Question

In the disproportionation reaction:
$H_{3}P{O}_2\to P{H}_3+H_{3}P{O}_3$
The equivalent mass of $H_{3}P{O}_2$ is ($m$ = molecular mass of $H_{3}P{O}_2)$.

• A. $\frac{3m}{4}$
• B. $\frac{3m}{5}$
• C. $\frac{5m}{6}$
• D. $\frac{3m}{2}$

Answer 1

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


As can be seen from the above reaction,
Equivalent weight of this reaction can be calculated as molecular mass divided by $n-$ factor.
where, $n-$ factor is the number of moles of electron lost or gained by per mole of the reactant. (during oxidation and reduction respectively).
For,
$H_3\stackrel{+1}{P}{O}_2\to \stackrel{-3}{P}{H}_3$
$\text{Equivalent weight}=\frac{m}{4}$
For,
$H_3\stackrel{+1}{P}{O}_2\to H_3\stackrel{+3}{P}{O}_3$
$\text{Equivalent weight}=\frac{m}{2}$
Hence,
$\text{The equivalent weight of}{H}_{3}P{O}_2=\frac{m}{4}+\frac{m}{2}=\frac{3m}{4}$