Some Basic Concepts of Chemistry - Result Question 40
42. A solution of 0.2 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 mL of the oxidant. The resultant
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 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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Answer:
Correct Answer: 42. (1:2)
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
$\begin{array}{ll} & \text { Meq of } \mathrm{C}_2 \mathrm{O}_4^{2-}=\mathrm{Meq} \text { of } \mathrm{MnO}_4^{-} \\ \Rightarrow \quad & 2 x=0.02 \times 5 \times 22.6 \quad \Rightarrow \quad x=1.13 \end{array}$
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 $\mathrm{Cu}^{2+}$.
$\begin{aligned} & \Rightarrow \quad \text { Meq of } \mathrm{Cu}^{2+}=\text { Meq of } \mathrm{I}_2=\text { meq of hypo } \\ & \Rightarrow \quad \quad x=11.3 \times 0.05=0.565 \\ & \Rightarrow \text { Moles of } \mathrm{Cu}^{2+}: \quad \text { moles of } \mathrm{C}_2 \mathrm{O}_4^{2-}=0.565: 1.13=1: 2 \end{aligned}$
BETA
