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Chemical Reactivity

Hydrogen pero...

Question

Hydrogen peroxide solution having volume of $20$ mL reacts quantitatively with a solution of $K M n O_{4}$ of volume $20$ mL acidified with dilute $H_{2} S O_{4}$ . The same volume of the $K M n O_{4}$ solution is just decolourised by $10$ mL of $M n S O_{4}$ in a neutral medium simultaneously forming a dark-brown precipitate of hydrated $M n O_{2}$ . The brown precipitate is dissolved in $10$ mL of $0.2 M$ sodium oxalate under boiling condition in the presence of dilute $H_{2} S O_{4}$ . If the molarity of $H_{2} O_{2}$ is $x$ , then find the value of $10 x$ .

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Solution

The reactions involved are given below:
$2 K M n O_{4} + 3 H_{2} S O_{4} + 5 H_{2} O_{2} \to K_{2} S O_{4} + 2 M n S O_{4} + 8 H_{2} O$
$2 K M n O_{4} + 3 M n S O_{4} + 2 H_{2} O \to 5 M n O_{2} + 2 H_{2} S O_{4} + K_{2} S O_{4}$
$M n O_{2} + N a_{2} C_{2} O_{4} + 2 H_{2} S O_{4} \to M n S O_{4} + N a_{2} S O_{4} + 2 H_{2} O + 2 C O_{2}$

Millimoles of $N a_{2} C_{2} O_{4} = 10 \times 0.2 = 2$
Milliequivalents of $N a_{2} C_{2} O_{4} = 4$
Milliequivalents of $M n O_{2} = 4$
Milliequivalents of $K M n O_{4} = 4$
Milliequivalents of $H_{2} O_{2} = 4$
Millimoles of $H_{2} O_{2} = 2$
Millimoles of $H_{2} O_{2} =$ Molarity $\times V_{m L}$
or $2 = Molarity \times 20$
or $Molarity = 0.1 M$
Hence, the value of $10 x$ is $10 \times 0.1 = 1$ .

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