# Introductory Chemistry Online/Aqueous Solutions
## Chapter 7. Aqueous Solutions

Water is the most remarkable solvent! The O—H bonds in water are polarized due to the differences in electronegativity between hydrogen and oxygen. When this uneven charge distribution is coupled with the fact that water has a “bent” molecular geometry, the two covalent bond dipoles combine to form a molecular dipole (shown in the electrostatic potential map on the right. This molecular dipole allows water to surround and stabilize ions in solution, making water a powerful solvent for the dissolution of polar and ionic compounds. If we know the amount of solute that we have dissolved in a given volume of solvent, we can define the term molarity, as the number of moles of solute in each liter of solution. Finally, by combining the concept of molarity with what we have learned about simple stoichiometric calculations, we can now approach quantitative chemical calculations in solution.

## 7.1 Dipole Moments and the Properties of Water

Water is an amazing solvent, and has remarkable physical and chemical properties that make it the essential ingredient to life as we know it. The special properties of water come from the fact that the elements hydrogen and oxygen have differing electronegativities. In Chapter 3 we learned that covalent bonds formed between atoms of differing electronegativity are polarized. Because electronegativity is a measure of how strongly a given atom attracts electrons to itself, the atom in the covalent bond with the highest electronegativity will tend to draw the bonding electrons towards itself, resulting in a bond that is electron-rich on one end and electron-poor on the other. Covalent bonds that are polarized are said to have a dipole, where the term dipole moment refers to the direction and magnitude of the charge separation.
Consider water. The electronegativities of hydrogen and oxygen are 2.20 and 3.44, respectively. That means that in each covalent bond, the electrons will be attracted towards the oxygen, leaving the hydrogen electron-poor. In Chapter 3, we used a calculated electrostatic potential map to visualize the electron density around molecules. The map for water is shown to the left and is colored using red to indicate a high electron density and blue to show electron-poor regions. Because electrons carry a negative charge, this also means that the red regions of the molecule are anionic (negative) and that the blue regions are cationic (positive).

## 7.2 Molecular Dipoles

Electrostatic potential maps are useful because they clearly show the electron distribution around covalent bonds within molecules. They must be calculated, however, using sophisticated computer programs, and then rendered in color for visualization. Because of this, the polarization of covalent bonds is typically shown using a special arrow (a dipole arrow) to indicate the direction in which the bond is polarized. A dipole arrow is crossed at the beginning (as in a plus sign) and points in the direction of the greatest electron density. Thus for hydrogen fluoride, the electronegativities are 2.20 and 3.98 for the hydrogen and fluorine, respectively. We would predict that the H—F bond would be polarized with the greatest electron density towards the fluorine. The resulting dipole moment is shown below.
A molecule such as water, with two covalent bonds, will have two local dipoles, each oriented along the covalent bonds, as shown below. Because water is asymmetric (it has a bend structure) both of these local dipoles point in the same direction, generating a molecular dipole, in which the entire molecule has a charge imbalance, with the “oxygen end” being anionic and the “hydrogen end” being cationic. The predicted molecular dipole is also shown in the electrostatic potential map in Figure 7.3.
Molecules with local dipoles do not necessarily possess a molecular dipole. Consider the molecule boron trihydride (BH$_3$). The BH$_3$ molecule is planar with all three hydrogens spaced evenly surrounding the boron (trigonal planar). The electronegativities of boron and hydrogen are 2.04 and 2.20, respectively. The bonds in BH$_3$ will therefore be somewhat polarized, with the local dipoles oriented towards the hydrogen atoms, as shown below. But because the molecule is symmetrical, the three dipole arrows cancel and, as a molecule, BH$_3$ has no net molecular dipole.
Exercise 7.1 Drawing Molecular Dipoles

