Is a negative pH level physically possible? A friend of mine was looking over the definition of pH, and wondering if it is possible to have a negative pH level. From the equation below, it certainly seems possible—just have a $1.1$ (or something $\gt 1$) molar solution of $\ce{H+}$ ions:
$$\text{pH} = -\log([\ce{H+}])$$
(Where $[\ce{X}]$ denotes the concentration of $\ce{X}$ in $\frac{\text{mol}}{\text{L}}$.)

If $[\ce{H+}] = 1.1\ \frac{\text{mol}}{\text{L}}$, then $\mathrm{pH} = -\log(1.1) \approx -0.095 $

So, it is theoretically possible to create a substance with a negative pH. But, is it physically possible (e.g. can we create a 1.1 molar acid in the lab that actually still behaves consistently with that equation)?

One publication for you: “Negative pH Does Exist”, K. F. Lim, J. Chem. Educ. 2006, 83, 1465. Quoting the abstract in full:

  The misconception that pH lies between 0 and 14 has been perpetuated in popular-science books, textbooks, revision guides, and reference books.

The article text provides some counterexamples:

  For example, commercially available concentrated HCl solution (37% by mass) has $\mathrm{pH} \approx -1.1$, while saturated NaOH solution has $\mathrm{pH} \approx 15.0$.

$\mathrm{pH}$ is essentially a convention. It is defined as $$-\log_{10} [\ce{H+}]$$ since the concentrations of the solutions commonly used lie in the interval $$[10^{-14}\ \mathrm{mol/L},1\ \mathrm{mol/L}]$$ and thus the $\mathrm{pH}$ lies in $$[0,14]$$ But nothing constrains an aqueous solution from having a $\mathrm{pH}$ that does not lie in this interval. The only constraints are:
$$[\ce{H+}]\lt[\ce{H2O}]_\text{liquid}$$ and $$[\ce{OH-}]\lt[\ce{H2O}]_\text{liquid}$$
The first limiting case is when you suppose all water has turned into $\ce{H+}$, which is not quite true, because there must be some water that has turned into $\ce{OH-}$ in order to $$K_\mathrm w=[\ce{H+}][\ce{OH-}]$$
But $$[\ce{H2O}]_\text{liquid}=\frac{1\ \mathrm{mol}}{18\ \mathrm g}\frac{1000\ \mathrm g}{1\ \mathrm L}=55.6\ \mathrm{mol/L}$$
And then we have $$[\ce{H+}]\lt55.6\ \mathrm{mol/L}$$
The last case implies
$$[\ce{OH-}]\lt[\ce{H2O}]_\text{liquid}$$
which means (considering $K_\mathrm w=10^{-14}$) $$[\ce{H+}]\gt\frac{10^{-14}}{55.6}\ \mathrm{mol/L}$$
Then $$\frac{10^{-14}}{55.6}\ \mathrm{mol/L}\lt[\ce{H+}]\lt55.6\ \mathrm{mol/L}$$
$$-\log_{10}(55.6)\lt-\log_{10} [\ce{H+}]\lt-\log_{10}\left(\frac{10^{-14}}{55.6}\right)$$
$$-1.74\lt\mathrm{pH}\lt15.74$$


Why does screening effect decrease due to d-orbital? In 13th group, atomic radius increases from boron to aluminium. From aluminium to gallium, atomic radii decreases. From gallium to indium, atomic radii increases. And from indium to thallium, atomic radii decreases.

The reason for this irregular trend given is screening effect. While explaining my teacher told that due to d-orbital screening effect decreases.

I can't get why?

The reason d-orbitals make a difference is that electrons in d-orbitals do not screen nuclear charge as effectively as those in s and p orbitals.  This is because of something called penetration .

The mathematical shapes of d-orbitals prevent them from allowing electrons to penetrate very closely to the nucleus, compared with electrons in s or p-orbitals.  In gallium, you have $10$ electrons in the filled $3$d-subshell, and each of these electrons is doing a slightly worse job (relatively speaking) of screening the nuclear charge than the electrons in the s and p orbitals. Therefore, the effective nuclear charge in gallium is slightly higher than that in aluminum, so the increase in the radius is a quite a bit smaller than would be expected based on the difference between boron and aluminum, or gallium and indium.
The trend goes:

$\pu{82 pm}$ ($\ce{B}$)
$\pu{118 pm}$ $\ce{(Al)}$
$\pu{126 pm}$ $\ce{(Ga)}$
$\pu{144 pm \ce{(In)}}$

[covalent radii from www.webelements.com].

This effect is generally known as the d-block contraction.  (It can be more or less pronounced depending on how you define the atomic radii.)

A similar thing happens (in principle) when you go from indium to thallium; except in this case you are now dealing with adding a filled f-subshell to the valence shell.

Electrons in f-orbitals are even worse at screening nuclear charge than those in d-orbitals, therefore again, the effective nuclear charge in thallium is a bit larger than it is in indium, so again the jump in radius is fairly small (from $144$ to $148$ pm).  This effect (of the filled f-subshell) is generally known as the lanthanide contraction.

d and f orbitals experience more  attraction towards nucleus due to lesser shielding effect than s and p orbitals. This is why the ionization energies of d and f block elements are high and so their electronegativities.


Which conformations of 2,3-dimethylbutane are more stable? I am not sure what the proper arrangement of these conformations are from least to most stable. I got the wrong answer, but my thinking was this:
 First, I thought that because 2 and 4 are in eclipsed conformations they are least stable.  From there I thought that 4 was less stable than 2 because of the steric interactions of the methyl groups.  Then of the two staggered conformations 1 and 3, it appears to me that 1 has more steric interaction of the methyl groups than 3.  So my ordering from least to most stable was 4,2,1,3.  Is there an error in my line of reasoning?

The correct arrangement (from least stable to most stable) is probably: $4 > 2 > 3 \gtrapprox 1$ based on my calculations for the four gas phase rotamers you can see in the figure below.

Your reasoning for 4 over 2 sounds good, that would also be my point. For the two minima 1 and 3 (with Boltzmann weights of 46%:54% at the highest calc. niveau (red), 33%:66% at the lowest (blue)) you might also be right. But from the really small energy difference I don’t want to argue too much about it.

My calculation methods seem a bit random, but at first I only wanted to make that fast wB97X-D3/def2-TZVP//HF-3c scan and then decided to calculate the bigger one.

The solution given by user16347 seems logical but is not really the correct solution. According to the answer the anti form is more stable since it has the least energy, this is actually false. Experimentally, the gauche form with 3 gauche interactions ( that is structure with energy 2.7 ) is more stable than the anti form. The reason being that in the anti form the geminal repulsions due to the methyl on adjacent carbon hakes the system more unstable than the gauche interactions. This for this particular molecule the gauche conformer is more stable.

For details into this concept of geminal repulsion, refered Chapter 2 - Strain and Staibilty from the book Physical Organic Chemistry by anslyn doherthy


Is an imine considered a secondary amine? I am looking at the difference between primary, secondary, and tertiary amines.

What about if the nitrogen is double bonded to a carbon?

Is this carbon counted twice, and is the molecule above a considered tertiary amine?

No. Imines cannot be considered an example of secondary amines any more than pyridine can be considered a tertiary amine.
The easiest way to think about this is with oxidation state: imines are at a higher oxidation level than amines, just as aldehydes are at a higher oxidation level than alcohols.
Aside from this 'intuitive' explanation, the IUPAC gold book (a source of chemical nomenclature) defines both amines and imines, clearly differentiating the two:

amines. Compounds formally derived from ammonia by replacing one, two or three hydrogen atoms by hydrocarbyl groups, and having the general structures $\ce{RNH2}$ (primary amines), $\ce{R2NH}$ (secondary amines), $\ce{R3N}$ (tertiary amines).
imines. Compounds having the structure $\ce{RN=CR2}$ (R = H, hydrocarbyl). Thus analogues of aldehydes or ketones, having NR doubly bonded to carbon; aldimines have the structure RCH=NR, ketimines have the structure $\ce{R'2C=NR}$ (R' ≠ H). Imines include azomethines and Schiff bases. Imine is used as a suffix in systematic nomenclature to denote the C=NH group excluding the carbon atom.

Generally NO. Tertiary amine is an ammonia derivative with 3 hydrogen atoms substituted to 3 alkyl/aryl substituents. Whereas imines exhibit some properties of tertiary amines (like pyridine). It can be protonated or alkylated and form iminium salts.


Reaction between Copper nitrate and ascorbic acid I am confused on the products of reaction between copper nitrate and ascorbic acid. When I searched this online, one of the reactions I got was: $$\ce{2Cu(NO3)2 + C6H8O6 <=> 2CuO + 4H2O + CO2}$$

I am really confused as to how they determined that to be the product. My thinking was that the ascorbic acid breaks into $$\ce{HC6H7O6 <=> H+ + C6H6O6-}$$ and so the hydrogen ion, whose activity is higher than copper, replaces the copper to form nitric acid. I was thinking that this could be a double displacement reaction and so the copper ascorbate would become a precipitate. What am I thinking wrong here?

It is always good to cite the link when you mention that you saw a redox reaction online. It is wrong anyway.

Aqueous Cu(II) is not an strong enough oxidizing agent that it will convert organic matter into carbon dioxide. In other words, it is a weak oxidizing agent. Ascorbic acid is oxidized to dehydroascorbic acid and Cu(II) should be reduced to Cu(I). If you heat the solution strongly, Cu(I) may convert into Cu$_2$O.

Ascorbic acid is indeed an acid, but there is nothing which prevents its conjugate base to behave like a reducing agent. Many organic acids can behave like that.

  I am really confused as to how they determined that to be the product.

By conducting the experiment. There is no other simple way to predict what will happen in a complex redox reaction.

In my assessment, an advanced reaction involving a cyclic REDOX reaction occurring with Copper ions, which I have previously discussed in this work 'Generation of Hydroxyl Radicals from Dissolved Transition Metals in Surrogate Lung Fluid Solutions' by Edgar Vidrio relating to disease (cancer) likely induced from progressive damage to  DNA (in the new equation this is now an attack on the nitrate ion,...) arising from the transition metal copper interacting with oxygen (in the new reaction this is replaced by HNO3,...) leading to the powerful hydroxyl radical.

That reaction becomes cyclic (at least, to a limited degree depending on the respective concentration of Ascorbic acid) per Vidrio's to quote:

“Similar reactions can occur with Cu, Cr and Ni. Furthermore, biological chelators and reductants can greatly enhance the production of ROS (Burkitt et al., 1991; Engelmann et al., 2003; Wenk et al., 2001). For example, in the presence of ascorbate (Asc), a biological reductant, the oxidized form of the transition metal produced by the Fenton reaction can be reactivated (R2 and R3), thus allowing additional ROS to be produced.

