Question

$100$ mL of $1MKMn{O}_4$ oxidised $100$ mL of $H_2{O}_2$ in acidic medium (when ${Mn{O}_4}^{-}$ is reduced to $M{n}^{2+}$). Volume of same $KMn{O}_4$ required to oxidise $100$ mL of $H_2{O}_2$ in neutral medium (when ${Mn{O}_4}^{-}$ is reduced to $Mn{O}_2$) will be:

• A. $\frac{100}{3}$ mL
• B. $\frac{500}{3}$ mL
• C. $\frac{300}{5}$ mL
• D. $100$ mL

Answer 1

Answer: (B)

$\frac{500}{3}$ mL

In acidic medium, the oxidation number of $Mn$ in $KMn{O}_4$ changes by $5$. In neutral medium, the oxidation number changes by $3$.
In acidic medium, $N_1{V}_1\left(H_2{O}_2\right)=N_2{V}_2\left({Mn{O}_4}^{-}\right)$
$2M_1\times 100=5\times 100$
$M_1\left(H_2{O}_2\right)=2.5M$
In neutral medium, $N_1{V}_1={N_2}^{{}^{\prime }}{V_2}^{{}^{\prime }}$
$2M_1\times 100=3\times {M_2}^{{}^{\prime }}{V_2}^{{}^{\prime }}$
${V_2}^{{}^{\prime }}=\frac{500}{3}$

Hence, volume required will be $\frac{500}{3}$ mL.