Calculate the equivalent weight of $H_{2} S O_{4}$ in the following reaction is ?
$N a_{2} C r_{2} O_{7} + 3 S O_{2} + H_{2} S O_{4} \to 3 N a_{2} S O_{4} + C r_{2} (S O_{4})_{3} + H_{2} O$

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Solution

$H_{2} S O_{4}$ is a dibasic acid i.e. basicity is $2$
In other way $1$ mole of $H_{2} S O_{4}$ will give $2$ moles of $H^{+}$ .
In the given it reacts with $N a_{2} C r_{2} O_{7}$ .
Equivalent mass of $H_{2} S O_{4}$ is $\frac{Molecular mass}{2} = \frac{98}{2} = 49$

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