d
Given : \(\left[ K _{ sp }\right]_{ PbI _{2}}=8 \times 10^{-9}\)

To calculate : solubility of \(PbI _{2}\) in \(0.1 \,M\) sol of \(Pb \left( NO _{3}\right)_{2}\)

\((I)\) \(Pb \left( NO _{3}\right)_{2} \rightarrow Pb _{\text {(aq) }}^{+2}+2 NO _{3}^{-}( aq )\)

\(0.1\,M\quad \quad \quad \quad \quad -\quad \quad \quad \quad \quad -\)

\(-\quad \quad \quad \quad \quad \quad \quad 0.1\,M\quad \quad \quad0.2\,M\)

\((II)\) \(PbI _{2}( s ) \rightleftharpoons Pb ^{+2}( aq )+2 I ^{-}( aq )\)

\(\quad \quad \quad \quad \quad \quad \quad \quad s\quad \quad \quad \quad \quad 2s\)

\(= s +0.1\)

\(\simeq 0.1\)

Now : \(K _{ sp }=8 \times 10^{-9}=\left[ Pb ^{+2}\right][ I^- ]^{2}\)

\(\Rightarrow 8 \times 10^{-9}=0.1 \times(2 s )^{2}\)

\(\Rightarrow 8 \times 10^{-8}=4 s ^{2} \Rightarrow s =\sqrt{2} \times 10^{-4}\)

\(\Rightarrow S =141 \times 10^{-6} \,M\)

\(\Rightarrow x =141\)