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Neutralisation Reaction

In acidic med...

Question

In acidic medium, 100 ml of 1 M $K M n O_{4}$ solution oxidizes 100 ml of $H_{2} O_{2}$ solution. The volume of 0.01 M $K M n O_{4}$ , required to oxidize the same volume of $H_{2} O_{2}$ in alkaline medium in ml is :

A

\frac{300}{2}

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B

\frac{300}{5}

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C

\frac{500}{3}

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D

\frac{500}{2}

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Solution

The correct option is A $\frac{500}{3}$
In acidic medium, the equivalent weight of $K M n O_{4}$ is one-fifth of its molecular mass.
$N_{1} V_{1} = N_{2} V_{2}$
$100 \times 1 \times 5 = 100 \times N_{2}$
Molarity of $H_{2} O_{2} = 5 M$

In alkaline medium, the equivalent weight of $K M n O_{4}$ is one-third of its molecular mass.
$N_{1} V_{1} = N_{2} V_{2}$
$0.01 \times 3 \times V_{2} = 5 \times 100$
$V_{2} = \frac{500}{3}$

The volume of 0.01 M $K M n O_{4}$ , required to oxidize the same volume of $H_{2} O_{2}$ in alkaline medium in ml $= \frac{500}{3} m L$ .

Hence, option C is correct.

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