4} \ \mathrm{cm}^{-1} \nonumber \] shows a single oscillating electrical field and, perpendicular to that, a single oscillating magnetic field. This is an example of plane polarized light in which oscillation of the electrical field occurs at just one angle. Normally electromagnetic radiation oscillates simultaneously at all possible angles. shows the difference in these two cases. If we observe the plane polarized light as it oscillates toward us, we see the single line at the top of the figure where blue indicates a positive amplitude and red indicates a negative amplitude, and where the opacity of the shading shows the change in the amplitudes. The vertical dashed lines show nodes where the amplitude is zero and where no light is seen. With ordinary light, we see a circular beam of radiation because the electrical field is oscillating at all angles. The amplitude's sign and magnitude, and the presence of nodes where the amplitude is zero, remain evident to us. Note that if we observe the source's intensity, then the each of the lines and circles in will appear blue (positive values as intensity is proportion to the square of the amplitude); we continue to observe fluctuations in the intensity and the presence of the nodes.We can describe the oscillations in the electric field as a sine wave\[A_{t}=A_{e} \sin (2 \pi \nu t+\Phi) \nonumber \]where At is the magnitude of the electric field at time t, Ae is the field’s maximum amplitude, \(\nu\) is the wave's frequency, and \(\Phi\) is a phase angle that accounts for the fact that \(A_t\) need not have a value of zero at time \(t = 0\). The identical equation for the magnetic field is\[A_{t}=A_{m} \sin (2 \pi \nu t+\Phi) \nonumber \]where Am is the magnetic field’s maximum amplitude.One of the important features of waves is that adding or subtracting together two (or more) gives a new wave. shows one example. The superposition of waves explains why two identical waves that are completely out-of-phase with each other produce a signal in which the amplitude is zero at all points.Another important consequence of the superposition of waves is that if we can add together a series of waves to produce a new wave, then there is a corresponding mathematical process that takes a complex wave and determines the underlying set of sine waves of which it is comprised. This process is called a Fourier transform, which we will revisit in later chapters.When light encounters matter—perhaps a particle, a solution, or a thin film—it can interact with it in several ways. In this section we consider two such interactions: refraction and reflection. Three additional types of interactions—the scattering of light, the diffraction of light, and the transmission of light—are considered in later chapters where they play an important role in specific instrumental methods of analysis.When light passes from one medium (perhaps air) into another medium (perhaps water) that has a different density, the light experiences a change in direction that is a consequence of a difference in its velocity in the two media. This bending of light is called refraction, the extent of which is given by Snell's law\[ \frac{\text{sin } \theta_1} {\text{sin } \theta_2} = \frac {\eta_2} {\eta_1} = \frac {v_1} {v_2} \nonumber \]where \(\eta_i\) is the refractive index of a medium and \(v_i\) is the velocity in a medium, and where the angles, \(\theta_i\), are shown in .In addition to refraction, when light crosses an interface that separates media with different refractive indexes, some of the light is reflected back. When then angle of incidence is 0° (that is, the light is perpendicular to the interface), then the fraction of light that is reflected is given by\[\frac{I_r}{I_0} = \frac{(\eta_2 - \eta_1)^2}{(\eta_2 + \eta_1)^2} \nonumber \]where \(I_r\) is the intensity of light that is reflected, \(I_0\) is the intensity of light from the source that enters the interface, and \(\eta_i\) is the refractive index of the media. If light crosses more than one interface—as is the case when light passes through a sample cell—then the total fraction of reflected light is the sum of the fraction of light reflected at each interface.This page titled 6.2: Wave Properties of Electromagnetic Radiation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

6.3: Quantum Mechanical Properties of Electromagnetic Radiation

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In the last section, we considered properties of electromagnetic radiation that are consistent with identifying light as a wave. Other properties of light, however, cannot be explained by a model that treats it as a wave; instead, we need to consider a model that treats light as a system of discrete particles, which we call photons.As shown in , in a photoelectric cell, a metal, such as sodium, is held under vacuum and exposed to electromagnetic radiation, which enters the cell through an optical window. If the frequency of the radiation is sufficient, electrons escape from the metal with a kinetic energy that we can measure; we call these photoelectrons. If the photocell's anode is held at a potential that is positive relative to the potential applied to the cathode, the photoelectrons move from the cathode to the anode, generating a current that is measured by an ammeter. If the voltage applied to the