Question

$K_{sp}$ of $Pb{I}_2$ is $7.1\times {10}^{-9}$. Amount in $mg$ of ${Pb}^{2+}$ per $mL$ in $0.05M$ $KI$ saturated with $Pb{I}_2$ is $X\times {10}^{-4}mg/mL$. The value of $X$ is

Answer 1

The expression for the solubility product is :

$K_{sp}=\left[P{b}^{2+}\right][I^{-}{]}^2$

$7.1\times {10}^{-9}=\left[P{b}^{2+}\right][0.05{]}^2$

$\left[P{b}^{2+}\right]=2.84\times {10}^{-6}M$.

Convert the unit from mole per liter to mg/mL.

$Molarity\times Molarmass=2.84\times {10}^{-6}\frac{mol}{L}\times 207\frac{g}{mol}$

$=5.9\times {10}^{-4}g/L=5.9\times {10}^{-4}mg/mL=X\times {10}^{-4}mg/mL$

Hence, $X=6$