Question #44090

A $100\text{ mL}$ aqueous buffer solution contains $0.16\text{ M}$ of a weak acid $\ce{HA}$ ($pK_a = 3.0$) and $0.04\text{ M}$ of its sodium salt $\ce{NaA}$. To this solution, $4.0 \times 10^{-3}\text{ moles}$ of $\ce{KMnO4}$ and $5.0 \times 10^{-3}\text{ moles}$ of $\ce{H2O2}$ are added at $298\text{ K}$. The $\ce{H2O2}$ completely reacts to reduce a portion of the $\ce{MnO4-}$ to $\ce{Mn^{2+}}$. Assuming the volume of the solution remains constant at $100\text{ mL}$ and no other side reactions (e.g., formation of $\ce{MnO2}$) occur, what is the reduction potential of the $\ce{MnO4- / Mn^{2+}}$ half-cell at the final equilibrium state?

(Given: $E^\circ_{\ce{MnO4- / Mn^{2+}}} = 1.510\text{ V}$, $\frac{2.303 RT}{F} = 0.059\text{ V}$, $\log 2 = 0.30$)