Question

Calculate the solubility of sparingly soluble salt $Pb{I}_2$ in water at ${25}^{\circ }C$ which is $90$% dissociated.
$K_{sp}\left(Pb{I}_2\right)=9\times {10}^{-9}$ at ${25}^{\circ }C$

(Take $\sqrt[3]{33.08}=1.45$)

• A. $0.145\times {10}^{-6}mol{L}^{-1}$
• B. $1.45\times {10}^{-4}mol{L}^{-1}$
• C. $1.45\times {10}^{-5}mol{L}^{-1}$
• D. $1.45\times {10}^{-3}mol{L}^{-1}$

Answer 1

The correct option is D $1.45\times {10}^{-3}mol{L}^{-1}$

Lets consider solid $Pb{I}_2$ is in contact with its saturated aqueous solution.The equilibrium is established between undissolved $Pb{I}_2$ and its ions.
Since $Pb{I}_2$ is $90$% dissociated, $\left[P{b}^{2+}\right]=0.9\times s$ and $\left[I^{-}\right]=2\times 0.9\times s=1.8s$

$Pb{I}_2\left(s\right)\rightleftharpoons P{b}^{2+}\left(aq\right)+2I^{-}\left(aq\right)Initial:C00Equilibrium:C-0.9s0.9s1.8s$
$K_{sp}=\left[P{b}^{2+}\right][I^{-}{]}^2$
$9\times {10}^{-9}=\left(0.9s\right)\times (1.8s{)}^2{s}^3=\frac{9\times {10}^{-9}}{0.9\times (1.8{)}^2}=3.08\times {10}^{-9}s=1.45\times {10}^{-3}mol{L}^{-1}$