Question

For the given disproportionation reaction:
$H_{3}P{O}_2\to P{H}_3+H_{3}P{O}_3$
The equivalent mass of $H_{3}P{O}_2$ in grams is :
(Given : $\text{Atomic mass of}P=31u$)

Answer 1

$H_3\stackrel{+1}{P}{O}_2\to \stackrel{-3}{P}{H}_3+H_3\stackrel{+3}{P}{O}_3$
n-factor calculation,
$n_f=(|O.S{.}_{Product}-O.S{.}_{Reactant}|)\times \text{number of atoms}$
$H_3\stackrel{+1}{P}{O}_2\to H_3\stackrel{+3}{P}{O}_3\text{oxidation}$
$(n_f{)}_{\text{oxidation}}=|3-1|\times 1=2$
$H_3\stackrel{+1}{P}{O}_2\to \stackrel{-3}{P}{H}_3\text{reduction}$
$(n_f{)}_{\text{reduction}}=|-3-1|\times 1=4$
valency factor in disproportionation reaction is given by
$\frac{1}{n_f}=\frac{1}{(n_f{)}_{\text{oxidation}}}+\frac{1}{(n_f{)}_{\text{reduction}}}$
$\frac{1}{n_f}=\frac{1}{2}+\frac{1}{4}$
$n_f=\frac{4}{3}$
$\therefore$ The equivalent weight of $H_{3}P{O}_2=\frac{\text{Molar mass of}{H}_{3}P{O}_2}{\frac{4}{3}}=\frac{3\times \text{Molar mass of}{H}_{3}P{O}_2}{4}$
$\text{Molar mass of}{H}_{3}P{O}_2=3+31+32=66gmo{l}^{-1}$
$\text{Equivalent weight}=\frac{3\times 66}{4}=49.5g$