Finding pH of tri-protic acid
amphoteric
  
  A chemical species that behaves both as an acid and as a base is called amphoteric. This property depends upon the medium in which the species is investigated: $\ce{H2SO4}$ is an acid when studied in water, but becomes amphoteric in superacids.


from the IUPAC goldbook.



For ampholytes like $\ce{NaH2PO4}$ the $\ce{pH}$ is independent on the concentration (in first approximation). Consider the reactions of that are possible:
\begin{align}\ce{
H2A-    + H3+O &amp;~&lt;=&gt; H3A     + H2O  \tag1\\
H2A-    + H2O  &amp;~&lt;=&gt; HA^{2-} + H3+O \tag2\\
HA^{2-} + H2O  &amp;~&lt;=&gt; A^{3-}  + H3+O \tag3\\
}\end{align}

As the first approximation, we will assume that everything happening in equation $(3)$ is negligible, as is will only very little influence the concentration of $\ce{HA^{2-}}$. Therefore we can focus on the acidity constants $K_{a_1}$ of $(1)$ and $K_{a_2}$ of $(2)$:
\begin{align}
K_{a_1}  &amp;= \frac{c(\ce{H3+O})\cdot c(\ce{H2A-})   }{c(\ce{H3A})}  \tag4\\
K_{a_2}  &amp;= \frac{c(\ce{H3+O})\cdot c(\ce{HA^{2-}})}{c(\ce{H2A-})} \tag5\\
\end{align}
We will just go ahead and multiply these equations, since they are coupled, simultaneously happening processes. Then we will cancel whatever cancels.
\begin{align}
K_{a_1} \cdot K_{a_2} &amp;= \frac{c(\ce{H3+O})\cdot c(\ce{H2A-})   }{c(\ce{H3A})} \cdot
  \frac{c(\ce{H3+O})\cdot c(\ce{HA^{2-}})}{c(\ce{H2A-})} \\
K_{a_1} \cdot K_{a_2} &amp;= c^2(\ce{H3+O})\cdot  \frac{c(\ce{HA^{2-}})}{c(\ce{H3A})} \tag7\\
\end{align}
Now another major assumption is that the reactions $(1)$ and $(2)$ are happening to the same extent and therefore we can approximate
$$c(\ce{HA^{2-}})=c(\ce{H3A})\tag8$$
and rewrite $(7)$ as
\begin{align}
c^2(\ce{H3+O}) &amp;= K_{a_1} \cdot K_{a_2}\\
c(\ce{H3+O})   &amp;= \sqrt{K_{a_1} \cdot K_{a_2}}\\
\ce{pH}        &amp;= \frac12\left(\mathrm{p}K_{a_1}+\mathrm{p}K_{a_2}\right).
\end{align}

However, this equation only works as a first approximation. Usually the coupled equilibria are very complex and in most instances a computer needs to be involved to calculate it accurately. More information on polyprotic acids.