VII. Precipitation Equilibria
VII-1. Introduction and Background VII-2. Sample Problem VII-3. List of Problems VII-4. Simple Problems VII-5. Advanced Problems


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 25oC?
Unbalance reaction: Al(OH)3 (s) <--> Al3+(aq) + OH-(aq)         Ksp = 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:   Ksp = [Al3+][OH-]3

3. Q = [Al3+][OH-]3

Since the Al3+ concentration is zero, Q is zero, Q < Keq, 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:

[ ]o0.01.0x10-7
Delta[ ]+x M+3x M
[ ]eq(0.0 + x) M(1.0x10-7 + 3x) M

Where [ ]o are the pre-equilibrium concentrations, Delta[ ] are the changes in concentrations, and [ ]eq are the equilibrium concentrations.

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

Ksp = [Al3+][OH-]3

2x10-32 = (x)(1.0x10-7 + 3x)3

Working this problem exactly would be very complicated. Since Ksp 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 = Keq, so the system is at equilibrium.

Equilibrium Practice Problems
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