$\ce{Fe(III) + Asc (n) → Fe(II) + Asc (n') (R2)}$

$\ce{Cu(II) + Asc (n) → Cu(I) + Asc (n') (R3)}$ “

which cycles the cupric back to cuprous to feed a REDOX, which, for our new reaction involves aqueous cupric nitrate and Ascorbic acid, that is detailed in this reference, "Fenton chemistry in biology and medicine" by Josef Prousek, to quote reaction (15) on page 2330, a general depiction of Fenton-type reactions, to quote:

"For Fe(II) and Cu(I), this situation can be generally depicted as follows [20,39],

$\ce{ Fe(II)/Cu(I) + HOX → Fe(III)/Cu(II) + ·OH + X- }$ (15)

where X = Cl, ONO, and SCN. "

In the current context, the hydrolysis of cupric nitrate introduces HNO3, which interacts with cuprous (created from the Ascorbic acid presence) reforming cupric and, importantly, always hydroxyl radicals. The latter radical can attack nitrogen oxides compounds which can further progressively be broken down. Some expected reactions:

$\ce{ .OH + NO3- -> OH-  + ·NO2 }$

$\ce{ .NO2 + .NO2 -> N2O4 }$

$\ce{ N2O4 + 2 H2O <=> HNO2 + HNO3}$

$\ce{ 2 Cu(I) + HONO -> 2 Cu(II) + ·OH + N2O2^{2-} }$

etc., so the nitrate, nitrite,..., could be progressively decomposed (even to NH3), which may account for its notable absence in the cited text.

Lastly, the REDOX consumes H+ and select radical reactions could add OH-, all of which could eventually lead to either a basic copper salt, or Cu(OH)2. The latter on warming, forms CuO, as was claimed.


Density of water increases from 273 to 277 K due to increase in entropy? In a set of questions that I was solving, the answer to one of the questions asking why the density of liquid water increases from 0 to 4 degrees C had the above answer: due to an increase in entropy.

This website does explain this a bit with two thermodynamic equations, but I am unable to make sense of how entropy comes into the picture here. Please explain how the increase in density of water is due to an increase in entropy.

At $0°$C, the water is not yet made of independent $H_2O$ molecules. It is still a mixture  of hexagones containing 6 molecules $H_2O$. And these hexagones are separated by independent $H_2O$ molecules. There is still a certain order in the water at 0°C. These hexagones are destroyed between $0°$C and $4$°C. These hexagones are more cumbersome than 6 molecules $H_2O$, because there is a hole in the center of the hexagones. That is why the density of water at $0$°C is lower than at $4°$C.


Why do heavier elements have smaller specific heat values. In the periodic table the trend is that as the mass of each element increase's the specific heat tends to go down. This seems to be counter-intuitive because if I am not mistaken as mass increase's the amount of heat that the substance can hold goes up. So why does the specific heat of these elements go down?

The more particles in a system, the more modes of vibration (and rotation), so one gram of $\ce{H2}$ has about $235 \times$ more particles than one gram of U. The heat is "stored" in the motions of the atoms and electrons, until reaching millions of kelvin, at which point dissociation of the nucleus would be significant in specific heat. See these links and  for more information.

As the amount of a substance increases the Heat capacity of that substance does go up. However, Specific Heat is a correction for this because it is the Heat Capacity divided by the amount of the substance.

For Specific Heat the amount of a substance is irrelevant and its value is largely determined by the degrees of freedom for the constituent particles. The more degree of freedom, other than transnational, the higher the Specific Heat. An example of another degree being rotational energy which does not contribute to Temperature changes.


How can I remove the bitartrate anion from an organic bitartrate salt? I have a compound, dimethylaminoethanol (DMAE) bitartrate, which I originally purchased as a supplement and possible smart drug. I bought an absurd amount of it. It didn’t do anything for me as a supplement. I understand that DMAE, not the bitartrate salt, is commonly used in cosmetics, & I’d like to play with it for that purposes, maybe mixing it with glycerin, etc. Is there a simple way to strip off the bitartrate anion and have it precipitate or sublimate or something, leaving the DMAE behind? Or otherwise separating the DMAE out from the salt?

I have a strong preference for any other reactants and reaction products being non-toxic

I see on Wikipedia: "Dimethylethanolamine (DMAE or DMEA) is an organic compound with the formula (CH3)2NCH2CH2OH. It is bifunctional, containing both a tertiary amine and primary alcohol functional groups. It is a colorless viscous liquid. It is used in skin care products."

Now the fact that it is used in skin care products does not make it safe: the SDS  (https://www.fishersci.com/store/msds?partNumber=AC116180010&productDescription=N%2CN-DIMETHYLETHANOLAMINE+1LT&vendorId=VN00032119&countryCode=US&language=en) shows these pictograms:

Uh oh! Oh, well.

Well, let's forget that for a minute. DMAE is a liquid, boiling at 134C. Bitartrate anion is very insoluble if attached to potassium (potassium bitartrate is cream of tartar). If you added a stoichiometric amount of potassium bicarbonate (KHCO3, quite safe) to an aqueous solution of DMAE, you would precipitate almost all of the bitartrate, leaving the DMAE in water. Filter off the ppt and distill off the water and then distill and collect the DMAE. It will probably be a little wet and you will hate the overwhelming fishy smell.

Oh, and use a respirator (from the SDS: "Follow the OSHA respirator regulations found in 29 CFR 1910.134 or European Standard EN 149. Use a NIOSH/MSHA or European Standard EN 149 approved respirator if exposure limits are exceeded or if irritation or other symptoms are experienced.")

Good luck. Actually, I think you will have better luck if you do something else.

This might work:
Make a solution of NaOH in absolute ethanol, e.g. at a 1 mol/L concentration. Suspend the DMAE bitartrate in ethanol, and add 2 equivalents of the NaOH solution slowly to the suspension, making sure it doesn't heat up too much. After everything is added, stir or agitate it for a while. Then filter off the solids and wash them with some ethanol. Add the washings to the ethanol solution. From this solution, let the ethanol evaporate, and only DMAE should remain. The solids are sodium tartrate.


Solubility and equilibrium Is equilibrium between dissolution and recrystallization of a salt in an aqueous solution reached only when the solution is saturated; or can this also be established in an unsaturated solution?

Will Le Chatelier's principle apply in this situation (as it is about chemical equilibria)?

I was actually studying purification of salt by common ion effect where salt crystallizes from saturated aqueous solution on addition of common ion. Will the salt still crystallize if solvent still can dissolve more of it in unsaturated solution on addition of little amount of common ion. Since this crystallization is consequence of Le Chateliers principle, will crystallization still occur if solution taken was unsaturated and the common ion added is also in very small amount?

The solubility of a solute in a solvent is determined by the chemical potentials of the solute for the solution and the pure solute. Substances move from high chemical potential to low chemical potential. At the concentration of solubility, these two chemical potentials are equal so there is no net transfer. Below this concentration, the chemical potential of the solute in the solution is lower than that of the pure solute so molecules will transfer from the pure solute to solvent until it is all dissolved or saturation is reached.

When you invoke the common ion effect, you are increasing the chemical potential of the common ion in solution which decreases the solubility. It is possible to have a nearly saturated solution and add enough common ions to reduce the solubility below the original concentration.

Beginning level Thermodynamics is about path independent processes near (or at) equilibrium. Almost no thermodynamic systems are left undisturbed long enough for true equilibrium to occur, so we speak about "near equilibrium" states of isolated, closed, or open systems. What exactly we mean by "near" is not obvious or easy. A solution in equilibrium is one in which the concentration of solute is constant (and stable). Visual observation of a solution to determine if it is stable is not optimum, although it is convenient. The problems with this approach are, I think, beyond the scope of your question. (They involve the actual molecular and atomic scale processes which occur during crystallization/dissolution.) So, to answer your first question: Equilibrium is reached when the concentration is constant. If sufficient solute is present to saturate the system, then it is saturated equilibrium, if less is present, then unsaturated equilibrium. The answer to your second question is a bit more complicated. If I take a saturated salt solution and add more solid salt to it, does the concentration in the solution change? No. So Le Chateliers fails here. You need to be aware of the difference between quantitative chemical relationships (for instance equilibrium constants) and qualitative ones (eg. Le Chats.Principle). I shouldn't have to tell you which one is more trustworthy. (OTOH, quantitative relationships are all specific, while qualitative ones can often be applied without knowing all of the details - they're more general because they're less specific. But along with generality comes exceptions to these "rules"....) Does Le Chatelier's Principle apply here? I suppose it could but I don't see any useful application of it. I guess the common ion effect could be looked at as an example of Le Chat's, but it seems to me easier to explain by looking at the equation for Ksp (quant wins over qual). As to your third question: It isn't coherent. "Will something crystallize out if it isn't at saturation?" is what you're asking. By definition, no. Can you reach saturation by adding a common ion rather that the solute of interest? Yes, as you suggest, this will be (generally) in solutions near saturation (and recall, saturation implies equilibrium.) But keep in mind that adding a common ion is actually adding part of the solute of interest - the [Anion] or [Cation] is increasing, and so the ratio of reactants to products also changes, moving closer to saturation.


Why does Devarda's alloy include copper? Devarda's alloy is used in the detection of nitrates. However, judging by the reaction equation, only aluminium plays the role of reducer:

$$\ce{3NaNO3 + 8Al + 5NaOH + 18H2O -> 8Na[Al(OH)4] + 3NH3(g)}$$

Why does it contain copper? Is copper also capable of acting as a reducer here?

I know that zinc, the third component, is capable of that.

Virtually every analytical chemistry textbook's chapter on anion determination mentions that Devarda's alloy is very brittle and can be easily pulverized in a mortar, unlike $\ce{Al}$ and $\ce{Zn}$ metals.
This is quite handy property as for qualitative analysis of nitrates in a heterogeneous media the reducing phase should be ground to fine powder in order to:

assure micro-scale precision of the added reactant, which is tricky to achieve with aluminium or zinc, as both metals are usually shipped in granules, and the fine metal powders are prone to oxidation, increasing the amount of unwanted oxides (plus, their powders are sticky);
increase the available reaction surface to speed up the analysis and increase its completeness (especially important for isotopic nitrate analysis).

On top of that, German Wikipedia
also supports both comments by Ivan Neretin

  Galvanic couple, I guess.

and aventurin

  Catalyst, I guess. ;-)

and suggests that copper facilitates catalytical formation of a local electrochemical cell.
A reduction with Devarda's alloy can be carried out in weakly alkaline solution or even in neutral solution, whereas reduction with aluminum often requires a strongly alkaline solution.
At the same pH, the reaction with Devarda's alloy is overall significantly faster than when using zinc or aluminum.

This was experimentally demonstrated by Mertens et al. [1]:

  The superiority of Devarda alloy compared to aluminum dust as a reductant in alkaline medium is illustrated best with the following experiment (see Figure 3).
  Twenty ppm $\ce{NO3-}$ is subjected to reduction, first with aluminum dust and the potential of the ammonia probe is recorded continuously.
  One observes a slowly rising potential.
  On addition of Devarda alloy to this reaction mixture, a potential jump is obtained and the potential stabilizes rapidly (after 5 minutes).
  This shows clearly that the kinetics of the reduction with Devarda alloy are much more rapid than with aluminum dust.

  Figure 3. Potential observed with the ammonia probe during the reduction of 20 ppm $\ce{NO3-}$ as a function of time
  From $t = 0$ to A: aluminum dust. At point A, Devarda alloy was added.

References

Mertens, John.; Van den Winkel, Pierre.; Massart, D. L. Determination of Nitrate in Water with an Ammonia Probe. Anal. Chem. 1975, 47 (3), 522–526. DOI: 10.1021/ac60353a009.

More recent science (as compared to 1975), the action of Al with NaOH is believed to produce surface absorbed hydrogen atoms on the aluminum. Related chemistry is commonly employed via the vigorous action of HCl on Zn (so one of the roles of zinc is to provide added surface area, which may be acted upon by NaOH also, to answer a posited question).

