Some Basic Concepts of Chemistry 2 Question 42
42. A solution of $0.2 \mathrm{~g}$ of a compound containing $\mathrm{Cu}^{2+}$ and $\mathrm{C}{2} \mathrm{O}{4}^{2-}$ ions on titration with $0.02 \mathrm{M} \mathrm{KMnO}{4}$ in presence of $\mathrm{H}{2} \mathrm{SO}_{4}$ consumes $22.6 \mathrm{~mL}$ of the oxidant. The resultant
Some Basic Concepts of Chemistry 7
solution is neutralised with $\mathrm{Na}{2} \mathrm{CO}{3}$, acidified with dilute acetic acid and treated with excess KI. The liberated iodine requires $11.3 \mathrm{~mL}$ of $0.05 \mathrm{M} \mathrm{Na}{2} \mathrm{~S}{2} \mathrm{O}{3}$ solution for complete reduction. Find out the mole ratio of $\mathrm{Cu}^{2+}$ to $\mathrm{C}{2} \mathrm{O}_{4}^{2-}$ in the compound. Write down the balanced redox reactions involved in the above titrations.
(1991, 5M)
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Solution:
- With $\mathrm{KMnO}_{4}$, oxalate ion is oxidised only as :
$5 \mathrm{C}{2} \mathrm{O}{4}^{2-}+2 \mathrm{MnO}{4}^{-}+16 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}{2}+8 \mathrm{H}_{2} \mathrm{O}$
Let, in the given mass of compound, $x$ millimol of $\mathrm{C}{2} \mathrm{O}{4}^{2-}$ ion is present, then
$$ \begin{array}{ll} & \text { Meq of } \mathrm{C}{2} \mathrm{O}{4}^{2-}=\text { Meq of } \mathrm{MnO}_{4}^{-} \ \Rightarrow & 2 x=0.02 \times 5 \times 22.6 \Rightarrow x=1.13 \end{array} $$
At the later stage, with $\mathrm{I}^{-}, \mathrm{Cu}^{2+}$ is reduced as :
$$ \begin{aligned} & 2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-} \longrightarrow 2 \mathrm{CuI}+\mathrm{I}{2} \ \text { and } \quad \mathrm{I}{2}+2 \mathrm{~S}{2} \mathrm{O}{3}^{2-} & \longrightarrow 2 \mathrm{I}^{-}+\mathrm{S}{4} \mathrm{O}{6}^{2-} \end{aligned} $$
Let there be $x$ millimol of $\mathrm{Cu}^{2+}$.
$\Rightarrow \quad$ Meq of $\mathrm{Cu}^{2+}=$ Meq of $\mathrm{I}_{2}=$ meq of hypo
$\Rightarrow \quad x=11.3 \times 0.05=0.565$
$\Rightarrow$ Moles of $\mathrm{Cu}^{2+}:$ moles of $\mathrm{C}{2} \mathrm{O}{4}^{2-}=0.565: 1.13=1: 2$
