Question #44820

The reduction of $\ce{MnO4^-}$ by a substrate $\ce{S}$ in an aqueous buffer proceeds via the rapid pre-equilibrium formation of a soluble Mn(IV) intermediate, $\ce{I(OH)4}$. This intermediate then undergoes two competing rate-determining pathways:

Pathway A (Acidic): $\ce{I(OH)4 + S ->[k_A] Mn^{2+} + P_{ox} + 4H2O}$ Pathway B (Neutral): $\ce{I(OH)4 ->[k_B] MnO2(s) + 2H2O}$

The rate constant $k_A$ is thermodynamically coupled to the reduction potential of the $\ce{I(OH)4 / Mn^{2+}}$ couple. According to a linear free-energy relationship at 298 K, $\log\left(\frac{k_A}{k_A^\circ}\right) = \frac{n}{0.059} (E - E^\circ)$, where $n=2$ is the number of electrons transferred in the half-reaction: $\ce{I(OH)4 + 2e^- + 4H^+ -> Mn^{2+} + 4H2O}$ (Assume pseudo-standard states, i.e., concentration = 1 M, for all Mn species when calculating the kinetic potential shift $E - E^\circ$).

Pathway B is independent of pH ($k_B = k_B^\circ$).

Given that $k_A^\circ [\ce{S}] = 10^8 \, k_B^\circ$, at what pH will the rate of formation of $\ce{Mn^{2+}}$ exactly equal the rate of formation of $\ce{MnO2(s)}$?