Some substances disproportionate. This means a species with an intermediate oxidation state yields two species with higher and lower oxidation state. For example:

$$\ce{Hg2Cl2 -> Hg + HgCl2}$$

My question is if there are substance that have the analogous acid/base behavior, where an intermediate protonation state yields the conjugate acid and the conjugate base. You could write it like this:

$$\ce{2AH -> A- + AH2+}\tag{1}$$

An example would be water: $$\ce{2H2O <- OH- + H3O+}$$

What I am looking for, however, is a case where the mixture of conjugate acid and the conjugate base (of the same amphoteric starting material) are the major species, i.e. a reaction of the type (1) with an equilibrium constant larger than 1. This would mean that the pKa values are in the opposite order than usual, i.e. it is "easier" to lose the second proton than it is to lose the first. For disproportionation, there is a similar feature of the reduction potentials (it is easier to accept the second electron than it is to accept the first).

Are there any such examples?

Update: To clarify my question, I am looking for substances that skip a protonation state, i.e. pKa2 < pKa1. If you could isolate the intermediate protonation state, it would yield the higher and lower protonation state. This is similar to species that skip an oxidation number (or have the corresponding species at very low concentration) because the reduction potentials favor disproportion of the (hypothetical or barely detectable) species with intermediate oxidation state.

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    $\begingroup$ So, you mean if autodissociation constant can be more then 1? In theory, yes, in practice, no. BTW it would be nice if you edited it for clarity. $\endgroup$ – Mithoron May 6 '20 at 14:23
  • $\begingroup$ Complete hydrolysis might be an example, though admittedly somewhat far-fetched. $\endgroup$ – Ivan Neretin May 6 '20 at 14:58
  • $\begingroup$ I don't think that such substance should exist in nature, though may be created in-situ. As clearly by your statement "...with equilibrium constant larger than 1..." should mean that equilibrium is shifted towards products side. But as you want them to be acid and base themselves, they will react with each other again resulting in the reactant, unlike products of disproportionation. So actually, this shouldn't be true. But observation and experiments is the final truth. Edit: Sorry, didn't see Mithoron had said the same thing, realised now. $\endgroup$ – ba-13 May 6 '20 at 17:05
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    $\begingroup$ @Mithoron How about dinitrogen pentoxide? At least in the solid phase, it seems the autodissociation constant is much greater than 1. Admittedly, such cases are very rare, and this case is also a Lewis acid-base, no protons involved. Also, Karsten, I know of a single example where a compound has a second proton which is more acidic than the first: the hydrated vanadyl complex. Does this get vaguely in the direction which you're thinking? $\endgroup$ – Nicolau Saker Neto May 7 '20 at 1:19
  • $\begingroup$ @NicolauSakerNeto Yes, that's exactly it. Thanks to your wikipedia reference, I now have a name for it: an amphoteric substance that does not follow "Pauling's first rule". I wonder what the graph for the speciation looks like, and how this affects buffering capacity. Anyway, I accept your answer, even if it is a comment. $\endgroup$ – Karsten Theis May 7 '20 at 11:46

As I mentioned in the comments, I know of one case where a polyprotic acid has a second proton which is more acidic than the first: the aqueous pervanadyl complex, $\ce{[VO_2(H_2O)_4]^+}$. According to Wikipedia,

$$ \begin{array}{rcl} \ce{[VO_2(H_2O)_4]^+ &<=>& H3VO4 + H+ + 2 H2O} & \mathrm{pK_{a_1}} = 4.2\\ \ce{H3VO4\ \ &<=>& H2VO4^- + H+} & \mathrm{pK_{a_2}} = 2.60\\ \ce{H2VO4- &<=>& HVO4^2- + H+} & \mathrm{pK_{a_3}} = 7.92\\ \ce{HVO4^2- &<=>& VO4^3- + H+} & \mathrm{pK_{a_4}} = 13.27 \end{array} $$

Though it certainly feels quite odd that $\mathrm{pK_{a_2}<pK_{a_1}}$, I suspect this feature is not absurdly uncommon in larger/more complex systems, where structural reorganisation upon deprotonation can be severe. Perhaps what makes the pervanadyl complex unusual is its comparative simplicity.

