Calcium fluoride dissolves only partially in water; it is an equilibrium reaction \[\mathrm{CaF}_2 (\mathrm{solid}) \rightleftharpoons \mathrm{Ca}^{2+}+2\mathrm{F}^{-}\] The solubility product \(K_s\) is equal to the product of the concentrations of the dissolved ions, with each concentration raised to a power with exponent equal to the stoichiometric coefficient of the ion involved in the ionization equation. In this case, we have: \[K_s=\left[\mathrm{Ca}^{2+}\right]\cdot \left[\mathrm{F}^{-}\right]^2\] Suppose that \(x\) mole per liter of calcium fluoride dissolves, then the following holds: \[\left[\mathrm{Ca}^{2+}\right]=x, \quad \left[\mathrm{F}^{-}\right]=2x\] and \[K_s=x\cdot (2x)^2=4x^3\] This shows that the solubility product \(K_s\) in this particular case is a third-degree power function in the amount of moles of solids that dissolves per liter. Given the solubility product \(K_s\) you can calculate with this formula how much substance dissolves per liter. Conversely, you can calculate the solubility product if you know the ion concentration in the saturated aqueous solution.

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Pay attention to the difference between the notions of solubility product and solubility: The latter term is about a physical property of a substance, namely, the degree to which the substance can dissolve (in terms of ion concentration in a saturated solution), while the first term is an equilibrium constant of a chemical reaction.