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Heat of Reaction

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Question

A solution of $0.2 g$ of a compound containing $C u^{2 +}$ and ${C_{2} O_{4}}^{2 -}$ ions on titration with $0.02 M K M n O_{4}$ in presence of $H_{2} S O_{4}$ consumes $22.6 m L$ oxidant. The resulting solution is neutralised by $N a_{2} C O_{3}$ , acidified with dilute acetic acid and titrated with excess of $K I$ . The liberated iodine required $1.3 m L$ of $0.05 M N a_{2} S_{2} O_{3}$ for complete reduction. Find out the whole ratio of $C u^{2 +}$ and ${C_{2} O_{4}}^{2 -}$ in the compound.

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Solution

$1 s t$ case : Only $C_{2} O_{4}^{2 -}$ ions are oxidised by $K M n O_{4}$ solution.
Normality of $K M n O_{4}$ solution $= 0.02 \times 5 = 0.1 N$
$22.6 m L$ of $0.1 N K M n O_{4} = 22.6 m L$ of $0.1 N C_{2} O_{4}^{2 -}$ soln.
Mass of $C_{2} O_{4}^{2 -} i o n s$ in the solution $= \frac{N \times E \times V}{1000} = \frac{N \times M \times V}{1000 \times 2}$
No, of moles of $C_{2} O_{4}^{2 -}$ ions in the solution $= \frac{N \times M \times V}{1000 \times 2 \times M}$
$= \frac{N \times V}{2000}$
$= \frac{0.1 \times 22.6}{2000}$
$= 11.3 \times 10^{- 4}$
$2 n d$ case : Only $C u^{2 +}$ ions are reduced by $K I$ and iodine liberated in neutralised by $N a_{2} S_{2} O_{3}$ solution.
$11.3 m L$ of $0.05 M N a_{2} S_{2} O_{3} \equiv 11.3 m L$ of $0.05 N N a_{2} S_{2} O_{3}$
$= 11.3 m L$ of $0.05 N I_{2}$
$= 11.3 m L$ of $0.05 N C u^{2 +}$
Mass of $C u^{2 +}$ ions in the solution $= \frac{N \times E \times V}{1000} = \frac{N \times M \times V}{1000}$
No, of moles of $C u^{2}$ ions in the solution $= \frac{N \times M \times V}{1000 \times M}$
$= \frac{N \times V}{1000}$
$= \frac{0.05 \times 11.3}{1000}$
$= 5.65 \times 10^{- 4}$
Molar ratio of $\frac{C u^{2 +}}{C_{2} O_{4}^{2 -}} = \frac{5.65 \times 10^{- 4}}{11.3 \times 10^{- 4}} = \frac{1}{2}$ .

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