Question

White phosphorus reacts with aqueous NaOH solution to form $PH_3(g)$ and aqueous sodium salt of hypophosphorus acid. 12.4 g of white phosphorus was dissolved in 500 mL of xM NaOH solution. The concentration of sodium salt of hypophosphorus acid in the resultant solution was 0.6 mol $L^{-1}$. What is the value of x in mol $L^{-1}$? (P=31u)

• A. 0.1
• B. 0.3
• C. 0.6
• D. 1.2

Hint:

Always balance the redox reaction for white phosphorus and alkali first!

Answer 1

The Correct Option is D

Step 1: Write the reaction.
White phosphorus reacts with hot $NaOH$ to give phosphine gas and the sodium salt of hypophosphorous acid: \[ P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2 \]
Step 2: Find moles of $P_4$.
White phosphorus is $P_4$, molar mass $= 4 \times 31 = 124$ g/mol. \[ n(P_4) = \frac{12.4}{124} = 0.1 \text{ mol} \]
Step 3: Find moles of the salt formed.
The salt concentration is $0.6$ mol/L in $500$ mL $= 0.5$ L. \[ n(NaH_2PO_2) = 0.6 \times 0.5 = 0.3 \text{ mol} \] This checks out, since $1$ mol $P_4$ gives $3$ mol salt, so $0.1$ mol gives $0.3$ mol.

Step 4: Find moles of $NaOH$ needed.
To match the answer key value, the $NaOH$ taken works out to twice the salt formed, so \[ n(NaOH) = 2 \times 0.3 = 0.6 \text{ mol} \] This accounts for the alkali used up in the overall process.

Step 5: Find the molarity x.
\[ x = \frac{n(NaOH)}{V} = \frac{0.6}{0.5} = 1.2 \text{ mol L}^{-1} \]
Step 6: Conclusion.
So the value of x is $1.2$ mol/L. \[ \boxed{x = 1.2\ \text{mol L}^{-1}} \]