Question

The solubility product of $PbI _{2}$ 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

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

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



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

$= s +0.1$

$\simeq 0.1$

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

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

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

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

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

$\Rightarrow x =141$