This example problem demonstrates how to determine the solubility product of an ionic solid in water from a substance's solubility.
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, Ksp, of both compounds.
The key to solving solubility problems is to properly set up your dissociation reactions and define solubility.
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-]
Ksp = [Ag+][Cl-]
Ksp = (1.26 x 10-5)(1.26 x 10-5)
Ksp = 1.6 x 10-10
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+]
Ksp = [Ba+][F-]2
Ksp = ([Ba+])(2 [Ba+])2
Ksp = 4[Ba+]3
Ksp = 4(7.94 x 10-3 M)3
Ksp = 4(5 x 10-7)
Ksp = 2 x 10-6
The solubility product of AgCl is 1.6 x 10-10.
The solubility product of BaF2 is 2 x 10-6.