How accurately does (pKa1 + pKa2) / 2 estimate the pH of an amphoteric salt? 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:

$$
\begin{align}
\ce{H2A &amp;&lt;=&gt; AH- + H+}\\
\ce{HA- &amp;&lt;=&gt; A^2- + H+}
\end{align}
$$

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 


on concentration of the amphiprotic salt (as the concentration approaches zero, the $\mathrm{pH}$ should approach 7), 
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
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.
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 )