The equivalent mass of $H_{3} P O_{4}$ and $N a H_{2} P O_{4}$ in the reaction are respectively :
$2 N a O H + H_{3} P O_{4} \to {N a}_{2} H P O_{4} + 2 H_{2} O$

49, 142

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49, 71

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98, 71

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98, 142

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Solution

The correct option is D 49, 71
$E_{H_{3} P O_{4}} = \frac{M}{2} (t w o H a r e r e p l a c e d) = \frac{98}{2} = 49 g {e q u i v}^{- 1}$

$E_{{N a}_{2} P O_{4}} = \frac{M}{2} (A s a l t {N a}_{2} H P O_{4} f o r m e d i n r e a c t i o n)$
= $\frac{142}{2} = 71$

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Similar questions

2 N a O H + H_{3} P O_{4} \to N a_{2} H P O_{4} + 2 H_{2} O

H_{3} P O_{4} + 2 N a O H \to N a_{2} H P O_{4}

H_{3} P O_{4}

2 N a O H + H_{3} P O_{4} \to N a_{2} H P O_{4} + 2 H_{2} O

H_{3} P O_{4}

H_{3} P O_{4} + N a O H \to N a H_{2} P O_{4} + H_{2} O

2 N a O H + H_{3} P O_{4} \to N a_{2} H P O_{4} + 2 H_{2} O

H_{3} P O_{4}

\frac{M}{3}

= \frac{Molar mass}{n-factor}

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