|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.
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:
|[ ]||+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:
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.