## 7.3 Dissolution of Ionic Compounds

A simple ionic compound, such as sodium chloride (NaCl) consists of a sodium cation and a chloride anion. Because these are oppositely charge ions, they are strongly attracted to each other. This attraction is non-specific and the sodium cation would also be strongly attracted to any anion. When an ionic compound dissolves in water, the individual cations and anions are completely surrounded by water molecules, but these water molecules are not randomly oriented. A sodium cation in water will be surrounded by water molecules oriented so that the negative end of the molecular dipole is in contact with the sodium cation. Likewise, the waters surrounding the chloride anion are oriented so that the positive end of the molecular dipole contacts the anion (Figure 7.5). When arranged like this, the charged poles of the water molecules neutralize, and thus stabilize the charges on the ions.
The ability of water to interact with and stabilize charge particles goes well beyond the water molecules that actually touch the ion. Surrounding the inner water shell is another shell of waters that will orient themselves so that their dipoles bind to the exposed dipoles from the inner shell. This is shown in Figure 7.6 using drawing showing part of the “cluster” of water molecules surrounding the sodium cation. As the subsequent layers of water surround each other, the positive charge from the cation is dispersed or spread out over the whole group of interacting molecules. The cluster then becomes effectively neutral allowing the charged ion to exist free in solution, removed from its counter-ion (the chloride). The dynamic collection of water molecules surrounding an ion in solution is referred to as the solvation shell and it is the ability of water to solvate and stabilize ions that makes water such an important solvent, both in chemistry and in biology.
In addition to ionic compounds, water will also dissolve and stabilize most molecules that are polar, that is, if they possess a molecular dipole. An example of this is shown in Figure 7.7. The organic compound, methyl propionate, contains a highly polar carbon-oxygen double bond. The electrostatic potential map in the figure clearly shows the resulting molecular dipole and methyl propionate is quite soluble in water; 6.2 grams of methyl propionate will dissolve in 100 mL of water. The organic molecule propane, also shown in the figure, does not possess a significant molecular dipole and is only very slightly soluble in water.

## 7.4 Concentration and Molarity

As described in the previous section, sodium chloride is quite soluble in water. At 25 ˚C (about room temperature), 359 grams of sodium chloride will dissolve in one liter of water. If you were to add more sodium chloride to the solution, it would not dissolve, because a given volume of water can only dissolve, disperse and stabilize a fixed amount of solute (the stuff that dissolves). This amount is different for every compound and it depends on the structure of the particular compound and how that structure interacts with the solvation shell. When a substance is dissolved in water to the point that no more will go into solution, we say the solution is saturated. For most compounds, heating the solution will allow more of the substance to dissolve, hence it is important to note the temperature when you are speaking of the solubility of a particular compound.
