$H_{3} {P O}_{2} + N a O H ⟶ {N a H}_{2} {P O}_{2} + H_{2} O$
What is the equivalent weight of $H_{3} {P O}_{2}$ ? (Mol. wt. is M)

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M/2

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M/3

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None of these

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Solution

The correct option is B M
For the reaction $H_{3} {P O}_{2} + N a O H ⟶ {N a H}_{2} {P O}_{2} + H_{2} O$ , only one H atom of $H_{3} {P O}_{2}$ is replaced.
Hence, the equivalent wight of $H_{3} {P O}_{2}$ is equal to molecular weight.
$E = M$
$Equivalent weight of an acid = \frac{Molecular weight of an acid}{The number of replaceable hydrogen atoms}$

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