Question #44103

A $1.0\text{ L}$ aqueous solution at $298\text{ K}$ initially contains $0.10\text{ mol}$ of an amphiprotic salt $\ce{NaHA}$. The acid dissociation constants of the diprotic acid $\ce{H2A}$ are $K_{a1} = 10^{-3}$ and $K_{a2} = 10^{-7}$. To this solution, $0.02\text{ mol}$ of $\ce{KMnO4}$ and an excess of $\ce{H2O2}$ are added. The $\ce{KMnO4}$ is completely reduced to $\ce{Mn^{2+}}$ by $\ce{H2O2}$, which is oxidized to $\ce{O2(g)}$. Assuming no change in volume, that all $\ce{O2}$ escapes, and no $\ce{MnO2}$ is formed, what is the final pH of the solution?

(Given: $\log 2 = 0.30$, $\log 3 = 0.48$)