anode is made sufficiently negative, the electrons eventually fail to reach the anode and the current decreases to zero. The voltage needed to stop the flow of electrons is called the stopping voltage.In a photoelectron spectrum we vary the frequency and intensity of the electromagnetic radiation and observe their effect on either the number of photoelectrons released (measured as a current) or the energy of the photoelectrons released (measured by their kinetic energy). A typical set of experiments are shown in using Na and in using Na, Zn, and Cu. The data show several interesting features. First, we see in that the intensity of the light source has no effect on the minimum frequency of light needed to eject a photoelectron from Na—we call this the threshold frequency—but that a high intensity source of electromagnetic radiation results in the release of a greater number of photoelectrons and, therefore, a greater current than for a lower intensity source. Second, we see in that different metals have different threshold frequencies, but that once we exceed each metal's threshold frequency, the change in the kinetic energy of the photoelectrons with increasing frequency yields lines of equal slopes.We can explain these experimental observations if we assume that the source of electromagnetic energy has an energy, \(E_\text{ER}\), that does two things: it overcomes the energy that binds the photoelectron to the metal, \(E_\text{BE}\), and it imparts the remaining energy into the photoelectron's kinetic energy, \(E_\text{KE}\), where ER means electromagnetic radiation, BE means binding energy and KE means kinetic energy.\[E_\text{KE} = E_\text{ER} - E_\text{BE} \nonumber \]A wave model for electromagnetic radiation is insufficient to explain the photoelectric effect because when it strikes the metal the radiation's energy would be distributed across all atoms on the surface, none of which would then receive an energy that exceeds the photoelectron's binding energy. Instead, the results in make sense only if we assume that light consists of discrete particles with energies that are a function of frequency or wavelength\[E_\text{ER} = h \nu = \frac{hc}{\lambda} \label{qm} \]where \(h\) is Plank's constant. This leave us with the following equation relating kinetic energy, the energy of the photon, and the binding energy of the electron.\[E_\text{KE} = h \nu - E_\text{BE} \nonumber \]Note that the slope of the lines in is Plank's constant.Equation \ref{qm} is central to the particle, or quantum mechanical model of the atom in which we understand that chemical species—atoms, ions, molecules—exist only in discrete states, each with a single, well-defined energy. A wave, on the other hand, can take on any energy. A simple image is the possible energies of a ball as it rolls down a ramp (wave) or a staircase (particle), as in .When an atom, ion, or molecule moves between two of these discrete states, the difference in energy, \(\Delta E\), is given by\[\Delta E = h \nu = \frac{hc}{\lambda} \nonumber \]In absorption spectroscopy a photon is absorbed by an atom, ion, or molecule, which undergoes a transition from a lower-energy state to a higher-energy, or excited state ). The reverse process, in which an atom, ion, or molecule emits a photon as it moves from a higher-energy state to a lower energy state (\(\PageIndex{4}b\)), is called emission.The types of energy states involved in emission and absorption depend on the energy of the electromagnetic radiation. In general, \(\gamma\)-rays involve transitions between nuclear states, X-rays probe the energies of core-level electrons, ultraviolet-visible radiation probes the energies of valence electrons, infrared radiation provides information on vibrational energy states, microwave radiation probes rotational energy levels and electron spins, and radio waves provide information on nuclear spins. While infrared spectroscopy may provide information on a molecule's vibrational energy states, the energies available in ultraviolet-visible spectroscopy provide information on both the molecule's electronic states and its vibrational states, as shown in .This page titled 6.3: Quantum Mechanical Properties of Electromagnetic Radiation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

6.4: Emission and Absorbance Spectra

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In the last section we considered the source of emission and absorption. In this section we consider the types of emission and absorbance spectra that we will form the basis for many of the chapters that follow.When an atom, ion, or molecule moves from a higher-energy state to a lower-energy state it emits photons with energies equal to the difference in energy between the two states. The result is an emission spectrum that shows the intensity of emission as a function of wavelength. The shapes of these emission spectra fall into two broad types: line spectra and band spectra.When the energy states are well separated from each other, and when there is just one type of transition between the energy states, the result is a line spectrum that consists of a small number of narrow bands. , for example, shows the emission spectrum from gas phase Cu atoms, which consists of seven lines, two of which are too close to each other to resolve them from each other. The