In acid/base chemistry, are there amphoteric substances that undergo something that resembles disproportionation in redox chemistry

Question

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.

Answer 1

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.

Answer 2

All you need to do is heat most metal hydroxides; they decompose according to reactions such as the following:

Magnesium hydroxide: $\ce{Mg(OH)2 -> MgO + H2O}~ (\pu{350^\circ C})$

Calcium hydroxide: $\ce{Ca(OH)2 -> CaO + H2O}~ (\pu{580^\circ C})$

and even with one of the alkali metals: $\ce{2LiOH -> Li2O + H2O}~(\pu{924^\circ C})$

The lithium hydroxide decomposition temperature is close to that of calcium carbonate.

In all these cases the protic disproportionation

$\ce{2OH^- -> O^{2-} + H2O}$

is driven by a combination of two factors: the more favorable lattice energy of the oxide and the entropy of the escaping water, which is converted to steam. Only the hydroxides of sodium and heavier alkali metals, formed from the most strongly basic anhydrides and with less favorable lattice energy of the oxides, appear to form gas under atmospheric pressure without decomposing.