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Neutralization

An impure sam...

Question

An impure sample of sodium oxalate $(N a_{2} C_{2} O_{4})$ weighing $0.20 g$ is dissolved in aqueous solution of $H_{2} S O_{4}$ and solution is titrated at $70^{\circ} C$ , requiring $45 m L$ of $0.02 M K M n O_{4}$ solution. The end point is overrun and back titration in carried out with $10 m L$ of $0.1 M$ oxalic acid solution. Find the $%$ purity of $N a_{2} C_{2} O_{4}$ in sample.

A

75

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B

83.75

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C

90.25

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D

None of these

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Solution

The correct option is D $83.75$
Meq. of $N a_{2} C_{2} O_{4}$ = Meq. of $K M n O_{4}$ reacted
Total Meq. of $K M n O_{4} -$ excess Meq. of $K M n O_{4}$ reacted with $H_{2} C_{2} O_{4} = 45 \times 0.02 \times 5 - 10 \times 0.1 \times 2 = 2.5$ .
$1000 \times \frac{W}{134} \times 2 = 2.5$
$W_{N a_{2} C_{2} O_{4}} = 0.1675 g$ .
$%$ purity of $N a_{2} C_{2} O_{4}$ in sample $= \frac{0.1675}{0.2} \times 100 = 83.75$ .

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