individual emissions lines are very narrow, as we might expect, because the atom's energy levels have precise values.The emission spectrum for a gas phase atom is relatively simple because the number of possible transitions is small and because their individuals energies are well-separated from each other. When a molecule in a solvent emits light, the number of possible changes in energy levels can be quite large if the molecule undergoes transitions between electronic, vibrational, and rotational energy levels. The resulting spectrum has so many emission individual emission lines that we see a single broad peak, or band, that we call a band spectrum. shows the emission spectrum for the dye coumarin 343, which is incorporated in a reverse micelle and suspended in cyclohexanol.When considering sources of electromagnetic radiation for spectroscopic instruments, we usually describe them as line sources and continuous sources depending on on whether they emit discrete lines, as is the case for the hollow cathode lamp in , or exhibit emission over a broad range of wavelengths without any gaps, as is the case for a green light-emitting diode (LED), whose spectrum is shown in .When an atom, ion, or molecule moves from a lower-energy state to a higher-energy state it absorbs photons with energies equal to the difference in energy between the two states. The result is an absorbance spectrum that shows the intensity of emission as a function of wavelength. As is the case for emission spectra, absorbance spectra range from narrow lines to broad bands. The atomic absorption spectrum for Na is shown in , and is typical of that found for most atoms. The most obvious feature of this spectrum is that it consists of a small number of discrete absorption lines that correspond to transitions between the ground state (the 3s atomic orbital) and the 3p and the 4p atomic orbitals.Another feature of the atomic absorption spectrum in is the narrow width of the absorption lines, which is a consequence of the fixed difference in energy between the ground state and the excited state, and the lack of vibrational and rotational energy levels. Natural line widths for atomic absorption, which are governed by the uncertainty principle, are approximately 10–5 nm. Other contributions to broadening increase this line width to approximately 10–3 nm.The absorbance spectra for molecules consists of broad bands for the same reasons discussed above for emission spectra. The UV/Vis spectrum for cranberry juice in shows a single broad band for the anthocyanin dyes that are responsible for its red color. The IR spectrum for ethanol in shows multiple absorption bands, some broader and some narrower. The narrow bands, however, are still much broader than the lines in the atomic absorption spectrum for Na.This page titled 6.4: Emission and Absorbance Spectra is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

7.1: General Design of Optical Instruments

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The spectroscopic techniques in the chapters that follow use instruments that share several common basic components: a source of energy; a means for holding the sample of interest to us; a device that can isolate a narrow range of wavelengths; a detector for measuring the signal; and a signal processor that displays the signal in a form convenient for the analyst. shows four common ways of stringing together these units.The remaining sections of this chapter provide general information on each of these units. More specific details appear in the chapters on individual methods.This page titled 7.1: General Design of Optical Instruments is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

7.2: Sources of Radiation

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All forms of spectroscopy require a source of energy to place the analyte in an excited state. In absorption and scattering spectroscopy this energy is supplied by photons. Emission and photoluminescence spectroscopy use thermal energy, radiant (photon) energy, or chemical energy to promote the analyte to a suitable excited state. In this section we consider the sources of radiant energy.A source of electromagnetic radiation must provide an output that is both intense and stable in the region of interest. Sources of electromagnetic radiation are classified as either continuum or line sources. Table \(\PageIndex{1}\) provides a list of some common sources of electromagnetic radiation.Continuum sources emits radiation over a broad range of wavelengths, with a relatively smooth variation in intensity ), and are used for molecular absorbance using UV/Vis and IR radiation. Further details on these sources are in Chapters 13 and 16, respectively.A line source, on the other hand, emits radiation at discrete wavelengths, with broad regions showing no emission lines ), and are used for atomic absorption, atomic and molecular fluorescence, and Raman spectroscopy. Further details on hollow cathode lamps are included in Chapter 9.An important line source of radiation is a laser, which is an acronym for light amplification by stimulated emission of radiation. Laser emission is monochromatic with a narrow bandwidth of just a few micrometers. As suggested by the term amplification, a laser provides a source of high intensity emission. The source