To determine the solubility product (\(K_{sp}\)) of calcium fluoride (CaF\(_2\)), we must first consider its dissociation in water. The dissociation equation for CaF\(_2\) is as follows:
\(\text{CaF}_2 (s) \leftrightarrow \text{Ca}^{2+} (aq) + 2\text{F}^- (aq)\)
Given that the solubility of CaF\(_2\) is \(s\) moles per liter, this indicates the concentration of CaF\(_2\) that dissolves to reach the equilibrium. Therefore, at equilibrium:
The solubility product (\(K_{sp}\)) expression for CaF\(_2\) is:
\(K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2\)
Substituting the equilibrium concentrations into the solubility product expression:
\(K_{sp} = (s)(2s)^2\)
Simplifying the expression:
\(K_{sp} = s \cdot 4s^2 = 4s^3\)
Thus, the solubility product of CaF\(_2\) is \(4s^3\).
Therefore, the correct answer is \(4s^3\).