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

In $0.05\,M\,NaF$ we have $0.05\,M$ of $F^-$ ion contributed by $NaF.$ If the solubility of $PbF_2$

in this solution is $S\,M$, then

total $[{F^ - }] = [2S + 0.05]\,M$

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

Assuming $2S <  < 0.05,$

$S \times 25 \times {10^{ - 4}} = 4 \times {10^{ - 9}}$

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

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