This example problem demonstrates how to determine the solubility of an ionic solid in water from a substance's solubility product.
The solubility product of AgCl is 1.6 x 10-10 at 25 °C.
The solubility product of BaF2 is 2 x 10-6 at 25 °C.
Calculate the solubility of both compounds.
The key to solving solubility problems is to properly set up your dissociation reactions and define solubility. Solubility is the amount of reagent that will be consumed to saturate the solution, or reach equilibrium of the dissociation reaction.
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-]
To find these concentrations, remember that
Ksp = [A]c[B]d
for the reaction AB ↔ cA + dB
Ksp = [Ag+][Cl-]
since [Ag+] = [Cl-]
Ksp = [Ag+]2 = 1.6 x 10-10
[Ag+] = (1.6 x 10-10)½
[Ag+] = 1.26 x 10-5 M
solubility of AgCl = [Ag+]
solubility of AgCl = 1.26 x 10-5 M
The dissociation reaction of BaF2 in water is
BaF2 (s) ↔ Ba+ (aq) + 2 F- (aq)
The solubility is equal to the concentration of the Ba ions in solution.
For every mole of Ba+ ions formed, 2 moles of F- ions are produced, therefore
[F-] = 2 [Ba+]
Ksp = [Ba+][F-]2
Ksp = [Ba+](2[Ba+])2
Ksp = 4[Ba+]3
2 x 10-6 = 4[Ba+]3
[Ba+]3 = ¼(2 x 10-6)
[Ba+]3 = 5 x 10-7
[Ba+] = (5 x 10-7)1/3
[Ba+] = 7.94 x 10-3 M
solubility of BaF2 = [Ba+]
solubility of BaF2 = 7.94 x 10-3 M
The solubility of silver chloride, AgCl, is 1.26 x 10-5 M at 25 °C.
The solubility of barium fluoride, BaF2, is 3.14 x 10-3 M at 25 °C.