AP Chemistry · Unit 7 · Topic 7.12
Common-Ion Effect
A salt can dissolve readily enough in pure water but much less in a solution that already contains one of its ions. The equilibrium constant has not changed; the starting denominator-free ion product has.
Identify the shared ion
For \(MX(s)\rightleftharpoons M^+(aq)+X^-(aq)\), adding a soluble sodium salt NaX supplies X⁻ before MX dissolves. Sodium is a spectator in this simplified system. The larger initial X⁻ activity means much less M⁺ can accumulate before the product of ion activities reaches Ksp.
Keep the initial ion in the expression
If the initial common-ion concentration is \(C_X\) and s mol/L of MX dissolves, then [M⁺] = s and [X⁻] = \(C_X+s\). The exact concentration model is
Writing \(K_{sp}=s^2\) would incorrectly treat the solution as initially ion-free.
Compare three original preparations
| Solvent composition | Estimated s / M | Calculation |
|---|---|---|
| Initially pure water | 2.0 × 10⁻⁵ | \(\sqrt{K_{sp}}\) |
| 0.010 M X⁻ | 4.0 × 10⁻⁸ | \(K_{sp}/0.010\) |
| 0.020 M X⁻ | 2.0 × 10⁻⁸ | \(K_{sp}/0.020\) |
The last two rows neglect the tiny extra X⁻ released by dissolution.
Work through the approximation check
For the 0.010 M common-ion solution, suppose \(s\ll0.010\). Then \(s=(4.0\times10^{-10})/0.010=4.0\times10^{-8}\ \mathrm M\). The fractional addition to the existing X⁻ pool is \(s/0.010=4.0\times10^{-6}\), or 0.00040%. The approximation is excellent within this idealized model. Compared with water, the dissolved amount is smaller by a factor of 500.
Predict a controlled comparison
Equilibrate equal excess amounts of MX with equal volumes of pure water and common-ion solution at the same temperature. Filter before determining dissolved M⁺. The common-ion sample should show less free M⁺. Using equal excess solid is convenient, but it is the presence of solid at equilibrium—not identical initial solid mass—that fixes saturation.
Do not generalize to every added salt
An unrelated ion is not automatically a common ion. It may still affect measured solubility through ionic-strength effects or complex formation. A ligand that binds M⁺ can lower free M⁺ activity and increase total dissolution, the opposite of the simple shared-ion result. Always inspect the chemical roles of the added species.
Connect the idea to weak electrolytes
The same competition appears when a weak acid is mixed with its conjugate-base salt: added conjugate base suppresses the weak acid's fractional ionization. In both cases, equilibrium responds to an initially supplied product. The numerical expressions differ, so carry the idea forward without copying the solubility formula into an acid–base problem.
Practice with answers
- For MX with Ksp = 9.0 × 10⁻¹² in 0.030 M X⁻, estimate s. 3.0 × 10⁻¹⁰ M.
- Why is Ksp unchanged after adding NaX? Temperature and the defined dissolution reaction are unchanged.
- Does adding a salt with no common ion always leave solubility unchanged? No; other interactions and activity effects may matter.
Factual reference: OpenStax Chemistry 2e, 15.1. The comparisons, numerical model, and practice are independently authored.
Official curriculum reference: College Board AP Chemistry course page. The explanation and worked example are independently written for this study site.