Question

$2.0$ g sample of $NaCN$ is dissolved in $50$ mL of $0.3M$ mild alkaline $KMn{O}_4$ and heated strongly to convert all the $C{N}^{⊝}$ to $OC{N}^{⊝}$. The solution after acidification with $H_{2}S{O}_4$ requires $500$ mL of $0.05M$ $FeS{O}_4$. The percentage purity of $NaCN$ in the sample is :

• A. $36.75$
• B. $36.95$
• C. $38.75$
• D. none of the above

Answer 1

The correct option is A $36.75$

The various redox changes are shown as follows:
$a$. $\stackrel{+2}{\underset{\underset{x=2}{x-3=-1}}{C{N}^{⊝}}}\to \stackrel{+4}{\underset{\underset{x=4}{x-3-2=-1}}{OC{N}^{⊝}}}+2e^{-}(n=2)$
$b$. $3e^{-}+\underset{\underset{x=7}{x-8=-1}}{Mn{O}_4^{⊝}}\to \underset{\underset{x=3}{x-4=0}}{Mn{O}_2}$ $(n=3$, in mild basic medium$)$
$c$. $5e^{-}+Mn{O_4}^{⊝}\to M{n}^{2+}(n=5$, in acidic medium$)$
$d$. $F{e}^{2+}\to F{e}^{3+}+e^{-}(n=1)$
$mEq$ of $KMn{O}_4$ added in basic medium $=50\times 0.3\times 3(n$-factor$)=45.0$
$mEq$ of $KMn{O}_4$ in acidic medium $(n=5)$ left after reaction with $NaCN=mEq$ of $FeS{O}_4$ used $=500\times 0.05\times 1(n$-factor $)$ $=25.0$
$mEq$ of $KMn{O}_4(n$-factor $=3)$ left $=\frac{25\times 3}{5}=15$
$mEq$ of $NaCN$ in sample $=mEq$ of $KMn{O}_4$ added $-mEq$ of $KMn{O}_4$ left $=45-15=30$
$\therefore \frac{Weight}{EwofNaCN}\times {10}^3=30$ $[EwofNaCN=\frac{49}{2}(n-factor=2\left)\right]$
$\therefore \frac{W}{\frac{49}{2}}\times {10}^3=30$
$W_{NaCN}=0.735$ g.
% of $NaCN=\frac{0.735}{2.0}\times 100=36.75$%