More precisely on the referenced surface chemistry, let me quote from this 2008 doctoral thesis " Alkaline dissolution of aluminum: surface chemistry and subsurface interfacial phenomena" by Saikat Adhikari, as an example:

  Hydrogen embrittlement, a process in which brittlement and subsequent cracking of metals is caused by ingress of atomic hydrogen into the metal, is also sometimes thought of as a type of stress corrosion cracking. Due to the detrimental effects of such corrosion processes, it is important to have a fundamental understanding of associated alkaline or acidic dissolution mechanisms.

And more directly:

  Apart from activation of the surface, dissolution of aluminum in alkaline solutions at open-circuit also leads extremely high rates of H-absorption into the metal, [9-14] and also formation of nanoscale voids or hydrogen bubbles [15, 16] near the aluminum surface.

H-absorption on the Aluminum surface is a source of hydrogen radicals, •H, which could engage in a stepwise breakdown of nitrate to ammonia.
For example, with the nitrate ion a radical reaction:

$\ce{•H + NO3- ⇌ •OH + NO2-}$

$\ce{•H + •OH -> H2O}$

As: e- + H+ ⇌ •H, the net of the above two reactions can be written as (source, see Section 8.28.2.1 Summary of Properties) the following cited reaction for the first stage creation of nitrite:

$\ce{NO3- + 2 H+ + 2 e- -> NO2- + H2O}$

My experience with just alkaline Al is that it is apparently difficult to stop the reaction at the point of just nitrite creation only, by say limiting the amount of aluminum.

So we have a complex reaction system involving surface chemistry, and in the presence of other metals (like copper), a galvanic cell (which implies two different metals, namely Zn and Cu, to answer a second question) introducing solvated electrons, which may push the equilibrium below to the right:
$$
e-(aq) +  H3O+(aq) ⇌ •H (aq)
$$
promoting hydrogen radicals. The latter could explain the acceleration in the breakdown of nitrate (as reported).

Interestingly, one may assume that the •H radical functional behaves (per its seemingly reversible formation reaction: e- + H+ = •H ) as apparently a (e-, H+) pair acting on ions. For an example from 'Hydrometallurgy 2008: Proceedings of the Sixth International Symposium', p. 818, a commercial reductive leaching application citing the reaction (based on surface-treated Aluminum) to quote:

" PbS + 2 •H = Pb + H2S (5) "

Source reference for the above.

which I view functionally as follows:

Pb(+2)S(2-) + 2 (e-, H+) = Pb + H2S (g)

Further, I do NOT recommend experimenting with Al/NaOH in the presence of impurities in a chemistry class. To quote from Bretherick, Volume 1, page 23-24:

  A student mixed aluminum foil and drain cleaner in a soft drink bottle, which started emitting gas. Another student carried the bottle outside and was claimed to have been overcome by the toxic fumes [1]. Most drain cleaners are alkalis, so that aluminum will dissolve to produce hydrogen. The bleach that it is is suggested may have been present will produce no toxic fumes in alkali, and one would be surprised to find arsenic or antimony compounds present. If the collapse was not purely hysterical, the remaining, though remote, possibility would be phosphine. The soft drink the bottle had contained was one of the many perhaps best described as impure dilute phosphoric acid [2]. "


What is the structural formula of alkali hypohalite: MOX or MXO? There seems to be an inconsistency in the formula. Somewhere it is written as MOX and in some places, it is written as MXO. This is what Google gives when you search sodium hypochlorite:

I think the formula of alkali hypohalite should be MOX because since the negative charge bears over the oxygen atom, it accepts the electron coming from the alkali metal ion and forms a duplet  thus forming ionic bond.

Is my assumption correct? What is the correct formula of alkali hypohalite?

I felt tempted to close this as a duplicate of the sodium acetate notation question, because of my answer there, but the question in itself does not ask the same thing, so I’ll just rewrite the relevant parts of my answer.

In short, there is no correct formula. Depending on which point you want to bring across, you can use either, and depending on the context, $\ce{ClNaO}$ may also be acceptable.

Structually, you are correct with your assumption; the oxygen bears the negative charge. But that doesn’t include any implication that the formula should be written that way; compare $\ce{SO4^2-}$ where the charge is also on the oxygen atoms and yet they come last.

In fact, that leads us to our first clue. By convention, oxygen atoms are typically written last; compare $\ce{ClO3- , ClO4- , BrO3-}$ and others. You could probably explain that with electronegativity, the convention being that electronegative elements come last. This assumption argues for $\ce{NaClO}$.

On the other hand, the corresponding acid $\ce{HOCl}$ is often understood as derived from water $\ce{HOH}$, replacing a hydrogen with a chlorine atom. From this starting point, it makes sense to write $\ce{NaOCl}$, because then you replaced the other hydrogen; this time with sodium. It also includes a resemblance to $\ce{NaOH}$.

Both ways can thus be supported by the reasons given above. In practice, I think I have always seen the substance being referred to as $\ce{NaOCl}$ but I could simply have ignored the occurances of $\ce{NaClO}$.

Both formulae are correct. NaOX is correct structurally but preferred form is NaXO in correspondence with other series salts like Oxo-halate(III) NaXO2, Oxo-halate(V) NaXO3 and  Oxo-halate(VII) NaXO4.


Calculation of proportion of zwitterion from this please. This is the question. I tried to post it in the title section but it was rejected because it was considered lengthy. Question: Valine is one of the 20 natural amino acids and has an acid -COOH (pka1=2.6) and an amine-NH2(pka2=9.6). If we consider a valine solution with pH=10.6, what is the proportion (%) of the zwitterion form of the solution? A) 2.4 B) 3.6 C) 5.7 D) 9.1 E) 10.2.
Sorry if this might be too elementary but I don't know how to solve this question. I need help, thanks.

(I shouldn't answer this because it is a homework question with no effort, but since someone posted a wrong answer I'm answering anyway)

Only the pK of 9.6 is relevant because the pH is 10.6; therefore, think of the situation as having just two species, zwitterion (acid) and negative (conjugate base).

Because pH is one unit above the pK, the ratio of acid to conjugate base is 1 to 10.

Ratio of acid to total is 1 to 11.

1/11 = 0.0909

Proportion of zwitterion = pka amine : pka acid
                                           =9.6 : 2.6
                                           =3.6923


Reaction of alkyne with sodium in ethanol What is the product obtained on the reaction of but-2-yne with $\ce{Na}$ in ethanol?

I know that $\ce{Na}$ in liquid $\ce{NH3}$ converts an alkyne to an alkene by anti addition, and $\ce{Na}$ in ethanol is the principle reagent in Bouveault Blanc Reduction, but what about this reaction.

The answer given in my book is a trans-alkene. Is this a standard reaction? Also does it depend if it is a terminal alkyne or not? (I guess yes, because the terminal alkyne will try to deprotonate)

the reduction of alkynes using Na, proceeds through the anion radical, which can obtain trans conformation for the stability. Now considering the usage of NH3 and ethanol are proton donors. So, there is no effect on trans conformation altering the proton donors.

That's actually an example of Birch reduction of internal alkynes.

It's given in detail on this website.

Also, you're completely right in your reasoning that a terminal alkyne will not show this reaction.


Can eggs boil indefinitely? At breakfast someone asked if the eggs were boiling. While obviously they meant the water for the eggs, but it lead to a discussion regarding if you can make an egg boil without it coagulating or even if you can heat a coagulated egg to the point that it starts boiling under special circumstances such as in an oxygen free environment to prevent combustion.

So would it be possible to make an egg boil without coagulation or otherwise destroying it? The shell would break either way so we can remove that for the experiment.

Boiling means dynamic evaporation of liquid.

Infinite boiling would mean to have infinite amount of that liquid and therefore the egg would have infinite mass.

Such an egg would collapse into a black hole and we would not be able to continue in our observation.

In reality, in inert atmosphere, with overlapping stages:

protein coagulation
water boiling
organic matter thermal dehydration
general organic matter decomposion.
carbon stage

If we kept temperature just above $\pu{100 ^\circ C}$ to keep boiling,  stage 3-4 would be very slow and we would not reach carbon.

While I agree some chemistry is involved in this operatory laboratorium experiment, I believe it may be a mere physics quest. Taking the theoretic existence of a blackhole into account, one can infer that this is a very hypothetical question. Scientists have discovered a way to test this finding in laboratoriums relating to physics.

Now.. it is essential to find a physics laboratory and ask questions for the laboratorium supervisors. An infinite mass of egg, such a computational integer exists not. Perhaps finding the forma for your hypothesis and simply taking the limit to infinity. The time and progress of an egg to boil indefinitely will be of answer to the equation. This will be an educated guess and a quite remarkable and extraordinary result.


Does H3CO3+ ion exist and what is its structure? Why doesn't $\ce{H3CO3+}$ ion exist and why doesn't the carbonyl group in carbonic acid $(\ce{H2CO3})$ seem reactive?

A protonated carbonic acid (trihydroxymethyl cation) $\ce{H3CO3+}$ does exist, and is in fact incredibly stable in the solutions of superacids [1] up to $\pu{0 °C}.$ One way of obtaining $\ce{H3CO3+}$ is dissolving inorganic carbonates and hydrogen carbonates in magic acid at $\pu{-80 °C}$ [2]:

  $$\ce{CO3^2- or HCO3- ->[FSO3H-SbF5/SO2][\pu{-80 °C}] H3CO3+ ->[\pu{-10 - 0 °C}] CO2 + H3O+}$$

As for the structure, the $C_\mathrm{3h}$-symmetrical molecule (triskelion shape, structure 7 below) was found to be the global minimum [2] (reference numbers updated):

  We also carried out ab initio and IGLO calculations […] on bicarbonate as well as free and mono-, and diprotonated carbonic acids to rationalize the observed experimental results. Initially the possible geometries were optimized at the HF/6-31G* level. The most stable isomers were subsequently determined from
  relative energies at the MP2/6-31G*//HF/6-31G* level. Finally, the lowest energy structures were further optimized at the MP2/6-31G* level. MP2/6-31G* optimized structures were used to calculate $\ce{^{13}C}$ NMR chemical shifts.

  […]

  Protonated Carbonic Acid. The symmetrical $C_\mathrm{3h}$ structure 7
  was found to be the global minimum. This structure was suggested for the observed species in superacid solution by Olah et al. [1] on the basis of $\ce{^1H}$ and $\ce{^{13}C}$ NMR spectroscopy. The ion at $\pu{-80 °C}$ shows single sharp peaks at $δ(\ce{^1H})~12.05$ and $δ(\ce{^{13}C})~165.4$ in the proton and $\ce{^{13}C}$ NMR spectra, respectively.

  Protonation of carbonic acid on the hydroxyl oxygen atom (8) was shown to be unfavorable by $\pu{23.4 kcal/mol}$ over protonation on the carbonyl oxygen (7). Frequency calculation at the HF/6-31G*//HF/6-31G* level showed that the structure 8 is not a minimum, as it contains two imaginary frequencies.

$\ce{H3CO3+}$ shares structural similarities with its triaza-analog, the guanidinium ion, as both are possessing resonance stabilization via their onium forms [3, p. 60].