of this intensity is embedded in the term stimulated emission, to which we now turn our attention.To understand how a laser works, we need to consider four key ideas: pumping, population inversion, stimulated emission, and light amplification.Emission cannot occur unless we first populate higher energy levels with electrons, which we can accomplish by, for example, the absorption of photons, as shown in . Emission occurs when an electron in a higher energy state relaxes back to a lower energy state by emitting a photon with an energy equal to the difference in the energy between these two states. The process of populating the excited states with electrons is called pumping and is accomplished by using an electrical discharge, by passing an electrical current through the lasing medium, or by absorption of high energy photons. The goal of pumping is to create a large population of excited states.Normally the majority of the species we are studying are in their ground electronic state with only a small number of species in an excited electronic state. For a laser to achieve a high intensity of emission, it is necessary to create a situation in which there are more species in the excited state than in the ground state, as shown in where the non-inverted population has four species in the ground state and two species in the excited state, and where the inverted population has four species in the excited state and two species in the ground state. shows emission of a photon following absorption of a photon of equal energy. No more than one photon is emitted for each photon that is absorbed, with some species in an excited state relaxing to the ground state through non-radiative pathways. This spontaneous emission is a random process, which means that the timing of emission and the direction in which emission occurs are random.Emission in a laser, as depicted in , is stimulated by a photon with an energy equal to that of the difference in energy between the excited state and the ground state. The interaction of the incoming photon with the excited state results in the excited state's immediate relaxation to the ground state by the emission of a photon. The original photon and the emitted photon are coherent, with identical energies, identical directions, and identical phases. Because two coherent photons are emitted, the amplitude of the emitted radiation is doubled, as we see in ; this is what we mean by light amplification.As the previous sections suggest, creating a population inversion is the limiting factor in generating radiation from a laser. The two-level system in , which involves a single excited state and a single ground state, cannot create a population inversion because when the ground state and excited state are equal in population, the rate at which excited states are produced through pumping equals the rate at which excited states are lost through emission. To achieve stimulated emission, laser systems use three-level or four-level systems, as outlined in .In a three-level system, pumping is used to populate the excited states in level two. From level two, an efficient pathway for non-radiative relaxation populates the excited state in level three, which is sufficiently stable to allow for a population inversion. In a four-level system, the population inversion is achieved between level three and level four.Lasers are categorized by the nature of the lasing medium: solid-state crystals, gases, dyes, and semiconductors. Solid-state lasers use a crystalline material, such as aluminum oxide, that contains trace amounts of an element, such as chromium or neodymium, which serves as the actual lasing medium. Gas lasers use gas phase atoms, ions, or molecules as a lasing medium. The lasing medium in a dye laser is a solution of an organic dye molecule. A dye laser typically is capable of emitting light over a broad range of wavelengths, but is tunable to a specific wavelength within that range. Finally, a semiconductor laser uses modified light-emitting diodes as a lasing medium.This page titled 7.2: Sources of Radiation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

7.3: Wavelength Selectors

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In Nessler’s original colorimetric method for ammonia, which was described at the beginning of the chapter, the sample and several standard solutions of ammonia are placed in separate tall, flat-bottomed tubes. As shown in , after adding the reagents and allowing the color to develop, the analyst evaluates the color by passing ambient light through the bottom of the tubes and looking down through the solutions. By matching the sample’s color to that of a standard, the analyst is able to determine the concentration of ammonia in the sample.In every wavelength of light from the source passes through the sample. This is not a problem if there is only one absorbing species in the sample. If the sample contains two components, then a quantitative analysis using Nessler’s original method is impossible unless the standards contains the second component at the same concentration as in the sample.To overcome this problem, we want to select a wavelength that only the analyte absorbs. Unfortunately, we can not isolate a single wavelength of radiation from a continuum source, although we can narrow the range of