Question

Equivalent mass of $H_{3}P{O}_2$ when it disproportionate into $P{H}_3$ and $H_{3}P{O}_3$ is:

• A. M
• B. M/2
• C. M/4
• D. 3M/ 4

Answer 1

The correct option is D 3M/ 4

The disproportionation reaction of $H_{3}P{O}_2$ is :

$3H_{3}P{O}_2\to P{H}_3+2H_{3}P{O}_3$

This reaction can also be written as:

$(P{O}_2{)}^{3-}+4H^{+}+4e^{-}\to P^{3-}+2H_{2}O$ --(1)

$(P{O}_2{)}^{3-}+H_{2}O\to (P{O}_3{)}^{3-}+2H^{+}2e^{-}$ --(2)

Multiply the second equation by 2 and then on adding equation (1)+ (2), we get

$3(P{O}_2{)}^{3-}\stackrel{4e^{-}}{\to }{P}^{3-}+2(P{O}_3{)}^{3-}$

Thus, in total $4e^{-}$ are exchanged.

Equivalent mass = $\frac{Molarmass}{Valencyfactor}$

As Valency factor is no. of moles of electron taken or given by $onemole$ of Oxidising/Reducing agent.

Valency factor = $\frac{4}{3}$

Let M be the mass of $H_{3}P{O}_4$.

Equivalent mass of $H_{3}P{O}_4$ in the given reaction=$\frac{3M}{4}$.

Hence, the correct option is (D).