Question

An impure sample of $P_2{O}_3$ weighing $1.5$ g was dissolved in water and warmed till whole of $P_2{O}_3$ undergoes disproportionation in to $P{H}_3$ and $H_{3}P{O}_4$. The solution was then boiled to remove $P{H}_3$ from it and finally cooled to room temperature and made $100$ mL. $10$ mL of this solution was mixed with $1O$ mL of $0.6$ M $NaOH$. Now, $\frac{1}{3}$ of this solution required $3.6$ mL of $0.05$ M $H_{2}S{O}_4$ for back titration. What is $\%$ purity of $P_2{O}_3$ sample ?

• A. $83\%$
• B. $65\%$
• C. $80\%$
• D. None of these

Answer 1

The correct option is C $80\%$

$12e+(P^{3+}{)}_2⟶2P^{3-}$
$\frac{(P^{3+}{)}_2⟶2P^{5+}+4e]\times 3}{4(P^{3+}{)}_2⟶2P^{3-}+6P^{5+}}$
or, $4P_2{O}_3⟶2P{H}_3+6H_{3}P{O}_4$
or, $4P_2{O}_3+12H_{2}O⟶2P{H}_3+6H_{3}P{O}_4$
Now, Meq. of $H_{3}P{O}_4$ formed in IO mL$=$ Meq. of $NaOH$ added Meq.of $H_{2}S{O}_4$ needed to neutralise whole $NaOH$ left $=10\times 0.6-3\times [3.6\times 0.05\times 2]$ $=6-1.08=4.92$
$\therefore$ Moles of $H_{3}P{O}_4$ formed in 100 mL $=4.92\times 10=49.2$
$\therefore$ Moles of $H_{3}P{O}_4$ formed $=\frac{49.2}{1000\times 3}=0.0164$
$\therefore$ Moles of $P_2{O}_3=\frac{4\times 0.0164}{6}=0.0109$
$\therefore$ Mass of $P_2{O}_3=0.0109\times 110=1.20g$
$\therefore \%$ of $P_2{O}_3=\frac{1.2}{1.5}\times 100=80\%$