pH calculation in case of amphiprotic ion


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> $100\ \pu{ml}$ of $0.1\ \pu{M}\ \ce{H3PO4}$ is titrated with $0.5\ \pu{M}\ \ce{NaOH}$ solution till 2nd equivalence point. Then resultant solution was mixed with $10\ \pu{ml}$ of $0.5\ \pu{M}\ \ce{HCl}$ solution. The pH at 2nd equivalence point will be?  
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>  Given: $K\_1 = 10^{–3}\ \pu{M}, K\_2 = 10^{–8}\ \pu{M},K\_3 = 10^{–13}\ \pu{M}$
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**My attempt:** 


I found the volume of $\ce{NaOH}$ (equivalents of $\ce{H3PO4} = $ equivalents of $\ce{NaOH}$). 


$$2\times 100\times 0.1 = 0.5 \times V\\
V=40\ \pu{ml}$$
Since the solution is diluted now, the concentration doesn't remain the same. So $[\ce{Na2HPO4}]=0.071\ \pu{M}$.


Now $\ce{Na2HPO4 -> 2Na+ + HPO4^2-}$


For this I tried to find the hydrolysis constant. I am getting it to be 


$$K\_\mathrm h=\frac{K\_\mathrm{w}^2}{K\_1 K\_2}$$


The formula given to evaluate the pH after the 2nd equivalence point is $$\frac{\mathrm{p}K\_2+\mathrm{p}K\_3}{2}$$ How to derive this?


