Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\mathrm{P}\mathrm{b}\mathrm{C}{\mathrm{l}}_2\rightleftharpoons \mathrm{P}{\mathrm{b}}^{2+}+2\mathrm{C}\mathrm{l}$
${\mathrm{K}}_{\mathrm{s}\mathrm{p}}=\left[\mathrm{P}{\mathrm{b}}^{2+}\right][\mathrm{C}{\mathrm{l}}^{-}{]}^2$
$=\left(0.01\right)(2\times 0.01{)}^2$
${\mathrm{K}}_{\mathrm{s}\mathrm{p}}=4\times 10^{-6}$
$\mathrm{P}\mathrm{b}\mathrm{C}{\mathrm{l}}_2\rightleftharpoons \underset{\underset{\mathrm{S}}{0}}{\mathrm{P}{\mathrm{b}}^{2+}\left(\mathrm{a}\mathrm{q}\right)}+\underset{\underset{(2\mathrm{S}+0.1)}{0.1}}{2\mathrm{C}{\mathrm{l}}^{-}\left(\mathrm{a}\mathrm{q}\right)}$
${\text{K}}_{\text{sp}}\text{=}[{\text{Pb}}^{\text{2+}}]{{\text{[Cl}}^{-}\text{]}}^{\text{2}}$
$\left[0.1+2\text{S}\approx 0.1\right]$
$4\times 10^{-6}=\mathrm{S}\times (0.1{)}^2$
$\mathrm{S}=4\times 10^{-4}\mathrm{M}$
Maximum concentration of$\mathrm{P}\mathrm{b}\mathrm{C}{\mathrm{l}}_2$in 0.1 M NaCl is$4\times 10^{-4}\mathrm{M}$.