# Solubility from Solubility Product Example Problem

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

## Problem

• The solubility product of silver chloride (AgCl) is 1.6 x 10-10 at 25 °C.
• The solubility product of barium fluoride (BaF2) is 2 x 10-6 at 25 °C.

Calculate the solubility of both compounds.

## Solutions

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 the equilibrium of the dissociation reaction.

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-]

To find these concentrations, remember this formula for solubility product:

Ksp = [A]c[B]d

So, 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

BaF2

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