The Common Ion Effect
In this section we will consider the acid-base properties of a solution with two dissolved solutes, that contain one ion in common. The presence of a common ion suppresses the ionization of a weak acid or a weak base. The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. The common ion effect plays an important role in determining the pH of a solution and the solubility of a slightly soluble salt, which we take up later. Keep in mind that despite its distinctive name the common ion effect is nothing more than a special case of Le Chatelier's principle.
Example 1
Calculate the percent ionization of (a) a 0.10 M solution of acetic acid, HC2H3O2, and (b) a 0.10 M solution of acetic acid, which is also 0.10 M in sodium acetate, NaC2H3O2.
a) First we construct our usual table.
| Concentrations(M) | HC2H3O2(aq) | H2O(l) | <==> | H3O+(aq) | C2H3O2-(aq) | ||
| Starting | 0.10 | 0 | 0 | ||||
| Change | -x | +x | +x | ||||
| Equilibrium | 0.10 - x | x | x |
The next step is to substitute the equilibrium concentrations into the equilibrium constant expression.
[H3O+][C2H3O2-] (x)(x)
Ka = -------------- = 1.8 x 10-5 --------
[HC2H3O2] 0.10 - x
Assume that we can neglect x compared to 0.10.
x2 = 1.8 x 10-6
x = [H3O+] = 1.3 x 10-3 (negligible compared to 0.10)
[H3O+] 1.3 x 10-3
% ionization = ------- x 100% = --------- x 100% = 1.3%
0.10 0.10
(b) Construct the usual table.
| Concentrations(M) | HC2H3O2(aq) | H2O(l) | <==> | H3O+(aq) | C2H3O2-(aq) | ||
| Starting | 0.10 | 0 | 0.10 | ||||
| Change | -x | +x | +x | ||||
| Equilibrium | 0.10 - x | x | 0.10 + x |
Next substitute the equilibrium concentrations into the equilibrium constant expression.
[H3O+][C2H3O2-] (x)(0.10 + x)
Ka = -------------- = 1.8 x 10-5 -------------
[HC2H3O2] 0.10 - x
Assume that we can neglect x compared to 0.10.
x = 1.8 x 10-5 = [H3O+] = 1.8 x 10-5 (negligible compared to 0.10)
[H3O+] 1.8 x 10-5
% ionization = ------- x 100% = --------- x 100% = 0.018%
0.10 0.10
Notice how the common ion (acetate) has suppressed the ionization.
Example 2
What is the pH of a 0.10 M solution of ammonia, NH3, which is also 0.20 M in ammonium nitrate, NH4NO3?
First we construct our usual table.
| Concentrations(M) | NH3(aq) | H2O(l) | <==> | NH4+(aq) | OH-(aq) | ||
| Starting | 0.10 | 0.20 | 0 | ||||
| Change | -x | +x | +x | ||||
| Equilibrium | 0.10 - x | 0.20 + x | x |
The next step is to substitute the equilibrium concentrations into the equilibrium constant expression.
[NH4+][OH-] (0.20 + x)(x)
Kb = ---------- = 1.8 x 10-5 = -----------
[NH3] 0.10 - x
Assume that we can neglect x compared to 0.10 and 0.20.
x = 9.0 x 10-6
x = [OH-] = 9.0 x 10-6 (negligible compared to 0.10 and 0.20)
pOH = -log[OH-] = -log(9.0 x 10-6) = 5.05
pH = 14.00 - 5.05 = 8.95