Finding pH of tri-protic acid

I am studying for my final and I am given this problem:

A solution of sodium phosphate is made from 10.5 g sodium phosphate in 150 mL of water. What is the pH of this solution

Given: Ka1, Ka2.

The solution says that we can simply calculate this pH with the following eqution: $$pH=\frac{pK_{a1}+pK_{a2}}{2}$$ However this makes absolutely no sense to me, as this is the pH at the equivalence point. What am I missing as towhy this equation can be used?

Answer

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.

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

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 $(2)$: $$\begin{align} K_{a_1} &= \frac{c(\ce{H3+O})\cdot c(\ce{H2A-}) }{c(\ce{H3A})} \tag4\\ K_{a_2} &= \frac{c(\ce{H3+O})\cdot c(\ce{HA^{2-}})}{c(\ce{H2A-})} \tag5\\ \end{align}$$ We will just go ahead and multiply these equations, since they are coupled, simultaneously happening processes. Then we will cancel whatever cancels. $$\begin{align} K_{a_1} \cdot K_{a_2} &= \frac{c(\ce{H3+O})\cdot c(\ce{H2A-}) }{c(\ce{H3A})} \cdot \frac{c(\ce{H3+O})\cdot c(\ce{HA^{2-}})}{c(\ce{H2A-})} \\ K_{a_1} \cdot K_{a_2} &= c^2(\ce{H3+O})\cdot \frac{c(\ce{HA^{2-}})}{c(\ce{H3A})} \tag7\\ \end{align}$$ 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 $$\begin{align} c^2(\ce{H3+O}) &= K_{a_1} \cdot K_{a_2}\\ c(\ce{H3+O}) &= \sqrt{K_{a_1} \cdot K_{a_2}}\\ \ce{pH} &= \frac12\left(\mathrm{p}K_{a_1}+\mathrm{p}K_{a_2}\right). \end{align}$$

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.