Use Appendix B to determine the${\mathbf{\text{K}}}_{\mathbf{\text{sp}}}$of${\mathbf{\text{CaF}}}_{\mathbf{\text{2}}}$

Solubility: The maximum amount of solute that can be dissolved in the solvent at equilibrium is defined as solubility.

Constant of soluble product: For equilibrium between solids and their respective ions in a solution, the solubility product constant is defined. In general, the term "solubility product" refers to water-equilibrium insoluble or slightly soluble ionic substances.

Considering the given information:

The equation for solubility of $\left({\text{CaF}}_{\text{2}}\right)$is as follows, ${\text{CaF}}_{\text{2}}\text{(s)}\rightleftharpoons {\text{Ca}}^{\text{2+}}{\text{(aq)+2F}}^{\text{-}}\text{(aq)}$

The expression for this reaction ${\text{ΔG}}_{\text{rxn}}^{\text{°}}$is,

The Gibbs free energy equation is as follows:

${\text{ΔG}}_{\text{rxn}}^{\text{°}}\text{=}\sum \text{m}{{\text{ΔG}}_{\text{f}}}^{\text{°}}\text{(Products)-}\sum \text{n}{{\text{ΔG}}_{\text{f}}}^{\text{°}}$(Reactants)

The reaction's free energy change is calculated as follows:

$\begin{array}{l}{\text{ΔG}}_{\text{rxn}}^{\text{°}}\text{=}[\left({\text{1molCa}}^{\text{2+}}\right)({\text{ΔG}}_{\text{f}}^{\text{o}}\text{of}{\text{Ca}}^{\text{2+}})\text{+}\left({\text{2molF}}^{\text{2-}}\right)({\text{ΔG}}_{\text{f}}^{\text{o}}\text{of}{\text{F}}^{\text{2-}})][\left({\text{1molCaF}}_{\text{2}}\right)({\text{ΔG}}_{\text{f}}^{\text{o}}\text{of}{\text{CaF}}_{\text{2}})]\\ {\text{ΔG}}_{\text{rxn}}^{\text{0}}\text{=}\left[\left({\text{1molCa}}^{\text{2+}}\right)\text{(-553.04kJ/mol)+}\left({\text{2molF}}^{\text{2-}}\right)\text{(-276.5kJ/mol)}\right]\text{[(1molCuF2)(-1162kJ/mol)]}\\ {\text{ΔG}}_{\text{rxn}}^{\text{°}}\text{=55.69kJ}\end{array}$

The value of ${\text{ΔG}}_{\text{rxn}}^{\text{0}}$is 55.69kJ and these values of ${\text{ΔG}}_{\text{f}}^{\text{o}}$are referred from the Appendix B.

K, the equilibrium constant, is calculated.

We are aware of the equilibrium equation.

${\text{ΔG=ΔG}}^{\text{o}}\text{+RTln(K)}$

Rearrange the equation above,

$\begin{array}{c}\mathrm{lnK}=-\frac{{\mathrm{\Delta G}}^{\mathrm{a}}}{\mathrm{RT}}\\ \mathrm{=}\left(\frac{55.96\mathrm{kJ}/\mathrm{mol}}{-(8.314\mathrm{J}/\mathrm{mol}\times \mathrm{K})(298\mathrm{K})}\right)\left(\frac{{\mathrm{10}}^{\mathrm{3}}\mathrm{J}}{1\mathrm{kJ}}\right)\mathrm{lnK}=-22.586629\end{array}$

$\begin{array}{l}\text{Hence,}\\ {\text{K=e}}^{\text{-22.586629}}\\ {\text{K=1.5514995×10}}^{\text{-10}}\text{(or)}\\ {\text{K=1.55×10}}^{\text{-10}}\end{array}$

Therefore, the required solubility product${\text{K}}_{\text{sp}}$ of Calcium fluoride$\left({\text{CaF}}_{\text{2}}\right)$is$\text{K=}\underset{\_}{{\text{1.55×10}}^{\text{-10}}}$.