$C a (O H)_{2} + H_{3} P O_{4} \to C a H P O_{4} + 2 H_{2} O$ the equivalent weight of $H_{3} P O_{4}$ in the above reaction is

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Solution

The correct option is D 49
The equivalent weight of $H_{3} P O_{4} = \frac{m o l e c u l a r w e i g h t}{2}$
$∵$ mole wt of $H_{3} P O_{4} = 3 + 31 + 64 = 98$
$∴ \frac{98}{2} = 49$

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