Ksp - Solubility equilibrium

Passage: The solubility of a solid compound such as a salt refers to the quantity of it that must be added to a particular volume of solvent to form a saturated solution. A solubility equilibrium results when a solid or gaseous compound is in chemical equilibrium with a solution of that compound. At equilibrium, the solution is saturated. The concept of solubility equilibrium is based on the assumption that when a solid dissolves in water or another solvent, it dissociates into the smaller, constituent units from which it was formed. Consider the salt calcium fluoride. When it dissociates, it produces three constituent atoms: one calcium atom (molar mass 40 grams/mol), and two fluorine atoms (each of molar mass 19 grams/mol).

Molar solubility is a term that refers to the maximum number of moles of solute that will dissolve in a liter of a solution before it becomes saturated. The molar solubility of calcium fluoride (CaF₂), for example, is 2.0 x 10^-4 moles per liter.

The solubility product constant (K_sp) is an important constant used to calculate molar solubility, and this constant refers to the product of the concentrations of ions in the equilibrium, each raised to the power of its coefficient in the equation.

Question 1: Calculate the K_sp of calcium fluoride.

(a) 1.6 x 10^-11
(b) 3.2 x 10^-11
(c) 4.8 x 10^-11
(d) 6.4 x 10^-11

Explanation: As stated in the text, the molar solubility of calcium fluoride is 2.0 x 10^-4moles per liter.

CaF₂ dissolves and dissociates as follows:

CaF₂ (s) ⇔ Ca²⁺ (aq) + 2 F^- (aq)
K_sp = [Ca²⁺] [F^-]²

The ratio between CaF₂ and Ca²⁺ is 1:1, but that between CaF₂ and F^- is 1:2. Consequently, dissolution of 2.0 x 10^-4 moles per liter of CaF₂ leads to production of 2.0 x 10^-4 moles per liter of Ca²⁺, but 4.0 x 10^-4 moles per liter of F^- in solution.

Insert the appropriate values into the K_sp equation:

K_sp = [Ca²⁺] [F^-]²
= (2.0 x 10^-4) (4.0 x 10^-4)²
= 3.2 x 10^-11