Question

The solubility product of $Pb{I}_2$ is $8.0\times 10^{-9}$. The solubility of lead iodide in 0.1 molar solution of lead nitrate is $x\times 10^{-6}mol/L$. The value of X is (Rounded off to the nearest integer)__________.
[Given : $\sqrt{2}=1.41$]

Answer 1

Answer: 141

Given : ${\left[K_{sp}\right]}_{Pb{I}_2}=8\times 10^{-9}$

To calculate : solubility of $Pb{I}_2$ in $0.1M$ sol of $Pb{\left(N{O}_3\right)}_2$



(II) $Pb{I}_2(s)\rightleftharpoons \underset{s}{P{b}^{+2}}(aq)+\underset{2s}{2I^{-}}(aq)$

$=s+0.1$

$\simeq 0.1$

Now $:K_{sp}=8\times 10^{-9}=\left[P{b}^{+2}\right][I{]}^2$

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

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

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

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

$\Rightarrow x=141$