If we had a saturated solution of sodium chloride at 25 ˚C, we could quote the concentration as 359 grams/L, but because we know the molar mass of sodium chloride (58.44 grams/mole), we could also express our concentration as:
$\left({\frac {\left(3{\text{59 g}}\right)\times \left({\frac {1{\text{ mole}}}{{\text{58}}{\text{.44 g}}}}\right)}{1{\text{ L}}}}\right)=6.14{\text{ }}^{\text{moles}}\!\!\diagup \!\!_{\text{L}}\;$
In chemistry, the units of moles/L are called molarity, with the abbreviation M. Thus we could say that our saturated solution of sodium chloride was 6.14 molar, or 6.14 M.
The advantage of expressing concentrations in terms of molarity is that these solutions can now be used in chemical reactions of known stoichiometry because any volume of the solution corresponds directly to a known number of moles of a particular compound. For example, the molar mass of potassium bromide is 119.0 g/mole. If we dissolved 119.0 grams of KBr in 1.000 L of water, the concentration would be 1.000 mole/L, or 1.000 M. If we now took half o this solution (0.500 L) we know that we would also have 0.500 moles of KBr.
We can determine the concentration of a solution using the problem-solving algorithm we introduced back in Chapter 1. For example, if you wan to find the molarity of a solution containing 42.8 grams of KBr in 1.00 L of water, you would identify the given and 42.8 g, your ratio is the molar mass (119 g/mole) and you want to find molarity (or moles/L). Remembering to set the equation up so that the units of given appear in the denominator of the ratio, the number of moles is:
$\left(4{\text{2}}{\text{.8 g}}\right)\times \left({\frac {1{\text{ mole}}}{\text{119 g}}}\right)=0.360{\text{ moles}}$
and, the molarity is: $\left({\frac {0.{\text{360 mole}}}{{\text{1}}{\text{.00 L}}}}\right)=0.360{\text{ }}^{\text{moles}}\!\!\diagup \!\!_{\text{L}}\;$, or 0.360 M
When you become comfortable with the simple two-step method, you can combine steps and simply divide your given mass by the given volume to get the result directly. Thus, if you had 1.73 grams of KBr in 0.0230 L of water, your concentration would be:
$\left({\frac {\left(1.{\text{73 g}}\right)\times \left({\frac {1{\text{ mole}}}{\text{119 g}}}\right)}{0.{\text{0230 L}}}}\right)=0.632{\text{ }}^{\text{moles}}\!\!\diagup \!\!_{\text{L}}\;{\text{, or, 0}}{\text{.632 M}}$
We can also solve these problems backwards, that is, convert molarity into mass. For example; determine the number of grams of KBr that are present in 72.5 mL of a 1.05 M solution of KBr. Here we are given a volume of 0.0725 L and our ratio is the molarity, or (1.05 moles/L). We first solve for moles,
${\text{0}}{\text{.0725 L }}\times {\text{ }}\left({\frac {1.{\text{05 mole}}}{{\text{1}}{\text{.00 L}}}}\right)=0.0761{\text{ moles}}$
and then convert to mass using: ${\text{0}}{\text{.0761 moles }}\times {\text{ }}\left({\frac {1{\text{19 grams}}}{\text{1 mole}}}\right)=9.06{\text{ grams of KBr}}$.
Example 7.1 Molarity Calculations
Exercise 7.2 Molarity Calculations

