Question

White phosphorus reacts with aqueous NaOH to form \(PH_3(g)\) and sodium hypophosphite. When 6.2g of white phosphorus reacted with 500 mL of xM NaOH solution, the concentration of sodium hypophosphite in the resultant solution was \(0.3 \text{ mol L}^{-1}\). What are x (in M) and weight (in g) of \(PH_3\) formed respectively? (\(P=31\) u; \(H=1\) u; \(O=16\) u)

• A. 0.6, 1.7
• B. 0.3, 3.4
• C. 0.3, 1.7
• D. 0.6, 3.4

Hint:

Disproportionation reactions require careful balancing of both atoms and charges; always verify the coefficients.

Answer 1

The Correct Option is C

Concept: White phosphorus (\(P_4\)) undergoes a disproportionation reaction in the presence of an aqueous alkali like NaOH to produce phosphine gas (\(PH_3\)) and sodium hypophosphite (\(NaH_2PO_2\)).

Step 1: Write the balanced chemical equation.
The balanced equation for the reaction is: \[ P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2 \]

Step 2: Calculate the moles of reactant and product.
Given the mass of white phosphorus (\(P_4\)) is 6.2 g. Molar mass of \(P_4 = 4 \times 31 = 124 \text{ g/mol}\). \[ \text{Moles of } P_4 = \frac{6.2}{124} = 0.05 \text{ mol} \] From stoichiometry, 1 mole of \(P_4\) produces 3 moles of \(NaH_2PO_2\). \[ \text{Moles of } NaH_2PO_2 = 0.05 \times 3 = 0.15 \text{ mol} \]

Step 3: Determine the Molarity (x) and weight of \(PH_3\).
Concentration of \(NaH_2PO_2 = 0.3 \text{ M}\) in 500 mL (0.5 L) gives \(0.15 \text{ mol}\). This confirms our stoichiometric calculation. Stoichiometry shows 3 moles of NaOH are required for 1 mole of \(P_4\). \[ \text{Moles of NaOH} = 3 \times 0.05 = 0.15 \text{ mol} \] \[ x = \frac{\text{moles}}{\text{Volume}} = \frac{0.15}{0.5} = 0.3 \text{ M} \] \[ \text{Moles of } PH_3 = 0.05 \text{ mol} \] \[ \text{Weight of } PH_3 = 0.05 \times 34 = 1.7 \text{ g} \] \[ \boxed{0.3, 1.7} \]