Question

Find the equivalent weight of $P_4$ in the following reaction:

$P_4+NaOH\to Na{H}_{2}P{O}_2+P{H}_3$

• A. 31/4
• B. 31/3
• C. 31/2
• D. 31 4/3

Answer 1

The correct option is D 31 4/3

Correct answer is option D.

Equivalent weight$=\frac{Mol.Wt}{Valence.factor}$

Given reaction:

$P_4+NaOH\to Na{H}_{2}P{O}_2+P{H}_3$

In this reaction $P_4$ is undergoing both addition and reduction reaction.

Therefore the equivalent mass would be equal to equivalent mass of reduction reaction + the equivalent weight of the oxidation reaction.

For oxidation reaction equivalent wt of $P_4$ $=4\times \frac{31}{4}=31$

For reduction reaction equivalent wt of $P_4$$=4\times \frac{31}{12}=\frac{31}{3}$

Adding both, we get:

Equivalent weight of $P_4$ $=31+\frac{31}{3}=31\times \frac{4}{3}$.

Correct answer is option D.