## 7.5 Solution Stoichiometry

As we learned in Chapter 5, double replacement reactions involve the reaction between ionic compounds in solution and, in the course of the reaction, the ions in the two reacting compounds are “switched” (they replace each other). As an example, silver nitrate and sodium chloride react to form sodium nitrate and the insoluble compound, silver chloride.
Because these reactions occur in aqueous solution, we can use the concept of molarity to directly calculate the number of moles of products that will be formed, and hence the mass of precipitates. In the reaction shown above, if we mixed 123 mL of a 1.00 M solution of NaCl with 72.5 mL of a 2.71 M solution of AgNO$_3$, we could calculate the moles (and hence, the mass) of AgCl that will be formed as follows:
First, we must examine the reaction stoichiometry. In this reaction, one mole of AgNO$_3$ reacts with one mole of NaCl to give one mole of AgCl. Because our ratios are one, we don’t need to include them in the equation. Next, we need to calculate the number of moles of each reactant:
${\text{0}}{\text{.123 L }}\times {\text{ }}\left({\frac {1.{\text{00 mole}}}{{\text{1}}{\text{.00 L}}}}\right)=0.123{\text{ moles NaCl}}$
${\text{0}}{\text{.0725 L }}\times {\text{ }}\left({\frac {2.{\text{71 mole}}}{{\text{1}}{\text{.00 L}}}}\right)=0.196{\text{ moles AgNO}}_{\text{3}}$
Because this is a limiting reactant problem, we need to recall that the moles of product that can be formed will equal the smaller of the number of moles of the two reactants. In this case, NaCl is limiting and AgNO$_3$ is in excess. Because our stoichiometry is one-to-one, we will therefore form 0.123 moles of AgCl. Finally, we can convert this to mass using the molar mass of AgCl:
${\text{0}}{\text{.0725 L }}\times {\text{ }}\left({\frac {2.{\text{71 mole}}}{{\text{1}}{\text{.00 L}}}}\right)=0.196{\text{ moles AgNO}}_{\text{3}}$
In a reaction where the stoichiometry is not one-to-one, you simply need to include the stoichiometric ratio in you equations. Thus, for the reaction between lead (II) nitrate and potassium iodide, two moles of potassium iodide are required for every mole of lead (II) iodide that is formed.
For example: 1.78 grams of lead (II) nitrate are dissolved in 17.0 mL of water and then mixed with 25.0 mL of 2.5 M potassium iodide solution. What mass of lead (II) iodide will be formed and what will be the final concentration of potassium nitrate in the solution? Again, we need to look at this as a limiting reactant problem and first calculate the number of moles of each reactant:
${\text{1}}{\text{.78 g}}\times \left({\frac {1.{\text{00 mole}}}{{\text{331}}{\text{.2 g}}}}\right)=5.37\times {\text{10}}^{\text{-3}}{\text{ moles Pb(NO}}_{\text{3}}{\text}}_{\text{2}}$
${\text{0}}{\text{.0025 L }}\times {\text{ }}\left({\frac {2.{\text{50 mole}}}{\text{1 L}}}\right)=6.25\times {\text{10}}^{\text{-3}}{\text{ moles KI}}$
The stoichiometry of this reaction is given by the ratios $\left({\frac {{\text{1 mole PbI}}_{\text{2}}}{2{\text{ mole KI}}}}\right)$, and $\left({\frac {1{\text{ mole PbI}}_{\text{2}}}{{\text{1 mole Pb(NO}}_{\text{3}}{\text}}_{\text{2}}}}\right)$, so the number of moles of product that would be formed from each reactant is calculated as:
$\left({\frac {1{\text{ mole PbI}}_{\text{2}}}{{\text{1 mole Pb(NO}}_{\text{3}}{\text}}_{\text{2}}}}\right)$
$6.25\times {\text{10}}^{\text{-3}}{\text{ moles KI}}\times {\text{ }}\left({\frac {1{\text{ mole PbI}}_{\text{2}}}{\text{2 moles KI}}}\right)=3.12\times {\text{10}}^{\text{-3}}{\text{ moles PbI}}_{\text{2}}$
Potassium iodide produces the smaller amount of PbI$_2$ and hence, is limiting and lead (II) nitrate is in excess. The mass of lead (II) iodide that will be produced is then calculated from the number of moles and the molar mass:
${\text{3}}{\text{.12}}\times {\text{10}}^{\text{-3}}{\text{ moles }}\times {\text{ }}\left({\frac {4{\text{61 grams}}}{\text{1 mole}}}\right)=1.44{\text{ grams of PbI}}_{\text{2}}$
To determine the concentration of potassium nitrate in the final solution, we need to note that two moles of potassium nitrate are formed for every mole of PbI$_2$, or a stoichiometric ratio of $\left({\frac {2{\text{ moles KNO}}_{\text{3}}}{{\text{1 mole PbI}}_{\text{2}}}}\right)$. Our final volume is (17.0 + 25.0) = 42.0 mL, and the concentration of potassium nitrate is calculated as:
${\frac {3.12\times {\text{10}}^{\text{-3}}{\text{ moles PbI}}_{\text{2}}\times {\text{ }}\left({\frac {2{\text{ mole KNO}}_{\text{3}}}{{\text{1 mole PbI}}_{\text{2}}}}\right)}{0.0420{\text{ L}}}}=^{0.148{\text{ moles KNO}}_{\text{3}}}\!\!\diagup \!\!_{\text{L}}\;{\text{, or, 0}}{\text{.148 M}}$
Example 7.2 Solution Stoichiometry Calculations
Exercise 7.3 Solution Stoichiometry Calculations
If 1.0 g of NaN$_3$ reacts with 25 mL of 0.20 M NaNO$_3$ according to the reaction shown below, how many moles of N$_2$(g) are produced?

