$K H C_{2} O_{4} \cdot H_{2} C_{2} O_{4}$ , the equivalent weight of the given substance as acid or base is :

E = \frac{M}{3}

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E = \frac{M}{4}

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E = \frac{M}{2}

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none of the above

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Solution

The correct option is A $E = \frac{M}{3}$
1 mole of $K H C_{2} O_{4} \cdot H_{2} C_{2} O_{4}$ contains 3 moles of replaceable hydrogen ions.
Hence, its equivalent weight is equal to one third of its molecular weight.
$Equivalent mass of an acid = \frac{Molar mass of the acid}{Number of replaceable H atoms}$
$∴ E = \frac{M}{3}$

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