References

Olah, G. A.; White, A. M. Stable Carbonium Ions. LXIV. Protonated Carbonic Acid (Trihydroxycarbonium Ion) and Protonated Alkyl (Aryl) Carbonates and Hydrogen Carbonates, and Their Cleavage to Protonated Carbonic Acid and Carbonium Ions. The Possible Role of Protonated Carbonic Acid in Biological Carboxylation Processes. J. Am. Chem. Soc. 1968, 90 (7), 1884–1889. https://doi.org/10.1021/ja01009a036.
Rasul, G.; Reddy, V. P.; Zdunek, L. Z.; Prakash, G. K. S.; Olah, G. A. Chemistry in Superacids. 12. Carbonic Acid and Its Mono- and Diprotonation: NMR, Ab Initio, and IGLO Investigation. J. Am. Chem. Soc. 1993, 115 (6), 2236–2238. https://doi.org/10.1021/ja00059a020.
Dewar, M. J. S., Hafner, K., Heilbronner, E., Itô, S., Lehn, J.-M., Niedenzu, K., Rees, C. W., Schäfer, K., Wittig, G., Boschke, F. L., Series Eds.; Topics in Current Chemistry; Springer Berlin Heidelberg: Berlin, Heidelberg, 1979; Vol. 80.

$\ce{H3CO3}$ does not exist. Why should it exist ? Where does this formula come from ? Of course you are allowed to write any formula you want. It is not forbidden to write $\ce{HC4O3}$ $\ce{H3CO5}$ $\ce{H3C2O3}$ or any other combination of C, H and O atoms that you may think of. Just for fun ! But Nature ignores your fantasy. Nobody has ever been able to make these substances. So these substances do not exist.

The second part of your question may be explained by the fact that $\ce{H2CO3}$ is extremely instable, and is always largely decomposed into $\ce{CO2}$ and $\ce{H2O}$. $\ce{H2CO3}$ exists only in very dilute solution in water, and in equilibrium with $\ce{CO2}$ and $\ce{H2O}$. If you try to make a reaction with the carbonyl group of $\ce{H2CO3}$, it will first be decomposed before any other reaction. The only exception is its reaction with the OH- ion. But both $\ce{H2CO3}$ and $\ce{CO2}$ can react with NaOH, producing the same $\ce{CO3^2-}$ ion. So you cannot be sure about the nature of the molecule that reacts with NaOH. It may be $\ce{H2CO3}$ or $\ce{CO2}$.


Why is 2s more stable than 2p, if 2p experiences greater effective nuclear charge? I'm referring specifically to this graph, which appeared in a chemistry lecture but which the lecturer could not explain. We know that 2s is lower energy than 2p. But surely those electrons in 2p orbitals, which apparently experience greater effective nuclear charge for Z > 10, would have lower energy! Can someone reconcile my understanding of orbital energy levels and this diagram?

One big issue here is that we generally don't talk about different ionization energies of anything but the outermost orbitals. Once electrons are in "core," we usually don't care since they generally do not participate in any kind of everyday chemistry.

We generally consider the $2s$ orbital to be lower in energy than the $2p$ because the $2s$ has some density inside the radial node that we think of as "nuclear penetration," i.e., the $2s$ electrons are able to get closer to the nucleus through the $1s$ core electrons. Now, once you add a $3s$ electron, I wonder if the innermost radial probability regions of the $3s$ actually repel the $2s$ electrons more than the $2p$ electrons which are generally farther away. The differential repulsion would have to happen in such a way so that it more than compensates for any nuclear penetration effects of the $2s$ electrons. This would certainly align with the observation that the $2p$ electrons are more stabilized only when we add a $3s$ electron.

Note that the x-axis is atomic number.  The graph does not allow you to compare the energy of Fluorine's 2s and 2p orbitals (Z=9) but rather shows the energy of Lithium's and Beryllium's 2s orbitals (Z=3 and 4) and the energy of Boron's through Fluorine's 2p orbitals (Z = 5 through 9).  The energies of the orbitals are not comparable on the graph because the atomic numbers are not the same.  To compare the energies of Fluorine's 2s orbitals to its 2p orbitals, you would need to calculate the effective nuclear charge felt by the 2s electrons in Fluorine and the effective nuclear charge felt by the 2p orbitals in Fluorine, both calculations using Z=9.  You can do this using Slater's rules for assigning shielding constants felt by the electrons in these orbitals.


Does a grain of salt have exactly equal amount of sodium and chlorine? Does a grain of salt have exactly equal amount of sodium and chlorine? I know that the formula unit of salt is 1:1, but for example if there is 10000000 atoms of sodium is it possible that that there is 9999999 atoms chlorine? Thanks.

Crystal solids, however small, have defects in them. These defects are basically irregularities in the arrangement of constituent particles. They are of two types:

Point defects
Line defects.

Point defects themselves are divided into other kinds of defects. Here, I'll explain only the relevant one.

One type of point defect is a vacancy defect in which some of the lattice sites in the crystal are vacant. This means that some atoms of Na or Cl or both are missing from their lattice sites.

So the answer to your question is, yes. That scenario is very much possible.

If it doesn't it isn't "salt" or at least it isn't NaCl, which requires one of each per molecule.


Simple Explanation of orbitals Imagine I was your little brother, how would you explain an orbital to him? (assuming he knows what atoms electrons and neutrons are, and the rest of the basics till that point.)

I have been reading up on this and can't make heads out tails of it.
As a side how do the different energy levels relate to them.

I hope I'm clear, I'm doing my best. I read several books but can't get a clear bearing on this.

Well, the first step is to stop thinking electrons as being very small balls that orbit around the nucleus in a circular path. This is known as the Bohr model. Despite this model being an excellent model at an introductory level, it doesn't tell the entire story.

There are several issues with the notion that electrons travel in a circular orbit around the nucleus, the main one being that they should eventually undergo orbital decay and hence eventually slow down and crash into the nucleus. Of course, this doesn't happen. So how do electrons move around the nucleus?

This is where quantum theory came to the rescue in the early 20th century. It stated that electrons could not be treated as a classical particle and didn't not have a definite position and momentum. So this basically means that we can't exactly know an electron's position and its momentum at the same time. This relationship is given by the Heisenberg Uncertainty Principle.

I won't go into detail about it since to keep this explanation basic, but if you are interested check out this video which provides a really good explanation.

So basically what quantum theory tells us that we actually can't specify where and what a path an electron is going to take. However the Bohr model assumes that you can precisely plot out where an electron is going to be, but we now know that due to the Heisenberg Uncertainty Principle you can’t precisely know the exact position and momentum of an electron. So we actually can't construct an orbit and say that the electron is always going to follow that path.

Therefore, it makes it impossible to plot an orbit for an electron as how can you? If you don’t know where the electron is going to be next, there is no way you can predict its path.

So how can we get around this problem? Well, if we can’t draw the orbit for an electron the next best thing we can do is to plot an electron density probability orbital for it. Also known as atomic orbitals.

An orbital is a well defined region of space. Therefore an atomic orbital is a region of space that shows where the electron will be 95% of the time (usually we take 95% but it could be any number really such as 90% or 75%).

A simple way of thinking of orbitals is that imagine that you have a magic camera that can take a sequence of photos of an electron in a hydrogen atom. The electron appears as a dot. Now if we superimpose all of these images, you will see something like this:

Since we generally take atomic orbitals to be the region of space that shows where the electron will be 95% of the time, the very outer sphere of the above image is said to be the orbital of the electron.

Hopes this helps you understand orbitals. If you have any question, feel free to ask.

The oribitals are the locations where electrons settle in a stable atom.  This gets based upon two hemispheres of the underlying magnetic-like "nucleomagnetics" weak force from the nucleus to each electron keeping electrons in a shell.  This force is repulsive where electrostatic charge is attractive.  That I see why electrons stay in the field.

As a result, you get the first shell balancing points at the two axis end caps.  Hence, Hyrogen with one and Helium with two then the subshell and shell are full.

After that, electrons fill in two hemispheres growing by squares so the next shell is two subshells.  This of a pyramid that builds 1/3/5/5/3/1 from pole to equator to pole.  Note that 1+3+5 =9 which since a Perfect square.  Yet, these are in two hemispheres so subshells as 2, 6, 10 in size.  And full shells are 2x1, 2x4, 2x9.

See my book on Scrunched Cube Atomic Model to explain shells and subshells for ordinary people.


Why is heating increasing the solubility of sodium hydroxide? \begin{align}
\ce{NaOH &-> Na+ + OH-} & \Delta H^\circ &= \pu{-44.51 kJ/mol}
\end{align}

The dissolution of sodium hydroxide in water is an exothermic process, and so, according to Le Chatelier’s principle, cooling the container should shift the reaction to the right. Shouldn’t this mean that cooling the container will increase the solubility?

According to a solubility chart for sodium hydroxide, heating the container will increase the solubility; why is this the case?

The question to the Newton - Ask a Scientist program by Danna C Griffiths (to be found via the Internet Archive) has an answer for it.

What I understood from the explanation on that site:

Solubility is defined as the concentration of the solute in a saturated solution. So when considering the increase in solubility with temperature, you have to check the enthalpy of solution of $\ce{NaOH}$ in a saturated solution of $\ce{NaOH}$.

When you dissolve $\ce{NaOH}$ in pure water, the process is highly exothermic, but as the concentration of $\ce{NaOH}$ increases, the process becomes less exothermic and eventually near the saturation point it becomes endothermic (it becomes less favourable for more $\ce{NaOH}$ to dissolve).

So according to Le Chatelier’s Principle, although the dissolution process is overall exothermic, since solubility is determined only at the saturation point, the solubility increases with increase in temperature.

According to

$$\Delta G = -RT \ln K$$

K is dependent on Temperature.

Le Chatelier's principle is explicitly a mass balance issue, and thus is not affected by temperature, although that distinction may be pedantic.


Balancing complex haloform redox reaction I just completed a haloform reaction experiment, using acetophenone and sodium hypochlorite to form benzoic acid and chloroform.

My lab report has a question regarding balancing the equation of the reaction.  I have been working on it for quite a while, and I can't figure out how to do it.  I've tried looking up methods but too many atoms are reliant on too many molecules!  Can someone help with this question and explain how to do it so I can really understand this process?

I need to be able to identify the limiting reactant as well, which relies on the molar ratio of the reactant to the product, so I really need to know how to do this.

As you noted, you need to find the limiting reactant. This is most of the time done by taking the one with a 1:1 ratio of educt:product.

So in this case, the $\ce{ROCH_3}$. So you take that once. Now you see immediately that you need 2 $\ce{Na}$ on the right side, so you can put a 2 in front of the $\ce{NaOH}$. You have two $\ce{X}$ on the left, but 4 on the right, so put a 2 in front of the left one. Now count all your atoms. Subtract them from each other:

You have $a * \ce{H}$ on the left and $b * \ce{H}$ on the right. if $a-b=0$ its balanced. Do that for every reactant and you're done.

There are of course more strictly formulated methods, but in my time I noticed that a straightforward approach with a little thinking can be applied most universally without having to memorize some formulas.

The coefficients should be

1 3 4 1 1 3 3, from left to right. The limiting reagent is acetophenone.

Used this website to find the answer.

http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch18/ch18-3-2b.html

-- a fellow owl ;)


Is it true that organic chemistry as an independent research field is no longer developing? One of my professors told me that organic chemistry has become a tool for biological research, rather than an active research field of its own.

Is this assertion valid? Does it describe correctly the general attitude towards organic chemistry in research institutions today?

The closest thing to organic chemistry would now be the shape and sequencing of protiens and or DNA. That field is best left alone as it takes too much time to pay anyone to get any applicable results.
Advances in things like machine learning are promising to the unraveling of these complex structures. Humams are better at recognising the patterns hence the "fold it" game.


What is the benefit of using Am-241 in smoke detectors even though it is a radioactive element? I see the labels denoting Am-241 in my smoke detectors. I am having a hard time understand why Am-241 is used in smoke detectors even though it is a radioactive element. It seems to me like a lot of trouble to use a radioisotope in consumer products and notably extra effort given that Am-241 is a synthetic isotope. Why would Am-241 be used in the manufacture of smoke detectors?