## 7.6 Dilution of Concentrated Solutions

In the laboratory, a chemist will often prepare solutions of known concentration beginning with a standard stock solution. A stock solution is generally concentrated and, of course, the molar concentration of the solute must be known. To perform a reaction, a measured amount of this stock solution will be withdrawn and added to another reactant, or will be diluted into a larger volume for some other use. The calculations involved in these dilutions are trivial and simply involve calculating the number of moles transferred and dividing this by the final volume. For example, 15.0 mL of a stock solution of 1.00 M hydrochloric acid (HCl) is withdrawn and diluted into 75 mL of distilled water; what is the final concentration of hydrochloric acid?
First, the number of moles of HCl is calculated from the volume added and the concentration of the stock solution:
${\text{0}}{\text{.0150 L}}\times \left({\frac {{\text{1}}{\text{.00 moles}}}{\text{1 L}}}\right)=0.0150{\text{ moles Hcl}}$
We have diluted this number of moles into (15.0 + 75.0) = 90.0 mL, therefore the final concentration of HCl is given by:
$\left({\frac {0.0150{\text{ moles HCl}}}{{\text{0}}{\text{.0900 L}}}}\right)=\left(^{{\text{0}}{\text{.167 moles HCl}}}\!\!\diagup \!\!_{\text{L}}\;\right){\text{, or 0}}{\text{.167 M}}$
An even simpler way to approach these problems is to multiply the initial concentration of the stock solution by the ratio of the aliquot (the amount withdrawn from the stock solution) to the final volume, using Equation 7.1,
$\left({\text{stock concentration}}\right)\times \left({\frac {\text{volume of the aliquot}}{\text{final volume}}}\right)={\text{final concentration}}$ Eq 7.1
Using this method on the previous problem,
$\left({\text{1}}{\text{.00 M}}\right)\times \left({\frac {{\text{15}}{\text{.0 mL}}}{{\text{90}}{\text{.0 mL}}}}\right)={\text{0}}.167{\text{ M}}$
Note that we did not have to convert our volumes (15.0 and 90.0 mL) into L when we use this approach because the units of volume cancel in the equation. If the units that are given for the aliquot and the final volume are different, a metric conversion ratio may be required. For example, 10.0 µL of a 1.76 M solution of HNO$_3$ (nitric acid) are diluted into 10.0 mL of distilled water; what is the final concentration of nitric acid?
In this problem, we need to convert µL and mL into a common unit. We can do this using the ratios, $\left({\frac {10^{-6}{\text{ L}}}{{\text{1 }}\mu {\text{L}}}}\right){\text{ and }}\left({\frac {10^{-3}{\text{ L}}}{\text{1 mL}}}\right)$. We need to multiply each of our volumes by the appropriate factor to get our volumes in terms of liters, and then simply multiply by the initial concentration. Thus,
$1.76{\text{ M}}\times \left\{{\frac {10.0{\text{ }}\mu {\text{L}}\left({\frac {10^{-6}{\text{ L}}}{{\text{1 }}\mu {\text{L}}}}\right)}{10.0{\text{ mL}}\left({\frac {10^{-3}{\text{ L}}}{\text{1 mL}}}\right)}}\right\}=1.76\times 10^{-3}{\text{ M}}$
The final volume in this problem is actually (1.00  10$^{-2}$ L) + (1.00  10$^{-5}$ L) = 1.001  10$^{-2}$ L, but because our calculation is only accurate to three significant figures, the volume of the aliquot is not significant and the final volume has been rounded.
The standard method we have used here can also be adapted to the type of problem in which you need to find the volume of a stock solution that must be diluted to a certain volume in order to produce a solution of a given concentration. For example, what volume of 0.029 M CaCl$_2$ must be diluted to exactly 0.500 L in order to give a solution that is 50.0 µM?
In order to solve this problem in the simplest of terms, we should re-examine Equation 7.1.
$\left({\text{stock concentration}}\right)\times \left({\frac {\text{volume of the aliquot}}{\text{final volume}}}\right)={\text{final concentration}}$
This equation can be re-written as:
$\left({\frac {\text{volume of the aliquot}}{\text{final volume}}}\right)=\left({\frac {\text{final concentration}}{\text{stock concentration}}}\right)$
$\left({\frac {V}{V_{f}}}\right)=\left({\frac {C_{f}}{C_{i}}}\right)$
where C$_i$ and C$_f$ are the stock and final concentrations, respectively, V is the volume of the aliquot and V$_f$ is the final volume of the solution. Stated another way, this is simply a set of ratios; aliquot to final volume, and final concentration to initial concentration (operationally, these ratios will always be “small value/larger value”). Working with this set of ratios, we can directly solve this type of problem as follows:
First, we need to convert our final concentration (50.0 µM) into M, to match the units of our stock solution. The metric multiplier for µ is 10$^{-6}$, making our final concentration 50.0  10$^{-6}$ M, or more properly, 5.00  10$^{-5}$ M. Our equation is therefore:
$\left({\frac {V}{0.500{\text{ L}}}}\right)=\left({\frac {5.00\times 10^{-5}{\text{ M}}}{0.029{\text{ M}}}}\right)$
$V=\left({\frac {5.00\times 10^{-5}{\text{ M}}}{0.029{\text{ M}}}}\right)\times 0.500{\text{ L}}$
The volume of the aliquot, V, is 8.62  10$^{-4}$ L, or using the conversion factor $\left({\frac {{\text{10}}^{\text{3}}{\text{ mL}}}{\text{1 L}}}\right)$, the required volume is 0.86 mL (there are only two significant figures in the concentration of the stock solution, 0.029 M).
Dilution problems can be solved directly using Equation 7.1, or, as you become more comfortable with the math, using the initial and final ratios like we did in this problem (remember, the numbers in the two ratios are “smaller/larger”).
Example 7.3 Serial Dilutions
Ex 7.4 Serial Dilutions

