The solubility product of $P b l_{2}$ is $8.0 \times 10^{- 9} .$ The solubility of lead iodide in 0.1 molar solution of lead nitrate is

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$\sqrt{2 \times 10^{- 8}} M$

$\sqrt{2 \times 10^{- 4}} M$

$2 \times \sqrt{10^{- 8}} M$

$4 \times \sqrt{10^{- 8}} M$

answer is B.

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Detailed Solution

given ${[K_{s p}]}_{P b l_{2}} = 8 \times 10^{- 8}$

To calculate solubility of $P b l_{2}$ in 0.1M solution of $P b {(N O_{3})}_{2} .$

$(i) P b {(N O_{3})}_{2} \to P b^{2 +} (a q) + 2 N {O_{3}}^{-} (a q)$

(ii) $\begin{array}{l} 0. \bar{1} M 0. \bar{1} M 0. \bar{2} M \\ P b l_{2} (s) ⇌ P b^{2 +} (a q) + 2 l^{-} (a q) \\ s 2 s \end{array}$

$\begin{array}{l} ∴ [P b^{2 +}] = S + 0.1 \approx 0.1 \\ ∵ S < < 0.1 \end{array}$

Now, $K_{s p} = 8 \times 10^{- 9}$

$[P b^{2}] {[I^{-}]}^{2} = 8 \times 10^{- 9}$

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

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

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