Question

For each of the following pairs of solids, determine which solid has the smallest molar solubility

1. $$\begin{gathered} \mathbf{}\mathit{C}\mathit{a}{\mathit{F}}_{\mathbf{2}}\left(s\right)\mathbf{,}{\mathit{K}}_{\mathbf{s}\mathbf{p}}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{}\mathit{o}\mathit{r} \\ \mathit{B}\mathit{a}{\mathit{F}}_{\mathbf{2}}\left(s\right)\mathbf{,}{\mathit{K}}_{\mathbf{s}\mathbf{p}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}} \end{gathered}$$
2. $$\begin{gathered} \mathit{C}{\mathit{a}}_{\mathbf{3}}{\left(P{O}_4\right)}_{\mathbf{2}}\left(s\right)\mathbf{,}\mathbf{}{\mathit{K}}_{\mathbf{s}\mathbf{p}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{32}}\mathbf{}\mathit{o} \\ \mathit{F}\mathit{e}\mathit{P}{\mathit{O}}_{\mathbf{4}}\left(s\right)\mathbf{,}{\mathit{K}}_{\mathbf{s}\mathbf{p}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{22}} \end{gathered}$$
   

Short answer

CaF₂ has the smallest molar solubility.

FePO₄ has the smallest molar solubility.

Step-by-step solution

Definition of molar solubility

The number of moles of the solute that can be dissolved per liter of solution before the solution becomes saturated is defined as molar solubility.

Calculations to find the smallest molar solubility

The expression for the solubility product constant for CaF₂ is

$$\begin{gathered} K_{sp}=\left(s\right){\left(2s\right)}^2 \\ 4.0\times {10}^{-11}=4s^3 \\ s=1.82\times {10}^{-4}mol/L \end{gathered}$$

The expression for the solubility product constant for BaF₂ is

$$\begin{gathered} K_{sp}=\left(s\right){\left(2s\right)}^2 \\ 2.4\times {10}^{-5}=4s^3 \\ s=1.82\times {10}^{-2}mol/L \end{gathered}$$

CaF₂has smallest molar solubility

The expression for the solubility product constant for $C{a}_3{\left(P{O}_4\right)}_2$is

$$\begin{gathered} K_{sp}={\left(3s\right)}^3{\left(2s\right)}^2 \\ 1.3\times {10}^{-32}=108s^5 \\ s=1.64\times {10}^{-7}mol/L \end{gathered}$$

The expression for the solubility product constant for BaF_2is

_{$$\begin{gathered} K_{sp}=\left(s\right)\left(s\right) \\ 1.0\times {10}^{-22}=s^2 \\ s=1.0\times {10}^{-11}mol/L \end{gathered}$$}

FePO₄has the smallest molar solubility.