The only mitigating factor I can see is that it has a relatively long half life.

The usage in ionizing smoke detectors requires a radioactive isotope to work.  In addition to a sufficient half-life to make a smoke detector with a suitable service life, Am-241 emits mostly alpha radiation (which is used for the ionizing mechanism) with relatively little of the more hazardous and useless gamma radiation.  See the Applications section in  https://en.m.wikipedia.org/wiki/Americium.

Photons are blocked by water vapour, but the alpha particle radiation from A-241 continues through only stopped by smoke.

It's therefore much more sensitive to a single particle of smoke. Water vapour on the other hand, is only able to stop the photons of a photo-electric smoke detector, which uses no Americium-241 at all.

Americium is also likely the most toxic substances on the planet around with a teaspoon able to kill hundreds of thousands if taken intravenous. It's mixed with Gold and beaten into an aluminium container. You could go around burnt buildings to collect them perhaps, with a radiation detector.

The batteries last longer.


Doubts on IUPAC name 1,1'-methylenebis(4-isocyanatobenzene) Wikipedia states  IUPAC name of the above compound as 1,1'-methylenebis(4-isocyanatobenzene) .

I have a confusion. Shouldn't isocyanate group be given preference over the methylene group? I assume the name as 4,4'-methylenebis(1-isocyanatobenzene) .

Similar is the case with DDT.  Wikipedia states its IUPAC  name as 1,1'-(2,2,2-Trichloroethane-1,1-diyl)bis(4-chlorobenzene) .

Why is the numbering of carbon with three chlorine atoms done $1$ ? Moreover,  can it be  $1,1'$ in the first place in place of $4,4'$ ? I guess the $\text{diyl}$ term indicates that it is connected to two different carbons(of benzene)  .

To sum up,  I want to know which chain should I select as parent chain and why .

The isocyanato group is named by prefix (like halogeno-), it's not a principal characteristic group named by suffix. It can be said that these groups have lower priority than the “joint” position.
See the different numberings (examples from the IUPAC nomenclature of organic chemistry, 2013):

P-15.3.2.4.1
(…)

1,1′-oxybis(4-bromobenzene) (PIN)

4,4’-oxybis(2-bromobenzoic acid) (PIN)

Therefore it is named 1,1′-methylenebis(4-isocyanatobenzene), not 4,4′-methylenebis(1-isocyanatobenzene).
(You will probably argue that here, the oxy might have higher priority than bromo group or methylene connector, but “connector priority” is not taken into account.)
See also more rigirous answer to Q: What is the rationale of the IUPAC name for DDT?

One rule of nomenclature is that you should get the simplest name and, in both of your examples, the simplest is to acknowledge the symmetry of each molecule. So the "parent hydride" is methane in your first example and ethane in the second one because there is no simple way to describe an "alkyl group with 2 substituted phenyls attached".


Polymerisation of methyl methacrylate - can the C=O group get involved? When methyl methacrylate undergoes free radical addition polymerisation, the C=C pi bond is broken during the process:

I'm wondering why the C=O pi bond doesn't get broken during the polymerisation. I suspect it may be because the C=O bond is stronger than the C=C bond, but are there any other ways to rationalise this?

C=O bond is strongly polar, and the cleavage of the double bond would rather involve heterolytic breaking with ions formed than homolytic with radicals.


What type of reaction is this? (Modified organic nucleoside synthesis) What type of reaction is shown in step F of this patent image?
I tried looking it up in SciFinder, but I didn't get any results, and I have not seen this before.

Thank you!
Edit 1: I'm also wondering if there is a specific name for this reaction, so that I can look it up.
Edit 2 Yes I'm sorry I forgot the reagents. The patent number is WO2014078463A1 (Bennett et al.) and I included a screenshot below of the relevant description.

My thoughts are along the lines of @Andrew and @Waylander. Step F doesn't tell us much, @orthocresol! I prefer a cyclization, 1 $\rightarrow$ 2a, followed by hydrolysis of ester 2a to alcohol 2b. Alternatively, N-O bond homolysis with light would probably involve a cage mechanism. .

EDIT: I miss read the image and saw a carbonyl group where there was an alkene. This explanation has been changed.
This is my guess on what is occurring (an electrophilic addition to an alkene):

The nitrogen-oxygen bond on the amide is severed, forming a carboxylic acid an an electron deficient amide. The electron pair of this carboxylic acid is stabilized by resonance. What is left is a nitrenium ion ($\ce{NR2^{+}}$), which is described on Wikipedia as

Nitrenium species have been exploited as intermediates in organic reactions. They are typically generated via heterolysis of N–X (X = N, O, Hal) bonds. For instance, they are formed upon treatment of chloramine derivatives with silver salts or by activation of aryl hydroxylamine derivatives or aryl azides with Brønsted or Lewis acids. The Bamberger rearrangement is an early example of a reaction that is now thought to proceed via an aryl nitrenium intermediate. They can also act as electrophiles in electrophilic aromatic substitution.

The mechanism for an electrophilic aromatic substitution is similar to that of an electrophilic addition to an alkene.

The nitrenium ion then performs an electrophilic attack on the alkene (second carbon of the oxolane), with beta carbon undergoing a nucleophilic attack by either water in neutral solution or a hydroxyl group in basic solution, forming the primary alcohol. These two processes are likely concerted (otherwise, there would be the formation of a primary carbonation).

Below is a scheme for the predicted mechanism.


Cyclic compound stereoisomer How can the compound 1,2-cyclopentanediol have 3 stereoisomers? Isn't the configuration of the OH groups fixed in a ring?

By definition, stereoisomers are the compounds which have same connectivity of atoms but different spacial orientation of atoms and two stereoisomers do not have same orientation from any point of view.
There are different possibilities can a molecule of cylclopentane-1,2-diol can have:
For simplicity, I have made two hydroxyl ($\ce{OH}$) into two methyl groups.
But by definition you can cross out two structures because there exist a vantage point from which two molecules look the same (rotate molecule 180 out of plane and then rotate it slightly in plane, you get 2 pairs of identical molecules i.e 1,4 and 2,3. Hence, there are only two stereoisomers possible for cylopentane-1,2-diol.
Hope that helps.


Difference between Reaction Rate and Rate Law? Lets say you have the reaction

$$\ce{$a$A + $b$B -> $c$C + $d$D}$$

The rate of the reaction is given by the change in initial concentration over the change in time.

$$-\frac{\Delta \left[\ce A \right]}{a\Delta t} = -\frac{\Delta \left[\ce B \right]}{b\Delta t} = \frac{\Delta \left[ \ce C \right]}{c\Delta t} = \frac{\Delta \left[\ce D \right]}{d\Delta t}$$

However, there is also the rate expression, which is of the form

$$\text{reaction rate} = k \left[\ce  A \right]^{m}\left[\ce  B \right]^{n}$$

What is the difference between these two, and what purpose does each one serve?

Further, what does the $k$ represent?

Let's use the following gaseous chemical reaction as an example:
$$\ce{H2 + I2 -> 2HI}$$
When we study reaction kinetics, we oftentimes can't observe the changes in concentration of each reactant and product.  Therefore the rate expression helps us to relate the rates of appearance/disappearance of the species present:
$$-\frac{1}{1}\frac{\Delta \left[ \ce{H2} \right]}{\Delta t}\;
= \; -\frac{1}{1}\frac{\Delta \left[ \ce{I2} \right]}{\Delta t}\;
=\; \frac{1}{2}\frac{\Delta \left[ \ce{HI} \right]}{\Delta t}\tag{rate expression}\label{rateex}$$
Note that each rate is modified by the inverse of the stoichiometric coefficient.  I explicitly put $\frac{1}{1}$ in the expression just to emphasize this point.  We can see that multiplying the coefficients of the reaction will not change the rate expression in the following example: let's multiply the coefficients in the chemical reaction by 2 and then write the corresponding rate expression.
$$\ce{2H2 + 2I2 -> 4HI}$$
$$-\frac{1}{2}\frac{\Delta \left[ \ce{H2} \right]}{\Delta t}\;
= \; -\frac{1}{2}\frac{\Delta \left[ \ce{I2} \right]}{\Delta t}\;
=\; \frac{1}{4}\frac{\Delta \left[ \ce{HI} \right]}{\Delta t}\tag{new expression}\label{newex}$$
The $\ref{rateex}$ and $\ref{newex}$ are numerically equivalent.

Full disclosure here: the term rate expression is ambiguous and is also used as a synonym for the rate law. I've also seen the rate expression referred to as the rate relationship.

The rate law or differential rate law relates the rate of a reaction to the concentration (or pressure) of the reactants.  The rate of a reaction is proportional to the concentration (or pressure) of the reactants modified by some experimentally determined number called the reaction order.
$$\begin{align}
-\frac{\Delta\left[ \ce{H2} \right]}{\Delta t}
&\propto\left[\ce{H2}\right]^m\left[\ce{I2}\right]^n\\
-\frac{\Delta\left[ \ce{H2} \right]}{\Delta t}
&= k\left[\ce{H2}\right]^m\left[\ce{I2}\right]^n\tag{rate law}
\end{align}$$
Here, $m$ and $n$ are the reaction orders for $\ce{H2}$ and $\ce{I2}$, respectively, and the sum, $m + n$ yields the overall order of the reaction.  The proportionality $a\propto b$ can be replaced with an equality $a = k b$.  In kinetics, this constant of proportionality is referred to as the rate constant.

In summary: the rate expression tells us how the appearance/disappearance rates of the products and reactants relate to one another; the rate law tells us how the rate is related to the concentrations (or pressures) of reactants and the rate constant is a constant of proportionality that comes out of the rate law.

This answer is a summary of information found at the boundless website and from Tro's most recent general chemistry textbook.

Reaction rate $\dfrac{dA}{dt}$ is the rate at a specific concentration and a specific time.

Rate law is an equation that shows how (velocity) a rate varies as concentration changes. It describes rates at ALL concentrations and NOT just one specific rate at one specific concentration.


How to simplify these system of equilibrium equations What is the pH when $0.025$ mol $\ce{NiCl2}$ is added to $\pu{1.00 L}$ of a $\pu{0.100 M}$ solution of $\ce{HCN}$, if the formation constant for $\ce{Ni(CN)4^2-}$ is $K_f=\pu{1.0e22}$, and the $K_a$ for $\ce{HCN}$ is $\pu{4.9e-10}$?

My approach is as follows:

1) $\ce{\frac{[H+][CN-]}{[HCN]}=4.9*10^{-10}}$

2) $\ce{\frac{[Ni(CN)4^{2-}]}{[Ni^2+][CN-]^4}}=1.0*10^{22}$

3) $\ce{[H+][OH-]}=10^{-14}$

4) $\ce{0.025=[Ni^{2+}] + [Ni(CN)4^{2-}]}$       .............  (mass balance for $\ce{Ni^{2+}}$)

5) $\ce{[CN-] + 0.05 + [OH-] + 2*[Ni(CN)4^{2-}] = 2*[Ni^{2+}] + [H+]}$ ........... (charge balance)

6) $\ce{0.1 = [CN-] + [HCN] + 4*[Ni(CN)4^{2-}]}$ ............. (mass balance for $\ce{CN^{-}}$)

Since the equilibrium constant for the formation of the complex ion is very large, I assume that $\ce{[Ni(CN)4^{2-}] >> [Ni^{2+}]}$, so $\ce{[Ni(CN)4^{2-}] = 0.025 M}$, from the mass balance for $\ce{Ni^{2+}}$.