wavelengths that reach the sample. As seen in , a wavelength selector always passes a narrow band of radiation characterized by a nominal wavelength, an effective bandwidth, and a maximum throughput of radiation. The effective bandwidth is defined as the width of the radiation at half of its maximum throughput.The ideal wavelength selector has a high throughput of radiation and a narrow effective bandwidth. A high throughput is desirable because the more photons that pass through the wavelength selector, the stronger the signal and the smaller the background noise. A narrow effective bandwidth provides a higher resolution, with spectral features separated by more than twice the effective bandwidth being resolved. As shown in , these two features of a wavelength selector often are in opposition. A larger effective bandwidth favors a higher throughput of radiation, but provide less resolution. Decreasing the effective bandwidth improves resolution, but at the cost of a noisier signal [Jiang, S.; Parker, G. A. Am. Lab. 1981, October, 38–43]. For a qualitative analysis, resolution usually is more important than noise and a smaller effective bandwidth is desirable; however, in a quantitative analysis less noise usually is desirable.The simplest method for isolating a narrow band of radiation is to use an absorption or interference filter.As their name suggests, absorption filters work by selectively absorbing radiation from a narrow region of the electromagnetic spectrum. A simple example of an absorption filter is a piece of colored glass or polymer film. A purple filter, for example, removes the complementary color green from 500–560 nm. Commercially available absorption filters provide effective bandwidths of 30–250 nm, although the throughput at the low end of this range often is only 10% of the source’s emission intensity. Interference filters are more expensive than absorption filters, but have narrower effective bandwidths, typically 10–20 nm, with maximum throughputs of at least 40%. The latter value suggests that an important limitation to an absorption filter is that it may significantly reduce the amount of light from the source that reaches the sample and the detector. shows an example of a filter holder with filters that pass bands of light centered at 440 nm, 490 nm, or 550 nm.An interference filter consists of a transparent dielectric material, such as CaF2, which is sandwiched between two glass plates, each coated with a thin, semitransparent metal film (\(\PageIndex{5}a\)). When a continuous source of light passes through the interference filter it undergoes constructive and destructive interference that isolates and passes a narrow band of light centered at a wavelength that satisfies Equation \ref{lambda}\[\lambda = \frac{2nb}{m} \label{lambda} \]where \(n\) is the refractive index of the dielectric material, \(b\) is the thickness of the dielectric material, and \(m\) is the order of the interference (typically first-order). shows the result of passing the emission from a green LED—a continuous source that emits light from approximately 500 nm to 650 nm—through an interference filter that produces an effective bandwidth of a few nanometers. In this case, a 210 nm thick film with a refractive index of 1.35 passes light centered at a wavelength of\[\lambda = \frac{2 \times 1.35 \times 210 \text{ nm}}{1} = 567 \text{ nm} \nonumber \]A filter has one significant limitation—because a filter has a fixed nominal wavelength, if we need to make measurements at two wavelengths, then we must use two filters. A monochromator is an alternative method for selecting a narrow band of radiation that also allows us to continuously adjust the band’s nominal wavelength. Monochromators are classified as either fixed-wavelength or scanning. In a fixed-wavelength monochromator we select the wavelength by manually rotating the grating. Normally a fixed-wavelength monochromator is used for a quantitative analysis where measurements are made at one or two wavelengths. A scanning monochromator includes a drive mechanism that continuously rotates the grating, which allows successive wavelengths of light to exit from the monochromator. A scanning monochromator is used to acquire a spectrum, and, when operated in a fixed-wavelength mode, for a quantitative analysis.The construction of a typical monochromator is shown in . Radiation from the source enters the monochromator through an entrance slit. The radiation is collected by a collimating mirror or lens, which focuses a parallel beam of radiation to a diffraction grating (left) or a prism (right), that disperses the radiation in space. A second mirror or lens focuses the radiation onto a planar surface that contains an exit slit. Radiation exits the monochromator and passes to the detector. As shown in , a monochromator converts a polychromatic source of radiation at the entrance slit to a monochromatic source of finite effective bandwidth at the exit slit. The choice of which wavelength exits the monochromator is determined by rotating the diffraction grating or prism. A narrower exit slit provides a smaller effective bandwidth and better resolution than does a wider exit slit, but at the cost of a smaller throughput of radiation.Polychromatic means many colored. Polychromatic radiation contains many different wavelengths