## Study Points

* Covalent bonds formed between atoms of differing electronegativity are polarized, resulting in a bond that is electron-rich on one end and electron-poor on the other. Covalent bonds that are polarized are said to have a dipole, where the term dipole moment refers to the direction and magnitude of the charge separation.
* If a molecule is asymmetric (such as a molecule with a bend structure) local dipoles along covalent bonds can combine, generating a molecular dipole, in which the entire molecule has an imbalance with regard to electron distribution. This can be shown with an dipole arrow (with a positive end) indicating the direction of the charge separation in the molecule.
* If a molecule is symmetrical (such as BH$_3$, which is trigonal planar), the individual dipoles associated with the covalent bonds cancel, leaving a molecule with no molecular dipole.
* Water has a significant molecular dipole, allowing it to strongly interact with other polar molecules and with individual ions from ionic compounds. Because of this, water is able to break the electrostatic attraction between ions in compounds and to move the ions into solution. In solution, cations will be surrounded by a solvation shell where the water molecules are oriented so that the negative end of the water molecule interacts with the cation. Likewise, the cationic end of water will surround and solvate anions.
* Molarity is simply defined as the number of moles of a solute dissolved in one liter of solvent, or $\left(^{moles}\!\!\diagup \!\!_{L}\;\right)$. The abbreviation for molarity is the uppercase M.
* You should remember that concentration multiplied by volume gives the number of moles of solute; $\left(^{moles}\!\!\diagup \!\!_{L}\;\right)\times L=moles$.
* When you are given the amount of solute in grams, remember, mass divided by molar mass gives moles. Dividing this by volume (in liters) gives molarity; ${\frac {\left({\frac {grams}{^{grams}\!\!\diagup \!\!_{mole}\;}}\right)}{L}}=molarity$.
* In a standard solution, we simply know the molarity of the solute(s). Because concentration (the molarity) multiplied by volume gives us moles, we can calculate the number of moles in given volume and use this value in standard stoichiometric calculations.
* A sample of a solution of known volume is called an aliquot. When an aliquot of a solution is diluted into a larger volume, the final concentration can be calculated as:
$\left({\frac {\text{volume of the aliquot}}{\text{final volume}}}\right)=\left({\frac {\text{final concentration}}{\text{stock concentration}}}\right)$
, or, $\left({\frac {V}{V_{f}}}\right)=\left({\frac {C_{f}}{C_{i}}}\right)$
where C$_i$ and C$_f$ are the stock and final concentrations, respectively, V is the volume of the aliquot and V$_f$ is the final volume of the solution. This relationship is also often stated as V$_1$C$_1$ = V$_2$C$_2$, where the subscripts refer to the initial and final concentrations and volumes.