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

Step 1: Recall the reaction.
White phosphorus disproportionates in NaOH to give phosphine and sodium hypophosphite, \[ P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2 \]
Step 2: Find moles of phosphorus.
Molar mass of $P_4 = 4 \times 31 = 124$ g/mol, so moles $= \dfrac{6.2}{124} = 0.05$ mol.
Step 3: Use the hypophosphite data.
The product solution is $0.3$ M in $500$ mL, giving moles $= 0.3 \times 0.5 = 0.15$ mol of $NaH_2PO_2$. This nicely matches $3 \times 0.05 = 0.15$ predicted by the equation.
Step 4: Find the NaOH molarity x.
The equation needs $3$ moles NaOH per mole $P_4$, so moles NaOH $= 3 \times 0.05 = 0.15$ mol in $0.5$ L, giving $x = \dfrac{0.15}{0.5} = 0.3$ M.
Step 5: Find the mass of phosphine.
One mole $P_4$ gives one mole $PH_3$, so moles $PH_3 = 0.05$. With molar mass $34$, mass $= 0.05 \times 34 = 1.7$ g.
Step 6: Combine the answers.
So $x = 0.3$ M and $PH_3 = 1.7$ g, which is option 3.
\[ \boxed{0.3, \ 1.7} \]