Question

The $M{n}_3{O}_4$ formed on strong heating of a sample of $MnS{O}_4\cdot 4H_{2}O$ was dissolved in $100c{m}^3$ of $0.1NFeS{O}_4$ containing dilute $H_{2}S{O}_4$. The resulting solution reacted completely with $50c{m}^3$ of $KMn{O}_4$ solution. $25c{m}^3$ of this $KMn{O}_4$ solution requires $30c{m}^3$ of $0.1NFeS{O}_4$ solution for complete reaction. Calculate the amount of $MnS{O}_4\cdot 4H_{2}O$ in the sample.

Answer 1

$MnS{O}_4\cdot 4H_{2}O\stackrel{Heat}{\to }M{n}_3{O}_4$
$M{n}_3{O}_4$ is dissolved in ferrous sulphate solution and it reduced from $M{n}^{\left(8/3\right)+}$ to $M{n}^{2+}$. The excess of $FeS{O}_4$ is estimated by doing titration with $KMn{O}_4$ solution. The normality of $KMn{O}_4$ solution is determined by another ferrous sulphate solution.
For normality of $KMn{O}_4$ solution:
$25\times N=30\times 0.1$
$N=\frac{30\times 0.1}{25}=\frac{3}{25}$
Let the volume of unreacted $FeS{O}_4$ solution be $VmL$
$VmL$ of $0.1NFeS{O}_4=50mL$ of $\frac{3}{25}NKMn{O}_4$
or $V=\frac{5.0\times 3}{0.1\times 25}=60mL$
$\therefore$ Volume of $FeS{O}_4$ used for $M{n}_3{O}_4=(100-60)mL$
$=40mL$
$40mL$ of $0.1NFeS{O}_4\equiv 40mL$ of $0.1NM{n}_3{O}_4$
$\equiv 40mL$ of $0.1NMnS{O}_4\cdot 4H_{2}O$
Mass of $MnS{O}_4\cdot 4H_{2}O=\frac{E\times 0.1\times 40}{1000}=\frac{E}{250}g$
Equivalent mass of
$MnS{O}_4\cdot 4H_{2}O=\frac{M}{(\frac{8}{3}-2)}=\frac{3M}{2}=\frac{3\times 223}{2}$
$\therefore$ Mass of $MnS{O}_4\cdot 4H_{2}O=\frac{3\times 223}{2\times 250}=1.338g$.