$H_{3} P O_{3}$ forms $N a H_{2} P O_{3}$ and $N a_{2} H P O_{3}$ . The ratio of the equivalent weights of acid at two different stages of salt formation is $x : y$ , then value of $x + y$ is :

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Solution

$H_{3} P O_{3} \to N a H_{2} P O_{3} \to N a_{2} H P O_{3}$
$H_{3} P O_{3} = 1 H^{+} (N a H_{2} H P O_{3})$
$E q u i v a l e n t w e i g h t = m o l e c u l a r w e i g h t$ .
$E_{1} = 82 g$ $e q u i v a l e n t^{- 1}$ .
$H_{3} P O_{3} = 2 H^{+} (N a_{2} H P O_{3})$ .
Thus, $e q u i v a l e n t w e i g h t = \frac{m o l e c u l a r w e i g h t}{2}$ .
$E_{2} = 41 g$ $e q u i v a l e n t^{- 1}$ .
So, ratio of $E_{1} / E_{2} = 82 / 41 = 2 / 1$ .
So, value of $x + y = 2 + 1 = 3$ .

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