of light. Monochromatic means one color, or one wavelength. Although the light exiting a monochromator is not strictly of a single wavelength, its narrow effective bandwidth allows us to think of it as monochromatic.Although prism monochromators were once in common use, they have mostly been replaced by diffraction gratings. There are several reasons for this. One reason is that diffraction gratings are much less expensive to manufacture. A second reason is that a diffraction grating provides a linear dispersion of of wavelengths along the focal plane of the exit slit, which means the resolution between adjacent wavelengths is the same throughout the source's optical range. A prism, on the other hand, provides a greater resolution at shorter wavelengths than it does a longer wavelengths.The inset in the diffraction grating monochromator in shows the general saw-toothed pattern of a diffraction grating, which consists of a series of grooves with broad surfaces exposed to light from the source. As shown in , parallel beams of source radiation (shown in blue) from the monochromator's collimating mirror strike the surface of the diffraction grating and are reflected back (shown in green) toward the monochromator's focusing mirror and the detector. The parallel beams from the source strike the diffraction grating at an incident angle \(i\) relative to the grating normal, which is a line perpendicular to the diffraction grating's base. The parallel beams bounce back toward the detector do so at a reflected angle \(r\) to the grating normal.Constructive interference between the reflected beams occurs if their path lengths differ by an integer multiple of the incident beam's wavelength (\(n \lambda\)), where \(n\) is the diffraction order. A close examination of shows that the difference in the distance traveled by two parallel beams of light, identified as 1 and 2, that strike adjacent grooves on the diffraction grating is equal to the sum of the line segments \(\overline{CB}\) and \(\overline{BD}\), both shown in red; thus\[n \lambda = \overline{CB} + \overline{BD} \]The incident angle, \(i\), is equal to the angle CAB and the reflected angle, \(r\), is equal to the angle DAB, which means we can write the following two equations\[\overline{CB} = d \sin{i} \]\[\overline{BD} = d \sin{r} \]where \(d\) is the distance between the diffraction grating's grooves. Substituting back gives\[n \lambda = d(\sin{i} + \sin{r}) \label{nlambda} \]which allows us to calculate the angle at which we can detect a wavelength of interest, \(r\), given the angle of incidence from the source, \(i\), and the number of grooves per mm (or the distance between grooves).At what angle can we detect light of 650 nm using a diffraction grating with 1500 gooves per mm if the incident radiation is at an angle of \(50^{\circ}\) to the grating normal? Assume that this is a first-order diffraction.The distance between the grooves is\[d = \frac{1 \text{ mm}}{1500 \text{ grooves}} \times \frac{10^6 \text{ nm}}{\text{mm}} = 666.7 \text{ nm} \nonumber \]To find the angle, we begin with\[ n \lambda = 1 \times 650 \text{ nm} = d(\sin{i} + \sin{r}) = 666.7 \text{ nm} \times (\sin{} + \sin{r}) \nonumber \]\[0.9750 = 0.7660 + \sin{r} \nonumber \]\[0.2090 = \sin{r} \nonumber \]\[ r = 12.1^{\circ} \nonumber \]The quality of a monochromator depends on several key factors: the purity of the light that emerges from the exit slit, the power of the light that emerges from the exit slit, and the resolution between adjacent wavelengths.The radiation that emerges from a monochromator is pure if it (a) arises from the source and if it (b) follows the optical path from the entrance slit to the exit slit. Stray radiation that enters the monochromator from openings other than the entrance slit—perhaps through small imperfections in the joints—or that reaches the exit slit after scattering from imperfections in the optical components or dust, serves as a contaminant in that the power measured at the detector has a component at the monochromator's analytical wavelength and a component from the stray radiation that includes radiation at other wavelengths.The amount of radiant energy that exits the monochromator and reaches the detector in a unit time is power. The greater the power, the better the resulting signal-to-noise ratio. The more radiation that enters the monochromator and is gathered by the collimating mirror, the greater the amount of radiation that exits the monochromator and the greater the power at the detector. The ability of a monochromator to collect radiation is defined by its \(f/number\). As shown in , the smaller the \(f/number\), the greater the area and the greater the power. The light-gathering power increases as the inverse square of the \(f/number\); thus, a monochromator rated as \(f/2\) gathers \(4 \times\) as much radiation as a monochromator rated as \(f/4\).To separate two wavelengths of light and detect them separately, it is necessary to to disperse them over a sufficient distance. The angular dispersion of a monochromator is defined as the change in the angle of reflection (see the angle \(r\) in ) for a change in wavelength, or \(dr/d\lambda\). Taking the derivative of Equation \ref{nlambda} for a fixed angle of incidence (see the angle \(i\) in ) gives the angular dispersion as\[ \frac{dr}{d \lambda} = \frac{n}{d \cos{r}} \label{angdisp} \]where \(n\) is the diffraction order. The linear dispersion of radiation, \(D\), gives the change in wavelength as a function of \(y\), the distance along the focal plane of the monochromator's exit slit; this is related to the angular dispersion by\[D = \frac{dy}{d \lambda} = \frac{F dr}{d \lambda} \label{lineardisp} \]where \(F\) is the focal length. Because we are interested in wavelength, it is convenient to take the inverse of Equation \ref{lineardisp}\[D^{-1} = \frac{d \lambda}{dy} = \frac{1}{F} \times \frac{d \lambda}{dr} \label{invlineardisp} \]where \(D^{-1}\) is the reciprocal linear dispersion. Substituting Equation \ref{angdisp} into Equation \ref{invlineardisp} gives\[D^{-1} = \frac{d \lambda}{dy} = \frac{d \cos{r}}{nF} \]which simplifies to\[D^{-1} = \frac{d}{nF} \]for angles \(r < 20^{\circ}\) where \(\cos{r} \approx 1\). Because the linear dispersion of radiation along the monochromator's exit slit is independent of wavelength, the ability to resolve two wavelengths is the same across the spectrum of wavelengths.Another way to report a monochromator's ability to distinguish between two closely spaced wavelengths is its resolving power, \(R\), which is defined as\[R = \frac{\lambda}{\Delta \lambda} = n N \nonumber \]where \(\lambda\) is the average of the two wavelengths, \(\Delta \lambda\) is the difference in their values and \(N\) is the number of grooves on the diffraction grating that are exposed to the radiation from the collimating mirror. The greater the number of grooves, the greater the resolving power.A monochromator has two sets of slits: an entrance slit that brings radiation from the source into the monochromator and an exit slit that passes the radiation from the monochromator to the detector. Each slit consists of two metal plates with sharp, beveled edges separated by a narrow gap that forms a rectangular window and which is aligned with the focal plane of the collimating mirror. shows a set of four slits from a monochromator taken from an atomic absorption spectrophotometer. From bottom-to-top, the slits have widths, \(w\), of 2.0 mm, 1.0 mm, 0.5 mm, and 0.2 mm.Suppose we have a source of monochromatic radiation with a wavelength of 400.0 nm and that we pass this beam of radiation through a monochromator that has entrance and exit slits with a width, \(w\), of 1.0 mm and a reciprocal linear dispersion of 1.2 nm/mm. The product of these two variables is called the monochromator's effective bandwidth, \(\Delta \lambda_\text{eff}\), and is given as\[\Delta \lambda_\text{eff} = w D^{-1} = 1.0 \text{ mm} \times 1.2 \text{ nm/mm} = 1.2 \text{ nm} \]The width of the beam in units of wavelength, therefore, is 1.2 nm. In this case, as shown in , if we scan the monochromator, our beam of monochromatic radiation will first enter the exit slit at a wavelength setting of 398.8 nm and will fully exit the slit at a wavelength setting of 401.2 nm. In between these limits a portion of the beam is blocked and only a portion of the beam passes through the exit slit and reaches the detector. For example, when the monochromator is set to 399.4 nm or 400.6 nm, half of ther beam reaches the detector with a power of \(0.5\times P\). If we monitor the power at the detector as a function of wavelength, we obtain the profile shown at the bottom of . The monochromator's bandwidth encompasses the range of wavelengths over which some portion of the beam of radiation passes through the exit slit.Suppose we have a source of radiation that consists of precisely three wavelengths—399.4 nm, 400.0 nm, and 400.6 nm—and we pass them through a monochromator with an effective bandwidth of 1.2 nm. Using the analysis from the previous section, the radiation with a wavelength of 399.4 nm passes through the monochromator's exit slit for any wavelength setting between 398.8 and 400.0 nm, which means it overlaps with the radiation with a wavelength of 400.0 nm. The same is true for the radiation with a wavelength of 400.6 nm, which also overlaps with the radiation with a wavelength of 400.0 nm. As shown in , we cannot resolve the three monochromatic sources of radiation, which appear as a single broad band of radiation. Decreasing the effective bandwidth to one-half of the difference in the wavelengths of the adjacent sources of radiation produces, as shown in , baseline resolution of the individual sources of wavelength. To resolve the sources of radiation with wavelengths of 399.4 nm and 400.0 nm using a monochromator with a reciprical linear dispersion of 1.2 nm/mm requires an effective bandwidth of\[\Delta \lambda_\text{eff} = 0.5 \times (400.0 \text{ nm} - 399.4 \text{ nm}) = 0.3 \text{ nm} \nonumber \]and a slit width of\[w = \frac{\Delta \lambda_\text{eff}}{D^{-1}} = \frac{0.3 \text{ nm}}{1.2 \text{nm/mm}} = 0.25 \text{ mm} \nonumber \]The choice of slit width always involves a trade-off between increasing the radiant power that reaches the detector by using a wide slit width, which improves the signal-to-noise ratio, and improving the resolution between closely spaced peaks, which requires a narrow slit width. illustrates this trade-off. Ultimately, the needs of the analyst will dictate the choice of slit width.This page titled 7.3: Wavelength Selectors is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.