Question #44674

An aqueous solution contains $0.02\text{ mol}$ of $\ce{K2Cr2O7}$ and $0.04\text{ mol}$ of $\ce{KMnO4}$. The solution is divided into two equal aliquots, A and B.

Aliquot A is strongly acidified with dilute $\ce{H2SO4}$ and an excess of $\ce{KI}$ is added. The liberated $\ce{I2}$ is titrated with a $C\text{ M}$ $\ce{Na2S2O3}$ solution, requiring $V_A\text{ mL}$.

Aliquot B is buffered to a faintly alkaline pH ($pH \approx 9$) and an excess of $\ce{KI}$ is added. Under these conditions, $\ce{I-}$ is oxidized to $\ce{IO3-}$. The unreacted $\ce{KI}$ is removed, and the $\ce{IO3-}$ formed is quantitatively isolated, dissolved in dilute acid, and treated with excess $\ce{KI}$ to liberate $\ce{I2}$. This liberated $\ce{I2}$ is titrated with the same $C\text{ M}$ $\ce{Na2S2O3}$ solution, requiring $V_B\text{ mL}$.

Assuming complete reactions, what is the ratio $V_A / V_B$?