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](https://chemistry.stackexchange.com/questions/114488/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.




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](https://chemistry.stackexchange.com/questions/114488/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 )


 1