Question

In basic medium CrO$_4^{2-}$ oxidises S$_2$O$_3^{2-}$ to form SO$_4^{2-}$ and itself changes into Cr(OH)$_4^-$. The volume of 0.154 M CrO$_4^{2-}$ required to react with 40 mL of 0.25 M S$_2$O$_3^{2-}$ is _________ mL.

Hint:

Always calculate the $n$-factor per molecule of the reactant. For $S_2O_3^{2-}$, there are two sulfur atoms increasing in oxidation state, so the total change is 8.

Answer 1

Correct Answer: 173

Step 1: $n$-factor for $CrO_4^{2-} \rightarrow Cr(OH)_4^-$: $Cr(+6) \rightarrow Cr(+3)$, change = 3.
Step 2: $n$-factor for $S_2O_3^{2-} \rightarrow 2SO_4^{2-}$: Average $S(+2) \rightarrow S(+6)$, change $= 2 \times (6-2) = 8$.
Step 3: Equivalents of Oxidant = Equivalents of Reductant.
Step 4: $(M_1 \times n_1 \times V_1) = (M_2 \times n_2 \times V_2)$.
Step 5: $0.154 \times 3 \times V_1 = 0.25 \times 8 \times 40$.
Step 6: $0.462 V_1 = 80 \Rightarrow V_1 = 80 / 0.462 \approx 173.16$ mL.