Question

The solubility of $Pb{F}_2$ in water at ${25}^{o}C$ is $\sim {10}^{-3}M$. What is its solubility in $0.05M$ $NaF$ solution?
Assume the latter to be fully ionised.

• A. $1.6\times {10}^{-6}M$
• B. $1.2\times {10}^{-6}M$
• C. $1.2\times {10}^{-5}M$
• D. $1.6\times {10}^{-4}M$

Answer 1

The correct option is A $1.6\times {10}^{-6}M$

Solubility of $Pb{F}_2\approx {10}^{-3}M$

$\therefore$ $K_{sp}=4S^3=4\times {10}^{-9}$

In $0.05M$ $NaF$ we have $0.05M$ of $F^{-}$ ion contributed by $NaF$. IF the solubility of $Pb{F}_2$ in this solution is $S$ $M$, then
total $\left[F^{-}\right]=[2S+0.05]M$

$\therefore$ $S{[2S+0.05]}^2=4\times {10}^{-9}$

Assuming $2S<<0.05$
$S\times 25\times {10}^{-4}=4\times {10}^{-9}$

$\therefore$ $S=0.16\times {10}^{-5}M$ $\Rightarrow$ $1.6\times {10}^{-6}M$

We observe that our approximation that $2S<<0.05$ is justified.