Solubility Product From Solubility Example Problem

This example problem demonstrates how to determine the solubility product of an ionic solid in water from a substance's solubility.

Problem:

The solubility of silver chloride, AgCl, is 1.26 x 10^-5 M at 25 °C.
The solubility of barium fluoride, BaF₂, is 3.15 x 10^-3 M at 25 °C.

Calculate the solubility product, K_sp, of both compounds.

Solution:

The key to solving solubility problems is to properly set up your dissociation reactions and define solubility.

AgCl

The dissociation reaction of AgCl in water is

AgCl (s) ↔ Ag⁺ (aq) + Cl^- (aq)

For this reaction, each mole of AgCl that dissolves produces 1 mole of both Ag⁺ and Cl^-. The solubility would then equal the concentration of either the Ag or Cl ions.

solubility = [Ag⁺] = [Cl^-]
1.26 x 10^-5 M = [Ag⁺] = [Cl^-]

K_sp = [Ag⁺][Cl^-]
K_sp = (1.26 x 10^-5)(1.26 x 10^-5)
K_sp = 1.6 x 10^-10

BaF₂

The dissociation reaction of BaF₂ in water is

BaF₂ (s) ↔ Ba⁺ (aq) + 2 F^- (aq)

This reaction shows that for every mole of BaF2 that dissolves, 1 mole of Ba⁺ and 2 moles of F^- are formed. The solubility is equal to the concentration of the Ba ions in solution.

solubility = [Ba⁺] = 7.94 x 10^-3 M
[F^-] = 2 [Ba⁺]

K_sp = [Ba⁺][F^-]²
K_sp = ([Ba⁺])(2 [Ba⁺])²
K_sp = 4[Ba⁺]³
K_sp = 4(7.94 x 10^-3 M)³
K_sp = 4(5 x 10^-7)
K_sp = 2 x 10^-6

Answer:

The solubility product of AgCl is 1.6 x 10^-10.
The solubility product of BaF₂ is 2 x 10^-6.