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# Solubility Product From Solubility Example Problem

## Calculate Solubility Product of an Ionic Solid From Solubility

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, Ksp, 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-]

Ksp = [Ag+][Cl-]
Ksp = (1.26 x 10-5)(1.26 x 10-5)
Ksp = 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+]

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.