However, I am not sure what is the next step and what other assumption to make.

Since the equilibrium constant for the formation of the complex ion is very large, I assume that $\ce{[Ni(CN)4^{2-}] >> [Ni^{2+}]}$

From the comments:

  [comments:] That assumption is incorrect. The equilibrium constant seems large, but the exponents are high, so it is misleading. In fact, more of the cyanide is in the form of HCN than in complex with nickel, and there is very little free cyanide.

OP replied:

  How may I make a better assumption to solve this problem?

In the absence of nickel, the pH is around 5 and HCN is the major species, with the concentration of cyanide in the micromolar range (standard weak acid problem). When you add nickel, the concentration of cyanide gets lowered further. Because the free cyanide concentration contributes little to the total, we can combine the two reactions to see whether there is more complex or more HCN and to determine the pH:

$$\ce{4HCN + Ni^2+ <=> Ni(CN)4^2- + 4H+}$$

The equilibrium constant is

$$K_\mathrm{combined} = \pu{5.8e−16}$$

So for this reaction, reactants are favored. HCN and nickel concentrations hardly decrease compared to the initial concentrations. If we call the concentration of the complex $x$, we get:

$$K_\mathrm{combined} = \frac{x (4x)^4}{[\ce{HCN}]^4 \cdot [\ce{Ni^2+}]}$$

Solving for $x$, we get $\pu{2.24e−5}$, confirming that reactants are favored. From this we can calculate the pH as

$$\mathrm{pH} = -\log 4x = 4.05 $$

Checking the solution

Upon adding nickel, the pH turned a bit more acidic. That makes sense as a micromolar concentration of nickel complex formed, releasing some more protons from HCN. We can calculate the cyanide concentration from either one of the given equilibrium constants, and verify that it is miniscule. Then, we can verify that all reactions are at equilibrium (at pH 4, hydroxide is a minor species as well, and has little impact on the exact solution).

I had a different approach to the problem which I have highlighted below

Notes :

Initially, I have considered the entire reagents to have reacted and
then considered the equilibrium to set in
I have considered two equilibria simultaneously, one involving the dissociation of the Nickel complex and also the hydrolysis of the Cyanide ion because it is a strong conjugate base of HCN which is a weak acid

Hope this answers your question.


Why are Earth's oceans salty? As we all know water is $\ce{H2O}$, it does not contain sodium chloride. The ocean is a solution composed of water and a solute which is salt. I was wondering how the massive amounts of $\ce{H2O}$ in our oceans first mixed with salt? I'm sure the water didn't contain salt when it first got to the oceans so how did it get mixed with salt? Thanks.

Early history of Earth is a question to paleohydrology and paleochemistry. Briefly, water wasn't pure on the early stages of Earth.

Brief history of water on Earth. The planet was hot and seismoactive (lots of vulcanos) in the beginning. Water (as steam) escaped from magma, cooled down and precipitated on the Earth's surface. The surface already had salts on it, so it became saline from the very beginning. A better question is: how come we have fresh water in rivers and saline water in oceans? This separation is due to water cycle on our planet. Rivers bring dissolved salts to the ocean, then evaporates and precipitates over the land. Salts cannot evaporate (at any practical rate), so they stay in the ocean. This is similar to washing dishes. Consider this model: dish gets cleaner - that is a river; drain water gets dirtier - that is an ocean; drain water goes through purification and becomes clean before going to the supply pipe - this is evaporation of water from the surface of an ocean and precipitation over the land in our model.

This reminds me of my school days when we had to watch this movie:

Hemo the magnificant

It's an old movie but worth the time to watch it.

Actually sea water contains traces of all known natural elements including sodium and chlorine ions and our blood is an analog of the sea water of which it evolved.  Salt does not exist in water until the water is evaporated away.


Gases produced by bacterial decomposition of human urine I know that when urine (more specifically, the urea within) is metabolised by bacteria, ammonia is released. However, because of the high solubility of ammonia, I assume it simply dissolves into the water.

A container of urine will develop gases inside. What are they?

Carbon dioxide and ammonia when treated with bacteria along with water vapour


Why does potassium react more violently with water than lithium? Recently, I was telling my friends about the violent reaction that takes place when you throw potassium into water. Soon after, a friend of mine claimed that lithium would react more violently than potassium.

I disagreed with him, because potassium is more electropositive than lithium and thus more reactive.

My friend claimed lithium to be more reactive than potassium due to its position in the reactivity series of metals:

$$\ce{Li > Cs > Rb > K > Ba > Sr > Ca > Na > Mg}$$

Then we found out that potassium reacts indeed more violently in water.
But what about his argument? Why isn't he right?

For the reaction,
$$\ce{M -> M+ + e-}$$

the heat liberated is highest for lithium owing to its high negative $E^\circ$ value so one would think that the reaction must be most vigorous.

The reason behind the more violent reactivity of potassium rather than lithium lies in kinetics and not in thermodynamics.

No doubt, maximum energy is evolved with lithium but the vaporization and ionization will also consume maximum energy (the melting point and ionization energy of lithium are the highest) and so the reaction proceeds gently. On the other hand, potassium has a lower melting point and ionization enthalpy. The heat of reaction is sufficient to melt it. The molten metal spreads over the water and exposes a larger surface to water. Also, the hydrated radius of lithium is the greatest out of all alkali metals. This reduces the ionic mobility which in turn reduces the speed of the molten metal.

That's why potassium gives a more violent reaction with water.

Reference:

Kumar, Prabhat Conceptual Inorganic Chemistry; Shri Balaji Publications: Muzaffarnagar, U.P., 2014.

Potassium is higher up in the reactivity series. Reactivity of metals increases as more shells are added. It becomes easier for electrons to be lost, which actually influences the reactivity of the metal.


Nucleophilicty order of various nucleophiles Which one is more nucleophilic in strength $\ce{Br-}$ or $\ce{OH-}$?
I am confused since $\ce{OH-}$ is more electronegative than $\ce{Br-}$ but also a strong  base which contradicts for a better  nucleophile .

Like most concepts that are not captured by a single number, nucleophilicity is not a simple (or at least, relatively simple) concept like acidity.

Nucleophilicity depends a lot on orbital alignment and interaction, in addition to thinking about charges. It's also confusing because nucleophility is not always married to the stability of the adduct. For example, iodide is a great nucleophile, but alkyl iodides are also highly reactive.

In your specific case, since the two nucleophiles have the same charge, you should look at other factors. Bromide is significantly larger than hydroxide, and it is a much better nucleophile, at least when we're considering carbon electrophiles. Indeed, we see bromide used in nucleophilic catalysis but not hydroxide.

Better nucleophile, as per definition, are those which can donate their electron pair to (and form a bond with) a CARBON.
Although, we could see usually, nucleophilicity and basicity (ability to donate LP to a proton) go side by side, there are cases where it is not true.

So, for OH-, it is better nucleophile than Br- in an aprotic medium, due to higher charge density on O atom. While when it come to basicity in aqueous medium, OH- ions are heavily hydrated due to extensive H-bonding, making it less likely to donate its LP to a Carbon, hence less basic than Br-.


carbon dioxide and its behaviour in acid base balance In human body acid base balance, why and how does the dissolved carbon dioxide increases the hydrogen ion concentration with increasing in its partial pressure..? At the same time bicarbonate ion is staying in same concentration while hydrogen ion are increasing?

The disolved carbon dioxide increases the acid content of the human body. This happens when a carbon dioxide molecule bumps into a water molecule. Carbon dioxide makes water break into hydrogen and hydroxide ion. Hydroxide merges with the carbon dioxide molecule creating bicarbonate ion. Sometimes the hydrogen ion can become part of increasing acidity. Sometimes the hydrogen ion combines with the bicarbonate molecule and make carbonic acid. Carbonic acid could turn into carbonate ion and two hydrogen ions. This is what makes the human body acid base balance. The amount of bicarbonate is staying the same because the bicarbonate that does get converted into carbonic acid or carbonate ion an equal amount of bicarbonate is created from water and carbon dioxide. This just results in a huge amount of hydrogen ions.


Can water be a Lewis acid? I asked this question to my teacher a few days ago. In a reaction such as

$$\ce{NH3 + H2O -> NH4+ + OH-},$$

$\ce{NH3}$ is a Lewis base, but is $\ce{H2O}$ a Lewis acid?

He said that it isn't. Now I fail to see the reason why. Can someone explain?

Strictly speaking, a Lewis acid is something that forms a bond by accepting an electron pair form another molecule (Lewis base).  Water as such does not do that, rather it is a hydrogen ion from the water that does so.  When ammonia acts as a proton acceptor in water, the Lewis theory calls it a displacement reaction where the acid, $\ce{H+}$, is initially combined with one base ($\ce{OH-}$) and ends up combined with another base ($\ce{NH3}$).

My friend, here is how i understand it:
H2O is a neutral medium, does not like to be acid or base, but if we mix acid with it, then the water will try to fight back by turning to Lewis base. Same thing goes with adding base to H2O, H2o will fight back by turning acid to naturalize the solution.
So yes, water can be both Lewis acid or base.

I hope that makes sense


Query in Chemical Kinetics I have been recently studying chemical kinetics. I've understood what the rate law and the rate constant mean, and also that, kinetics depends completely on experimental observations, due to which we have zero order reactions as well as fractional order reactions.

Take for example, this: $$\text{Rate of a certain chemical reaction} = k [A]^p [B]^q [C]^r $$

Here, $[A]$, $[B]$ and $[C]$ are the molar concentrations of the three reactants involved, while $p$, $q$ and $r$ are the number of molecules of each respectively taking part in the reaction.

So, $\text{the order of the reaction} =p+q+r $.

Now, this $p$, $q$ and $r$ are completely determined experimentally, and cannot be determined theoretically.

How are these values determined experimentally? How does one find out how many molecules of each reactant are taking part in the reaction? Is there any special instrument to do this? If not, then how do scientists find the order of a reaction?

As an attempt to shorten out the answering area, consider this reaction: $$\ce {NO2 +CO -> NO +CO2} $$ Now, experimentally it has been determined that $\text {the rate of this reaction} = k[\ce{NO2}]^2$.  I want to know how have scientists determined that the rate depends on the square of the concentration of $\ce {NO2} $ and not on the other reactant.

N.B.: My question is how it can be experimentally determined, and not why it has to be experimentally determined.

Experimetally :- one reactant is kept constant and concentration of other is changed  and accordingly rate of reaction is measured...then dependence with thta reactant is measured.....and then other one  is kept  constant and ist one is changed then  similarly rate of reaction and thus dependence is measured....... We cant determine the actual dependence of rate of reaction on the concentration of reactants therotically ......we have tk calculate it and then we can formulate the rate law or order of reaction


What is pairing energy of electrons? What exactly is pairing energy when electrons are paired in orbitals and what is responsible for it.

Imagine a sky scraper where everyone wants to live on as low floor as possible, but landlord prohibits then to live more than 2 people per floor. These people will occupy lowest floors in pairs and the highest occupied level might have a single person or a pair. If someone moves out of the apartment on the 1st or 2nd floor, then someone from higher floors will move down to the vacant state. Electrons do the same. Floors are energy levels and the evil landlord is the Pauli exclusion principle.

Electrons are fermions. An electron's spin can be either +1/2 or -1/2. Two options. Two fermions (and, hence, electrons) cannot be the same. Electrons are the same if their spins are the same and they occupy the same orbital (Pauli Exclusion Principle). At lower temperatures (below red heat temperatures) electrons occupy the lowest energy level possible.

