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

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

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

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

$\Rightarrow x =141$