Question

The${\mathbf{K}}_{\mathbf{sp}}$ for${\mathbf{PbI}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$ is $\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}$. Calculate the solubility of${\mathbf{PbI}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$ in 0.048 M NaI.

Short answer

The solubility of${\mathrm{PbI}}_2$ in 0.048 M NaI solution is $6.07\times {10}^{-6}\mathrm{mol}/\mathrm{L}$.

Step-by-step solution

Definition of solubility

The maximum amount of a substance that will dissolve in a given amount of solvent at a specified temperature is defined as solubility

Calculations

The dissolution equilibrium of${\mathrm{PbI}}_2$ is

$\begin{array}{l}{\mathrm{PbI}}_2\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{I}}^{-}\left(\mathrm{aq}\right)\\ {\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{Pb}}^{2+}\right]}^3{\left[{\mathrm{I}}^{-}\right]}^2\end{array}$

Let ‘s’ mol/L represent the molar solubility of ${\mathrm{PbI}}_2$.

$\begin{array}{c}\left[{\mathrm{Pb}}^{2+}\right]=\text{'}\mathrm{s}\text{'}\mathrm{mol}/\mathrm{L}\\ \left[{\mathrm{I}}^{-}\right]=\text{'}2\mathrm{s}\text{'}\mathrm{mol}/\mathrm{L}\\ \left[{\mathrm{I}}^{-}\right]=2\mathrm{s}+0.048\mathrm{M}\\ \simeq 0.048\mathrm{M}\end{array}$

The solubility of${\mathrm{PbI}}_2$ is

$\begin{array}{l}{\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{Pb}}^{2+}\right]}^3{\left[{\mathrm{I}}^{-}\right]}^2\\ 1.4\times {10}^{-8}=\left[\mathrm{s}\right]{\left[0.048\right]}^2\\ \mathrm{s}=6.07\times {10}^{-6}\mathrm{M}\end{array}$