As a result, each new electron occupies the lowest allowed level. Because their spins can take two values 2 electrons are occupying each level. If you populate each energy level by a single electron you will get a high energy excited state that well emit light and go to lower energy two electrons per level state. If you try to squeeze more than two electrons it will not work because of Pauli principle. So, since they want to stay in lower energy levels and have to obey the exclusion principle we have electron pairs.

Bonus: here is why radicals recombine. For instance, why $\ce{2 H^. -> H2:}$

If an electron is the last one it is occupying the energy level alone.
So, electrons would like to all occupy the lowest level, but only two per level is allowed. In radicals only one electron occupies a level. When radicals A$^.$ and b$^.$ meet each of them sit on half empty orbital attracted to a single nucleus.  It is energetically favorable to share the same orbital (form a pair, which they are OK with) and orient fragments so that they "fly" (have the highest probability) between two nucleus (A and B). Staying between two opposed charged ions is better than staying next to just one charged ion. And electrons are allowed to stay in pairs (Pauli excluded 3+ electrons staying together).


Which functional groups are not reduced by DIBAL.H? According to the table here, diisobutyl aluminium hydride can reduce amides to aldehydes but according to the data given to me by my teacher DIBAL.H cannot reduce:  amides, acids, isocyanides and nitro groups.

It's I also given in my notes that DIBAL.H is  parallel to $\ce{LiAlH4}$ as a reducing agent but it is more selective.

I would like to know if DIBAL.H is able to reduce these functional groups or not.

DIBAL-H does the following:

Acid chlorides to aldehydes (Fast)
$3^\circ$ amides to aldehydes
Nitriles to aldehydes
Ketones to $2^\circ$ alcohols
Aldehydes to $1^\circ$ alcohols. (Slower)

Also, it works best in these solvents: THF, toluene.
We need to avoid solvents like alcohols, amines, strong acids and halogenated compounds.


How to calculate partition coefficients? Below is a transcript of this doubtful question:

  A weak monobasic organic acid $\ce{HA}$ is soluble in both water
  and $\ce{CHCl3}$.  $\pu{500.0 cm3}$ of a solution of $\ce{HA}$ in $\ce{CHCl3}$, with a concentration of
  $\pu{0.057 mol dm-3}$,
  is shaken well with $\pu{500.0 cm3}$ of water and allowed
  to attain equilibrium at $\pu{27°C}$.  An aqueous layer and
  $\ce{CHCl3}$ layer then separate out; the pH of the aqueous layer is
  found to be 3.21 under these conditions.

  The dissociation constant of $\ce{HA}$ in water at $\pu{27°C}$
  is $\pu{1E-5 mol dm-3}$.

  Calculate the partition coefficient at $\pu{27°C}$
  for the partitioning of $\ce{HA}$ between water and $\ce{CHCl3}$.
  In a second experiment, a further $\pu{500.0 cm3}$ portion of the
  same $\ce{HA}$ solution in $\ce{CHCl3}$ in which the concentration
  of $\ce{HA}$ is $\pu{0.057 mol dm-3}$, is shaken well with
  $\pu{500.0 cm3}$ of a $\pu{0.027 mol dm-3}$ aqueous $\ce{NaOH}$
  solution and allowed to reach equilibrium at $\pu{27°C}$.
  Calculate the $\mathrm{pH}$ of the aqueous layer under these conditions.
  State the assumptions you make, if any, in the above calculations.

My solution for Question 1.

Since pH is given, $\ce{[H^+](aq)}$ is determined by

\begin{align}
\mathrm{pH} &=-\log[\ce{H^+~(aq)}]\\
[\ce{H^+}~(\ce{aq})] &= 10^{-3.21}\\
    &= 6.16 \cdot 10^{-4}~\mathrm{mol\,dm^{-3}}\\
\end{align}

By appropriately applying $K_\mathrm{a}$, the $[\ce{HA}~({\ce{aq}})]$ present in aqueous layer is $0.038~\mathrm{mol\,  dm^{-3}}$.
Next, $[\ce{HA}~(\ce{aq})]$ present in the organic layer can be determined by subtracting the $[\ce{HA}~({\ce{aq}})]$ present in aqueous layer
from the initial concentration of $[\ce{HA}~({\ce{aq}})]$, which results in $0.019~\mathrm{mol\,  dm^{-3}}$.
Finally, I got $K_\mathrm{D} =
   \dfrac{[\ce{HA}_\mathrm{aqueous~layer}]}{[\ce{HA}_\mathrm{organic~layer}]}
   = \dfrac{0.038~\mathrm{mol\,  dm^{-3}}}{0.019~\mathrm{mol\,  dm^{-3}}} = 2$.

I hope that this is correct, but  I am not quite sure.

My incomplete solution for Question 2.

The reaction between $\ce{HA}$ and $\ce{NaOH}$ is

$$\ce{HA}~(\ce{aq}) + \ce{NaOH}~(\ce{aq})  \longrightarrow \ce{NaA}~(\ce{aq}) + \ce{H2O}~(\ce{l})$$

The number of moles of $\ce{NaOH}$ present is $0.0135~\mathrm{mol}$.

I am confused with the concentration of $\ce{HA}$ that I needed to take for this calculation. Initially I have found the concentration of $\ce{HA}$ present in organic layer as $0.019~\mathrm{mol~dm^{-3}}$. Do I need to use that here or $0.057~\mathrm{mol~dm^{-3}}$?

What would be the assumptions that I need to assume to take out this calculation?

Equilibrium are really dependent on the chemical activities of the species not their concentrations. So an overriding assumption for both parts is that the activities of the chemical species is the same as the concentration. That is a reasonable for these solutions. As a rule of thumb, for solutions with concentrations more than 0.1 molar the assumption is dicey.

I think you're very close to the right solution for part 1, but I'd state it a bit differently.

Since pH is given, $\ce{[H^+]}$ in the aqueous phase is determined by

    $[\ce{H^+}] = 10^{-\mathrm{pH}} = 10^{-3.21} = 6.2 \cdot 10^{-4}~\mathrm{mol\,dm^{-3}}$

      The mantissa of the pH only has two significant figures, so should the $\ce{[H^+]}$

We know that in the aqueous layer the charges have to balance so $\ce{[H^+] = [A^-] + [OH^-]}$, but $\ce{[A^-] >> [OH^-]}$ so we can assume that $\ce{[H^+] = [A^-]}$
Using $\ce{[H^+] = [A^-]}$ the $K_\mathrm{a}$ equation can be solved for the $\ce{[HA]}$ present in aqueous layer which gives $0.038~\mathrm{mol\,  dm^{-3}}$.

$$\ce{[HA]} = \dfrac{\ce{[H^+][A^-]}}{1\times10^{-5}} = \dfrac{(6.2\times10^{-4})^2}{1\times10^{-5}} = 0.038$$

Next, the $\ce{[A-]}$ in the aqueous phase is insignificant compared to the $\ce{[HA]}$ in the aqueous phase, so the moles of $\ce{HA}$ present in the organic layer can be determined by subtracting the moles of $\ce{HA}$ present in aqueous layer (0.038*0.5=0.019)from the initial moles of $\ce{HA}$ (0.057*0.50 = 0.029), which results in a final concentration of $0.020~\mathrm{mol\,  dm^{-3}}$ of $\ce{HA}$ in the organic layer.
Finally, $K_\mathrm{D} =
   \dfrac{[\ce{HA}_\mathrm{aqueous~layer}]}{[\ce{HA}_\mathrm{organic~layer}]}
   = \dfrac{0.038~\mathrm{mol\,  dm^{-3}}}{0.020~\mathrm{mol\,  dm^{-3}}} = 1.9$

    The $\ce{[HA]}$ in both layers is known to two significant figures so the $K_D$ value should have two significant figures too.

For Question 2.

The reaction between $\ce{HA}$ and $\ce{NaOH}$ in aqueous solution is, as you noted,

$$\ce{HA + NaOH -> Na^+ + A^- + H2O}$$

Assume none of $\ce{NaA}$ migrates into the organic phase.
Assume that no $\ce{NaA}$ forms in the aquaeous phase either, which is to say that both the $\ce{Na^+}$ and $\ce{A^-}$ ions are completely solvated.
Since $\ce{[Na^+]}= 0.027$ and the volume is 0.5000 $\mathrm{dm}^{-3}$, the number of moles of $\ce{NaOH}$ present is $0.0135$.
Started with 0.0295 moles of HA total, so the solution will be acidic.
Since the charges must balance $\ce{[Na^+] + [H^+] = [A^-] + [OH^-] }$, but $\ce{[A^-] >> [OH^-] }$ and $\ce{[Na^+] >> [H^+]}$ so $\ce{[Na^+] \approx [A^-] }$ which means that there are $0.0135$ moles of $\ce{[A^-]}$ in the aqueous solution, and 0.016 moles of $\ce{HA}$ split between the aqueous phase and the organic phase.
From above $0.016 = \ce{0.5[HA]_{aq} + 0.5[HA]_{org}}$ and using $K_D$ we can determine $\ce{[HA]_{aq} = 0.021}$ and $\ce{[HA]_{org} = 0.011}$
We can use the $K_a$ to solve for $\ce{[H^+]}$

\begin{align}
1\times10^{-5} &= \dfrac{\ce{[H^+][A-]}}{\ce{[HA]}}\\
\ce{[H^+]} &= \dfrac{1\times 10^{-5}\ce{[HA]}}{\ce{[A^-]}} = \dfrac{(1\times 10^{-5})(0.021)}{0.027} = 7.7\times10^{-6}\\
\mathrm{pH} &= 5.11
\end{align}

I think the question is a bit clever and may needs deeper analysis, a monobasic organic acid may have carboxylic acid group and amine group, both can exist in aqueous form by dissociation or by hydrogen bond. At the aforementioned equilibrium and the dilute initial concentration, there will be ionic forms in water and molecular forms in aqueous form. The sum HA and ions in that equilibrium is in aqueous form, and could be some molecules in organic phase if their sum is less than the initial moles. So here seems no significant quantity is in $\ce{CHCl3}$ phase. On my point of view.


What are examples of group 14 elements forming dπ - pπ bonding? While studying the Group 14 elements, I came across the anomalous behaviour of carbon. In that there was a point:

  Carbon forms only p$\pi$-p$\pi$ type bonding while other elements in the group form p$\pi$-p$\pi$, as well as d$\pi$-p$\pi$ type bonding.

I am searching for molecules in which  group 14 elements form d$\pi$-p$\pi$ bonds.

For $d\pi-p\pi$ bonding we need to consider other elements of group 14 except $\ce{C}$. (i.e, $\ce{Si, Ge ,Sn}$ etc.) which have got a vacant $d$ orbital of proper symmetry and energy to come into play bonding with attached group  having lone pair of electron in p orbital of comparable energy.
    For example $\ce{SiF4}$. $\ce{GeCl4}$ etc.
In tetrahedral geometry the orientation of vacant  d orbital of the central atom (dxy. dyz,dxz) may overlap with filled p orbital of the ligating atom  and hence $d\pi -p\pi$ bonding is possible.


Tungsten 4+ ion magnetism? I just had a quiz and struggled on a particular question:

  Tungsten forms a $+4$ ion. It is observed to be diamagnetic. The electron configuration for this ion may not be what you would expect. Write the electron configuration for tungsten(IV) that is diamagnetic.

