Question

$0.5$ g mixture of $K_{2}C{r}_2{O}_7$ and $KMn{O}_4$ is treated with excess of $KI$ in acidic medium. $I_2$ liberated requires $100$ mL of $0.15N$ $N{a}_2{S}_2{O}_3$ solution for titration. $\%$ of $K_{2}C{r}_2{O}_7$ and $\%$ of $KMn{O}_4$ are:

• A. $14.6\%$ and $85.6\%$
• B. $15.6\%$ and $85.6\%$
• C. $14.6\%$ and $87.6\%$
• D. none of the above

Answer 1

Answer: (A)

$14.6\%$ and $85.6\%$

The various redox changes that occur is shown below:
$a$. $5e^{-}+Mn{O}_4^{⊝}\to M{n}^{2+}(n=5)$
$b$. $6e^{-}+C{r}_2{O}_7^{2-}\to 2C{r}^{3+}(n=6)$
$c$.$2I^{⊝}\to I_2+2e^{-}(n=2)$
$d$. $2S_2{O}_3^{2-}\to S_4{O}_6^{2-}+2e^{-}(n=\frac{2}{2}=1)$
Let the weight of $K_{2}C{r}_2{O}_7$ and $KMn{O}_4$ are $a$ and $b$ g, respectively.
$\therefore a+b=0.5$ $.......\left(i\right)$
$mEq.$ of $K_{2}C{r}_2{O}_7+mEq.$ of $KMn{O}_4=mEq.$ of $KI$ $=mEq.$ of $I_2$ liberated $=mEq.$ of $N{a}_2{S}_2{O}_3$
$\therefore (\frac{a}{\frac{294}{6}}+\frac{b}{\frac{158}{5}})\times {10}^3=100\times 0.15$ $.......\left(ii\right)$
From equations $\left(i\right)$ and $\left(ii\right)$, $a=0.073,b=0.427$
% of $K_{2}C{r}_2{O}_7=\frac{0.073\times 100}{0.5}=14.6$%
% of $KMn{O}_4=\frac{0.427\times 100}{0.5}=85.6$%