Question

In the disproportion reaction:
$H_{3}P{O}_2\to P{H}_3+H_{3}P{O}_3$
The equivalent mass of $H_{3}P{O}_2$ is (M = molar mass of $H_{3}P{O}_2$)

• A. $\frac{3M}{4}$
• B. $\frac{3M}{5}$
• C. $\frac{5M}{6}$
• D. $\frac{4M}{3}$

Answer 1

The correct option is A $\frac{3M}{4}$

$H_3\stackrel{+1}{P}{O}_2\to \stackrel{-3}{P}{H}_3+H_3\stackrel{+3}{P}{O}_3$
$H_3\stackrel{+1}{P}{O}_2\to H_3\stackrel{+3}{P}{O}_{3}oxidation$
$(n_f{)}_{\text{oxidation}}=|3-1|\times 1=2$
$H_3\stackrel{+1}{P}{O}_2\to \stackrel{-3}{P}{H}_{3}reduction$
$(n_f{)}_{\text{reduction}}=|-3-1|\times 1=4$
valency factor in disproportion 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 $H_{3}P{O}_2=\frac{\text{Molar mass of}{H}_{3}P{O}_2}{\frac{4}{3}}=\frac{3M}{4}$