So if tungsten has a configuration of $\ce{[Xe] 6s^1 4f^{14} 5d^5}$ and it loses 4 electrons wouldn't it be $\ce{[Xe] 6s^0 4f^1^4 5d^2}$? And this would be paramagnetic since there are 2 unpaired electrons in the 5d subshell. How could it be diamagnetic?

The electron configuration for Tungsten is [Xe] 6s2 5d4 4f14. If it were to lose its electrons, and the answer is that it is diamagnetic, then it would lose its 4 d-electrons.

References

https://www.quora.com/What-are-the-differences-between-paramagnetic-and-diamagnetic-elements
https://www.tabulka.cz/english/elements/configuration.asp?id=74
https://socratic.org/questions/what-are-columbic-and-exchange-energy-and-how-it-it-determined


How does the inductive effect, explain the reactivity of tertiary alcohols with halogen acids? The reactions of primary and secondary alcohols with halogen acids ($\ce{HX}$, where $\ce{X}$=halogen) require the presence of a catalyst, $\ce{ZnCl_2}$. With tertiary alcohols, the reaction is conducted by simply shaking with concentrated $\ce{HCl}$ at room temperature. Which mean that, tertiary alcohols are more reactive than primary and secondary alcohols. The reason is said to be, due to the inductive effect.

I don't know how the inductive effect would explain that, tertiary alcohol is more reactive than primary and secondary.

Secondary and tertiary alcohols react with strong acid (and bases) such $\ce{HX}$ following SN1 mechanism.

$$\ce{ROH <=> HX + ROH_2^+ + X^-} \tag{1}$$
$$\ce{ROH_2^+ <=> R^+ +H_2O} \tag{2} $$
$$\ce{R+ +X- \rightarrow  RX} \tag{3} $$

In these reactions, the rate determining step is the formation of the carbocation (2). So the successive attack of the nucleophilic chlorides is dependent on that.

Here is how the inductive effect comes into play. The inductive effect of $\ce{R}$ groups of the alcohols stabilize the carbocation (figure show the stability of carbocations from the more stable to the less) and then increase the chances of a nucleophilic attack!

Primary alcohols react via $S_N2$ mechanism which involves the carbocation formation
but the secondary and tertiary react via $S_N1$ which is elementary and no carbocation is formed.
hence it appears that both primary and tertiary alcohols should react rapidly but it is not so.

the $\ce{C-O}$ bond weakens as the no of electron donating groups on the alpha-carbon increases this is due to corresponding increase in negative charge developed on $\ce{O}$ atom **

  thats why tertiary alcohols are more reactive.

**
further $\ce{ZnCl2}$ is a Lewis acid which coordinates with $\ce{O}$ atom and weakens the bond thus it need to be used for primary and secondary alcohols.


The Krieger-Li-Iafrate Approximation I am having trouble understanding what KLI approximation is. Could someone explain it to me in an intuitive/simple way?

The optimized effective potential (OEP) is a natural connection between local density-functional theory and many-body perturbation theory, which is derived from the Sham-Schlüter equation. Converging the full set of OEP equations is quite a challenge. Therefore an approximation was made in order to reduce the integral equations to an analytically solvable one via a dominant orbital approximation, called The Krieger-Li-Iafrate (KLI) approximation.

$$\sum_{i,j\neq i}\frac{\phi_j^*(\mathbf{r})\phi_i^*(\mathbf{r})}{\epsilon_i-\epsilon_r}\left(f_i \langle \phi_i|v_x|\phi_j\rangle - \sum_{k, \alpha}\frac{d_{ik}^{\alpha}d_{kj}^{\alpha}}{2}(\epsilon_i-\epsilon_k)\left[\frac{f_i(1-f_k)}{\epsilon_i-\epsilon_k+\epsilon_{\alpha}}+\frac{f_k(1-f_i)}{\epsilon_i-\epsilon_k+\epsilon_{\alpha}}\right]\right) + c.c.$$

$$=\frac{1}{2}\sum_{\alpha,i,k}\omega_a\frac{\left|d_{ik}^{\alpha}\right|^2 f_i(1-f_k)}{(\epsilon_i-\epsilon_k-\epsilon_{\alpha})^2}\left(\left|\phi_k(\mathbf{r})\right|^2-\left|\phi_i(\mathbf{r})\right|^2\right)$$

Source: Maybe they can see comments above.


Sodium + water reaction, noxious fumes? I was reacting pure sodium and tap water, and at one point I put some splattered sodium back into the water. I observed fumes that reacted with my respiratory system, producing a burning sensation.

I read that it is possible to produce sodium oxide fumes from sodium hydroxide at high temperatures (the sodium did burn at the end, producing an orange flame).

My question is what could be the chemical composition of these fumes?

I agree.  This is a dangerous reaction and should not be tried without proper equipment, including a lab hood and fire extinguishing equipment.

The fumes could be "just" droplets of sodium hydroxide solution.  I've seen similar fumes involving hydrochloric acid, just by adding water to rinse out traces of concentrated acid (in a hood, of course).  They may be carried by increased evaporation even without boiling and without need for a further "chemical reaction".

Well sodium hydroxide(NaOH) is solid at room temperature but at high temperatures it can be converted to gases.This happens due to the fact that the molecules of NaOH gain so much energy that the intermolecular bonds (dipole-dipole) between 2 sodium hydroxide molecules are broken.The reaction of sodium with water is exothermic and produces lot of heat and  sodium hydroxide may have absorbed some of the heat.

What you did was very dangerous and I wouldnt recommend anyone to try it at home without laboratory equipment.


Nucleophilic aromatic substitution on pyrimidines: C2 vs C4 selectivity [There are many similar questions on Chemistry.SE (e.g. 1, 2) and a quick search will reveal all of them, but nothing on this in particular.]

The reaction of 2,4-halopyrimidines with nucleophiles generally leads to selective displacement of the halide at the 4-position:1

In their textbook, Joule and Mills don't explain why, though, simply saying that it "is the normal situation for nucleophilic – 4-chloro > 2-chloro".

I don't see any obvious difference in the stability of the anionic intermediates, either (the so-called Meisenheimer complexes).

What can explain the regioselectivity of the above reaction?

Reference

Joule, J. A.; Mills, K. Heterocyclic Chemistry, 5th ed.; Wiley: Chichester, U.K., 2010, p 257.

An explanation can be provided by frontier molecular orbital theory.  The LUMO coefficient is higher on the C4 vs C2.  This is a general effect meaning that an EWG para to the leaving group is more reactive than ortho.  See 4-halopyridines vs 2-halo and 4-nitrobenznes vs 2-nitro.  There could also be an effect from the repulsion between the nitrogen lone pairs and the approaching electron rich nucleophile, which would be greater in proximity to 2 nitrogens rather than 1.

Try drawing the resonance hybrides out. You'll soon realise that the intermediate formed when substitution accours at the c4 position can be stabilised by three resonance hybrids!

:)


Chemical reaction that can produce lots of heat from 2-3 simple liquid ingredients? Have an experiment I want to try out utilizing thermoelectric power and I want to generate the heat via a simple chemical reaction. Ideally I'd like to mix 2 liquids together (one being water or alcohol would be great) that can get to around boiling water temperature for a few minutes or longer. Byproducts need to be only other liquids that can be drained easily or gasses that aren't dangerous to humans. No solids wastes and non corrosive / carcinogen / caustic.

So far I've found one that's close to my needs but still produces a solid waste product and isn't particularly safe, which is Calcium Oxide + Water -> Calcium Hydroxide + Heat. This might work if there's another liquid I could add after the main reaction to turn the Calcium Hydroxide into a liquid for disposal.

The first thing that comes to mind, if you have access to a stock-room, is mixing two solutions: (1) NaOH; (2) HCl. This can release a lot of heat if your solutions are concentrated enough, and it forms salt-water if your NaOH and HCl are of the same molarity.

Edit: I'm on my lunch break, so I did some of the math...

The chemical reaction of interest in this case is:

$$\ce{OH- + H+ -> H2O}$$

The standard enthalpies of formation ($\Delta H_\mathrm f^\circ$) of these species are:

$$\begin{array}{lr}
\hline
\text{Species} & ΔH_\mathrm f^\circ/\pu{kJ mol-1} \\
\hline
\ce{OH-} & -229.99 \\
\ce{H+}  & 0.00       \\
\ce{H2O} & -288.83 \\
\hline
\end{array}$$

Thus, the change in enthalpy for the reaction is:

$$\Delta H^\circ = -288.83\ \mathrm{kJ/mol} - (-229.99\ \mathrm{kJ/mol}+0\ \mathrm{kJ/mol}) = -58.84\ \mathrm{kJ/mol}$$

Therefore, for $1\ \mathrm{mol}$ of $\ce{NaOH}$ + $1\ \mathrm{mol}$ of $\ce{HCl}$, you get $58.84\ \mathrm{kJ}$ of heat. Say you want to release enough heat to get the net solution up to $100\ \mathrm{^\circ C}$.

Water has a heat capacity of $4.18\ \mathrm{J/(g\ ^\circ C)}$. Say you have $1\ \mathrm L$ of $\ce{NaOH}$ + $1\ \mathrm L$ of $\ce{HCl}$, you'll need enough heat to raise the temperature of $2\ \mathrm L$ of water to $100\ \mathrm{^\circ C}$. I'll assume the water starts off at $25\ \mathrm{^\circ C}$, so you have $2\,000\ \mathrm g$ and $75\ \mathrm{^\circ C}$ to go.

$$4.18\ \mathrm{J/(g\ ^\circ C)} \cdot 2\,000\ \mathrm g \cdot 75\ \mathrm{^\circ C} = 62\,700\ \mathrm J = 62.7\ \mathrm{kJ}$$

How many moles of $\ce{NaOH}$ + $\ce{HCl}$ do you need for that much heat?

$$\frac{q}{\Delta H^\circ} = \frac{62.7\ \mathrm{kJ}}{58.84\ \mathrm{kJ/mol}} = 1.065\ \mathrm{mol}$$

That would mean that you can mix $1\ \mathrm L$ of $1.065\ \mathrm M$ $\ce{NaOH}$ + $1\ \mathrm L$ of $1.065\ \mathrm M$ of $\ce{HCl}$, and would theoretically expect to get a temperature close to $100\ \mathrm{^\circ C}$.

This might be off a bit because I've made some assumptions:

$\Delta H^\circ$ is constant with respect to temperature from $25\ \mathrm{^\circ C}$ to $100\ \mathrm{^\circ C}$. This may not be true.
the dissolved salts in water dont significantly affect its heat capacity
You have $2\,000\ \mathrm g$ of water in $1.065\ \mathrm M$ $\ce{NaOH}$ + $1.065\ \mathrm M$ $\ce{HCl}$

I think this could get you close though? I'm a bit surprised that the molarities aren't higher... It's a starting point at least.

Disclaimer of course: be careful with the $\ce{NaOH}$ + $\ce{HCl}$ solutions, they can be dangerous. Use proper chemistry hygeine protocols. The mixture should be benign, but you should confirm this with pH paper.

As Zhe says, this isn't safe. Your question mentions "draining" chemicals as if it's OK to dump chemicals down the drain or out on the ground. It's not. What possible connection is there between an exothermic chemical reaction and an experiment in thermoelectrics, I can't imagine. Use a hot plate. Here's a hint: heating water to its boiling point requires a lot of energy (per gram). Another name for a chemical reaction that produces a lot of energy is "an explosion". I can't think of any which someone who thinks "if its liquid, I can just dump it down the drain" would be safe to implement.