Question

Ionic solids dissolve in water and break up into their ions. However, some ionic solids only partially dissolve, leaving a significant amount of solid undissolved. In some cases, the amount that dissolves is very small and is almost zero.
A mathematical formula that indicates the extent to which an ionic solid dissolves in water is called the solubility product constant. The solubility product constant, commonly referred to as Ksp, indicates the extent to which an ionic solid dissolves.
Which of the following is the correct solubility product constant for the reaction shown below?
CaF₂ ⇌ Ca²⁺(aq) + 2F⁻
A. Ksp = [Ca²⁺][F⁻]
B. Ksp = [Ca²⁺]2[F⁻]
C. Ksp = [Ca²⁺]2[F⁻]²
D. Ksp = [Ca²⁺][F⁻]²

Answer 1

Answer: The expression of $K_{sp}$ for calcium fluoride is $K_{sp}=[Ca^{2+}][2F^-]^2$

Explanation:

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The given chemical equation follows:

$CaF_2\rightleftharpoons Ca^{2+}(aq.)+2F^-(aq.)$

1 mole of calcium fluoride produces 1 mole of calcium ions and 2 moles of fluorine ions.

The expression of $K_{sp}$ for above equation follows:

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

Hence, the expression of $K_{sp}$ for calcium fluoride is $K_{sp}=[Ca^{2+}][2F^-]^2$