Question

Calculate the solubility (in $\mathbf{mol}.{\mathbf{L}}^{-1}$) of ${\mathbf{CaF}}_2\mathbf{\left(}\mathbf{s}\mathbf{\right)}$ at 25°C in a 0.040 M aqueous solution of NaF.

Short answer

Solubility is $2.43\times {10}^{-8}\mathrm{mol}/\mathrm{L}$.

Step-by-step solution

Ionic compounds

Ionic compounds undergo complete ionization in their aqueous solutions.

For example, when NaF is dissolved in water, it ionizes into sodium and fluoride ions, with each ion’s concentration equal to NaF concentration.

Common ion effect

In the case of sparingly soluble salts, their dissociation decreases further by adding a solution that contains a common ion to it.

$\begin{array}{l}\mathrm{NaF}\left(\mathrm{aq}\right)\to {\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+{\mathrm{F}}^{-}\left(\mathrm{aq}\right)\\ 0.040\mathrm{M}0.040\mathrm{M}0.040\mathrm{M}\end{array}$

Explanation

Solubility of ${\mathrm{CaF}}_2$ is as shown below:

$\begin{array}{l}{\mathrm{CaF}}_2\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Ca}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{F}}^{-}\left(\mathrm{aq}\right)\\ \mathrm{s}(2\mathrm{s}+0.040)\\ {\mathrm{K}}_{\mathrm{sp}}=\mathrm{s}\times {(2\mathrm{s}+0.040)}^2\\ 3.9\times {10}^{-11}=\mathrm{s}\times {\left(0.040\right)}^2\\ \mathrm{s}=\frac{3.9\times {10}^{-11}}{{\left(0.040\right)}^2}\\ =2.43\times {10}^{-8}\mathrm{mol}/\mathrm{L}\end{array}$