The solubility product of PbI _ 2 is 8 . 0 × 10 ^ - 9 . The...

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

To calculate : solubility of ${PbI}_{2} in 0 . 1M$ solution of $Pb{\left({NO}_{2}\right)}_{2}$

\left(I\right)Pb{\left({NO}_{3}\right)}_{2} \to {Pb}_{( aq )}^{+ 2} + 2{NO}_{3}^{-}\left(aq\right)$</p><p>$\begin{cases}0 . 1M & - & - \\ - & 0 . 1M & 0 . 2M\end{cases}$</p><p>\left(II\right){PbI}_{2}\left(s\right)$ ⇌ {Pb}^{+ 2}$\left(aq\right)$ + 2{I}^{-}\left(aq\right)$</p><p>$\begin{cases}s & 2s\end{cases}

$[ {Pb}^{+ 2} ] = s + 0 . 1$

$≃ 0 . 1$

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

$⇒ 8 \times {10}^{- 9} = 0 . 1 \times {\left(2s\right)}$^{2}$</p><p>$⇒ 8 \times {10}^{- 8} = 4{s}^{2} ⇒ s = \sqrt{2} \times {10}^{- 4}$</p><p>$⇒ S = 141 \times {10}^{- 6}M$</p><p>$⇒ x = 141$</p>$