Question

A solution of 0.2 g of a compound containing $C{u}^{2+}$ and $C_2{O}_4^{2-}$ ions on titration with 0.02 M $KMn{O}_4$ in the presence of $H_{2}S{O}_4$ consumes 22.6 mL of the oxidant. The resultant solution is neutralized with $N{a}_{2}C{O}_3$, acidified with dilute acetic acid and treated with excess of $KI$. The liberated iodine requires 11.3 mL of 0.05 N $N{a}_{2}S{O}_4$ solution for complete reduction . Find out the molar ratio of $C{u}^{2+}$ to $C_2{O}_4^{2-}$ in the compound.

• A. 1 : 2
• B. 2 : 1
• C. 1 : 4
• D. 4 : 1

Answer 1

The correct option is A 1 : 2

The mixture of $C{u}^{2+}$ and $C_2{O}_4^{2-}$ are reacting seperately first with $KMn{O}_4$ solution and then with $KI$ to liberate $I_2$. It can be seen that $C{u}^{2+}$ cannot be oxidised and $C_2{O}_4^{2-}$ cannot be reduced. So, reactions involved are:
$2Mn{O}_4^{-}+5C_2{O}_4^{2-}+16H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$
$2C{u}^{2+}+4I^{-}\to C{u}_2{I}_2+I_2$
$I_2+2S_2{O}_3^{2-}\to 2I^{-}+S_4{O}_6^{2-}$

Equivalents of $KMn{O}_4$ solution,
= $\frac{0.02\times 5\times 22.6}{1000}=2.26\times {10}^{-3}$

$\therefore$ moles of $C_2{O}_4^{2-}$,
= $\frac{2.26\times {10}^{-3}}{2}=1.13\times {10}^{-3}$

Equivalents of $N{a}_2{S}_2{O}_3$ solution
= $\frac{0.05\times 11.3}{1000}=5.65\times {10}^{-4}$

$\therefore$ moles of $C{u}^{2+}$
= $\frac{5.65\times {10}^{-4}}{1}=5.65\times {10}^{-4}$

$\therefore$ molar ratio of $C{u}^{2+}$ to $C_2{O}_4^{2-}$
= $\frac{5.65\times {10}^{-4}}{1.13\times {10}^{-3}}=\frac{1}{2}$