a
Solubility of \(Pb{F_2} \approx {10^{ - 3}}\,M\)

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

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.