However, Wikipedia provides no source for those acidity constants. I looked around for an article discussing the speciation of vanadium ions in aqueous solution, and discovered this nice reference:

Huang, J.-H.; Huang, F.; Evans, L.; Glasauer, S. Vanadium: Global (Bio)Geochemistry. Chemical Geology 2015, 417, 68–89. DOI:10.1016/j.chemgeo.2015.09.019

The relevant portion in this article is section 4.1. (Aqueous speciation chemistry of vanadium). The authors use thermodynamic data obtained elsewhere to establish the following equilibria:

$$ \begin{array}{rcl} \ce{[VO_2(H_2O)_4]^+ &<=>& H3VO4 + H+ + 2 H2O} & \mathrm{pK_{a_1}} = 3.67\\ \ce{H3VO4\ \ &<=>& H2VO4^- + H+} & \mathrm{pK_{a_2}} = 3.40\\ \ce{H2VO4- &<=>& HVO4^2- + H+} & \mathrm{pK_{a_3}} = 8.06\\ \ce{HVO4^2- &<=>& VO4^3- + H+} & \mathrm{pK_{a_4}} = 13.28 \end{array} $$

As you can see, the first two constants are rather different from the values quoted on Wikipedia, and now they've become difficult to distinguish. However, clearly this is still an unusual case. They take these data and then calculate speciation curves to obtain the following graph:

Evidently, the speciation curves don't really look anything out of the ordinary. It's likely this is partly due to the proximity of the values of $\mathrm{pK_{a_1}}$ and $\mathrm{pK_{a_2}}$ used, and also because the solution is highly dilute. It would be nice to know what happens in more concentrated solutions, but unfortunately we get something like this:

Like some other transition metals, vanadium has a tendency to oligomerise in solution, forming complex mixtures of polyoxometalates. Therefore, it's basically impossible to get a good visualisation of the effects of $\mathrm{pK_{a_2}<pK_{a_1}}$.

Fortunately, we can forget the chemistry and just study how the mathematics behaves. For simplicity, let's consider a diprotic acid $\ce{H2A}$, where we choose the values of $\mathrm{pK_{a_1}}$ and $\mathrm{pK_{a_2}}$. Also for simplicity, let us safely ignore the autodissociation of water by working with relatively concentrated solutions. Therefore, the mole fractions of the species $\ce{H2A}$, $\ce{HA-}$ and $\ce{A^2-}$ will be:

$$X_{H_2A} = \frac{[H^+]^2}{[H^+]^2 + K_{a_1}[H^+] + K_{a_1}K_{a_2}}$$

$$X_{HA^-} = \frac{K_{a_1}[H^+]}{[H^+]^2 + K_{a_1}[H^+] + K_{a_1}K_{a_2}}$$

$$X_{A^{2-}} = \frac{K_{a_1}K_{a_2}}{[H^+]^2 + K_{a_1}[H^+] + K_{a_1}K_{a_2}}$$

Let's keep $\mathrm{pK_{a_1}}=5$ and decrease $\mathrm{pK_{a_2}}$. Here are the individual speciation curves for $\mathrm{pK_{a_2}=10}$ to $\mathrm{pK_{a_2}=-1}$, in decrements of 1 unit of $\mathrm{pK}$ (click on graphs for a larger version).

As you can see, there is no sudden catastrophe as the second proton becomes more acidic than the first. As the second ionisation becomes more favourable, it smoothly and continuously suppresses the formation of $\ce{HA-}$ . Something nice does happen at the crossover point, where $\mathrm{pK_{a_1} = pK_{a_2}}$, namely all species are present in the same concentration when $\mathrm{pH = pK}$. However, this not surprising if you look at the equations above.

For good measure, here are all the plots once again, now with a logarithmic y-axis. It's nice to see that even with the suppression of $\ce{HA-}$, it still exists as a trace in the background, and still behaves smoothly.

And to finish things off, now here are graphs showing the mole fraction of each unique species as $\mathrm{pK_{a_2}}$ varies from 10 to -6, in decrements of 1 unit of $\mathrm{pK}$. I rather like the subtle way the curves for the fully protonated and fully ionised species change, as the intermediate ion stops being a significant contributor in the equilibria.

  • $\begingroup$ Thank you for looking beyond wikipedia as a source. $\endgroup$ – Andrew May 11 '20 at 11:57

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