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Hydrogen peroxide solution $(20 m L)$ reacts quantitatively with a solution of $K M n O_{4} (20 m L)$ acidified with dilute $H_{2} S O_{4}$ . The same volume of the $K M n O_{4}$ solution is just decolourised by $10 m L$ of $M n S O_{4}$ in neutral medium simultaneously forming a dark brown precipitate of hydrated $M n O_{2}$ . The brown precipitate is dissolved in $10 m L$ of $0.2 M$ sodium oxalate under boiling condition in the presence of dilute $H_{2} S O_{4}$ . Write the balanced equations involved in the reactions and calculate the molarity of $H_{2} O_{2}$ .

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Solution

(i) $M n O_{2} p p t . 1 m o l e + N a_{2} C_{2} O_{4} 1 m o l e + 2 H_{2} S O_{4} \to M n S O_{4} + N a_{2} S O_{4} + 2 C O_{2} + 2 H_{2} O$
$m M$ of $M n O_{2} = m N$ of $N a_{2} C_{2} O_{4} = 10 \times 0.2 = 2$
(ii) $2 K M n O_{4} 2 m o l e + 3 M n S O_{4} + 2 H_{2} O \to 5 M n O_{2} p p t . 5 m o l e + K_{2} S O_{4} + 2 H_{2} O$
$m M$ of $K M n O_{4} = m N$ of $M n O_{2} \times \frac{2}{5} = 2 \times \frac{2}{5} = \frac{4}{5}$
(iii) $2 K M n O_{4} 2 m o l e s + 3 H_{2} S O_{4} + 5 H_{2} O_{2} 5 m o l e \to K_{2} S O_{4} + 2 M n S O_{4} + 8 H_{2} O + 5 O_{2}$
$∴ m M$ of $H_{2} O_{2} = m M$ of $K M n O_{4} \times \frac{5}{2} = \frac{4}{5} \times \frac{5}{2} = 2$
$∴ M \times 20 = 2$
or $M_{H_{2} O_{2}} = \frac{2}{20} = 0.1 M$ .

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20 m L

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