Calculate the equivalent weight of $K M n O_{4}$ in acidic medium.

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Solution

$K M n O_{4}$ as an oxidizer in acidic media:

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$ (gained 5 electrons from reductant)

To find out the Eq.wt of any element we use the formula, Eq.wt = molar mass/ no. Of electrons lost or gained.

Therefore, Eq. wt. of grams/equivalent
$K M n O_{4} = \frac{158.04}{5} = 31.61$

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