$H_{3} P O_{4}$ forms $N a H_{2} P O_{4}$ , $N a_{2} H P O_{4}$ and $N a_{3} P O_{4}$ . Equivalent weight of acid at each stage of salt formation are $98$ , $49$ and $32.66$ respectively.
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Solution

As we know,
$E q u i v a l e n t w e i g h t = \frac{M o l e c u l a r w e i g h t}{V a l e n c y}$
equivalent weight
$H_{3} P O_{4} = N a H_{2} P O_{4} = 1 H^{+}$ $= M = 98$
$H_{3} P O_{4} = N a_{2} H P O_{4} = 2 H^{+}$ $= \frac{M}{2} = 49$
$H_{3} P O_{4} = N a_{3} H P O_{4} = 3 H^{+}$ $= \frac{M}{3} = 32.66$

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