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This question came up when answering Find the pH and amphiprotic salt is added to water at room temperature. If we dissolve $\ce{NaHA}$ in water, the species $\ce{HA-}$ can act as acid or base:

$$ \begin{align} \ce{H2A &<=> AH- + H+}\\ \ce{HA- &<=> A^2- + H+} \end{align} $$

The $\mathrm{pH}$ of the solution can be estimated as

$$\mathrm{pH} = \frac{\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2}}{2}$$

My question is how accurate this estimate is, depending

  1. on concentration of the amphiprotic salt (as the concentration approaches zero, the $\mathrm{pH}$ should approach 7),
  2. on the average of the $\mathrm{p}K_\mathrm{a}$ values (the closer these are to neutral, the smaller the difference between $[\ce{H2A}]$ and $[\ce{A^2-}]$), and
  3. on the difference between the $\mathrm{p}K_\mathrm{a}$ values (the bigger the difference, the lower the percentage of $\ce{NaHA}$ undergoing acid or base reactions).

So depending on these three variables, how accurate is the estimate? My guess is that high concentration, average of $\mathrm{p}K_\mathrm{a}$ near neutral and difference between $\mathrm{p}K_\mathrm{a}$ low would give good estimates. When any of these parameters a very different from those "ideal" conditions, the estimate probably gets worse.

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    $\begingroup$ I guess you can find some answers in my answer in the original question, where is estimation quantified. $\endgroup$
    – Poutnik
    Commented Apr 28, 2019 at 18:50
  • $\begingroup$ Could you elaborate on point 2? I'm not sure what you mean. $\endgroup$
    – MaxW
    Commented Apr 28, 2019 at 18:51
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    $\begingroup$ @MaxW if pH =about 7, [H2A]=[A^2-]. the farther pH goes away from 7, ion concentration is disbalanced by creation H+ or OH-. $\endgroup$
    – Poutnik
    Commented Apr 28, 2019 at 19:36
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    $\begingroup$ @MaxW Point 2: I could have written how close the actual pH is to 7. When the pH is very different from 7, the extra protons or hydroxide ions come from an excess of A(2-) over H2A or an excess of H2A over A(2-), respectively, breaking assumptions of the approximation. $\endgroup$
    – Karsten ♦
    Commented Apr 28, 2019 at 19:37
  • $\begingroup$ $$\mathrm pH =\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2} +\log \left( 1 + \frac { K_\mathrm{a1}} {[\ce{HA-}]}\right)}{2} $$ Therefore, if Ka1=[HA-], pH is by cca 0.15 higher than by simplified formula of averaging pKa. $\endgroup$
    – Poutnik
    Commented Apr 29, 2019 at 4:35

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Not to make duplicate, I will refer to my answer in the mentioned Find the ph and amphiprotic salt is added to water at room temperature question.

Of the buffering

If $\mathrm{p}K_\mathrm{a1}$ is very close to $ \mathrm{p}K_\mathrm{a2}$, then the acidic salt solution is a good double buffer $$ \ce{H2A/HA-}, \ce{HA^-/A^2-}$$

But if $\mathrm{p}K_\mathrm{a1} \lt\lt \mathrm{p}K_\mathrm{a2}$, then both buffer systems are far from their $\mathrm{p}K_\mathrm{a}$.

As the consequence, they have just fraction of the buffer capacity, compared to the case with close $\mathrm pK_\mathrm a$ values.

We have then a mix of 2 inferior buffers.

If e.g.the $\mathrm{p}K_\mathrm{a}$ difference = 4, the buffer capacity is roughly 25 times smaller, compared to $\mathrm pH=\mathrm pK_\mathrm {a1}$ or $\mathrm pH=\mathrm pK_\mathrm {a2}$.

( 2 orders=100, but 2 buffers ->50, and 0.5/0.5 vs 0.99:0.01 --> 25 )

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