Questions: Complete the following solubility constant expression for CaF₂. Ksp=

Solution Steps

The solubility product constant, \( K_{\mathrm{sp}} \), is an equilibrium constant for the dissolution of a sparingly soluble ionic compound. For calcium fluoride, \(\mathrm{CaF}_{2}\), the dissolution reaction in water can be written as:

\[ \mathrm{CaF}_{2}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + 2\mathrm{F}^{-}(aq) \]

The solubility product expression is derived from the equilibrium concentrations of the ions produced in the dissolution reaction. For the reaction above, the expression for \( K_{\mathrm{sp}} \) is:

\[ K_{\mathrm{sp}} = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \]

Here, \([\mathrm{Ca}^{2+}]\) is the molar concentration of calcium ions, and \([\mathrm{F}^{-}]\) is the molar concentration of fluoride ions. The fluoride concentration is squared because two fluoride ions are produced for each formula unit of \(\mathrm{CaF}_{2}\) that dissolves.

Final Answer