Question

A solution of $0.2g$ of a compound containing $C{u}^{2+}$ and $C_2{O}_4^{2-}$ ions on titration with $0.02M$ $KMn{O}_4$ in the presence of $H_{2}S{O}_4$ consumes $22.6$ mL of the oxidant.

The resultant solution is neutralised with $N{a}_{2}C{O}_3$, acidified with dilute acetic acid and treated with excess $KI$. The liberated iodine requires $11.3$ mL of $0.05$ M $N{a}_2{S}_2{O}_3$ solution for complete reduction. The mole ratio of $C{u}^{2+}$ and $C_2{O}_4^{2-}$ in the compound will be :

• A. $1:2$
• B. $3:2$
• C. $2:3$
• D. $2:1$

Answer 1

The correct option is A $1:2$

$KMn{O}_4$ reacts only with oxalate.

$2Mn{O}_4^{⊝}+5C_2{O}_4^{2-}+16H^{⨁}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$

mEq of $KMn{O}_4=M\times V\times n$ factor$=0.02\times 22.6\times 5$

mEq of oxalate $=0.02\times 22.6\times 5$

Millimoles of oxalate $=\frac{0.02\times 22.6\times 5}{2}$
($n$ factor for oxalate is $2$)

$KI$ reacts only with $C{u}^{2+}$.

$2C{u}^{2+}+4I^{⊝}\to C{u}_2{I}_2+I_2$
$I_2+2S_2{O}_3^{2-}\to 2I^{⊝}+S_4{O}_4^{2-}$

mEq of thiosulphate$=M\times V\times n$-factor $=0.05\times 11.3\times 1$

mEq of iodine$=0.05\times 11.3\times 1$

mEq of $C{u}^{2+}=0.05\times 11.3\times 1$

mmoles of $C{u}^{2+}=0.05\times 11.3\times 1$
($n$ factor for $C{u}^{2+}$ is $1$)

$\frac{mmolesofC{u}^{2+}}{mmolesofoxalate}=\frac{0.05\times 11.3\times 1}{\frac{0.02\times 22.6\times 5}{2}}=\frac{1}{2}$