Introduction

The following problem is an example of determining the equilibrium concentrations in a precipitation equilibrium problem. For a reminder of how to begin, see the document on the general solution of equilibria problems.

Sample Problem

What are the equilibrium concentrations of Al³⁺ and OH^- when solid Al(OH)₃ is added to water at 25^oC?
Unbalance reaction: Al(OH)_{3 (s)} Al³⁺_(aq) + OH^-_(aq)         K_sp = 2x10^-32

1. The pre-equilibrium concentration of Al³⁺ is zero, and [OH^-] = 1.0x10^-7 (for pure water).

2. The balanced chemical equilibrium is:   Al(OH)_{3 (s)} Al³⁺_(aq) + 3 OH^-_(aq)

and the equilibrium constant expression is:   K_sp = [Al³⁺][OH^-]³

3. Q = [Al³⁺][OH^-]³

Since the Al³⁺ concentration is zero, Q is zero, Q < K_eq, and the reaction will proceed in the forward direction to reach equilibrium.

4. For each mol of Al(OH)_{3 (s)} that dissolves, 1 mole of Al³⁺_(aq), and 3 moles of OH^-_(aq) will form. The pre-equilibrium, changes, and equilibrium concentrations are given in the following table:

	Al3+(aq)	OH-(aq)
[ ]o	0.0	1.0x10-7
[ ]	+x M	+3x M
[ ]eq	(0.0 + x) M	(1.0x10-7 + 3x) M

Where [ ]_o are the pre-equilibrium concentrations, [ ] are the changes in concentrations, and [ ]_eq are the equilibrium concentrations.

5. We can now calculate the equilibrium concentrations using the equilibrium constant expression:

K_sp = [Al³⁺][OH^-]³

2x10^-32 = (x)(1.0x10^-7 + 3x)³

Working this problem exactly would be very complicated. Since K_sp is such a small number, we can predict that the dissolution of Al(OH)_{3 (s)} will be very small, and that 3x < < 1.0x10^-7. We can therefore neglect the 3x, and the problem becomes:

2x10^-32 = (x)(1.0x10^-7)³

x = 2x10^-11 (which is < < 1.0x10^-7)

[Al3+] = x = 2x10-11 M

[OH-] = 1.0x10-7 M

Check results: Q = (2x10^-11)(1.0x10^-7) =2x10^-32. Q = K_eq, so the system is at equilibrium.