Equivalent weight of the underlined $2 H_{3} P O_{2} - ------ - ⟶ H_{3} P O_{4} + P H_{3}$ :

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Solution

The correct option is A 33
$E_{1} (o x i d a t i o n) = \frac{M}{4}$

$E_{2} (r e d u c t i o n) = \frac{M}{4}$

Eq. wt. of (x) when it undergoes disproportionation = $E_{1}$ (oxidation)+ $E_{2}$ (reduction)

$E_{1} + E_{2} = \frac{M}{4} + \frac{M}{4} = \frac{M}{2}$

$E_{1} + E_{2} = \frac{66}{2} = 33$

Hence option A is correct.

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H_{3} P O_{2}

2 H_{3} P O_{2} \to H_{3} P O_{4} + P H_{3}

M

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P_{4})

{P - -}_{4} \to P H_{3} + H_{2} P O_{2}^{-}

P - - H_{3}

S - - O_{2}

H N - - O_{3}

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H_{3} P O_{4} - ------ - + C a (O H)_{2} ⟶ C a H P O_{4} + 2 H_{2} O

H_{3} P O_{2} - ------ - + N a O H ⟶ N a H_{2} P O_{2} + H_{2} O

N a H C_{2} O_{4} - ---------- - + N a O H ⟶ N a_{2} C_{2} O_{4} + H_{2} O

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