Question

$0.5g$ mixture of $K_{2}C{r}_2{O}_7$ and $KMn{O}_4$ was treated with an excess of $KI$ in acidic medium. Iodine liberated required $100c{m}^3$ of $0.15N$ sodium thiosulphate solution for titration. Find the percentage amount of each in the mixture.

Answer 1

Let ${}^{\prime }{a}^{\prime }g$ of $K_{2}C{r}_2{O}_7$ be present in the mixture.
Mass of $KMn{O}_4=(0.5-a)g$
Eq. mass of $K_{2}C{r}_2{O}_7=\frac{Mol.mass}{6}=\frac{294}{6}=49.0$
Eq. mass of $KMn{O}_4=\frac{Mol.mass}{5}=\frac{158}{5}=31.6$
No. of equivalents of $K_{2}C{r}_2{O}_7=\frac{a}{49.0}$
No. of equivalents of $KMn{O}_4=\frac{(0.5-a)}{31.6}$
No. of equivalents of $N{a}_2{S}_2{O}_3$ in $100c{m}^3$ of $0.15N$ solution
$=\frac{100\times 0.15}{1000}=0.015$
Equivalents of $K_{2}C{r}_2{O}_7+$ Equivalents of $KMn{O}_4$
$\equiv$ Equivalents of iodine
$\equiv$ Equivalents of $N{a}_2{S}_2{O}_3$
$\frac{a}{49.0}+\frac{(0.5-a)}{31.6}=0.015$
$17.4a=1.274$
$a=0.0732$
% of $K_{2}C{r}_2{O}_7=\frac{0.0732\times 100}{0.5}=14.64$
% of $KMn{O}_4=85.36$.