7.12 Common Ion Effect

Common Ion Effect

Add a common ion to a saturated solution and watch Q exceed Ksp as precipitate forms. Side-by-side beakers compare solubility in pure water vs. common ion solution.

Common IonKspLe Châtelier4 Salts
Topic 7.12

Common-Ion Effect

Identify the solubility of a salt, and/or the value of Ksp for the salt, based on the concentration of a common ion already present in solution.

The solubility of a salt is reduced when it is dissolved into a solution that already contains one of the ions present in the salt. The impact of this common-ion effect can be understood qualitatively using Le Châtelier's principle, or calculated from the Ksp for the dissolution process.

Qualitative reasoning. For AgCl(s) ⇌ Ag⁺ + Cl⁻, adding NaCl introduces extra Cl⁻. That pushes Qsp above Ksp, so the system shifts left, precipitating AgCl until Qsp = Ksp again. Less AgCl remains dissolved: solubility has dropped.

The point that trips people up: Ksp itself is unchanged. Only temperature changes K. What changed is the distribution of the ions — one is now much more abundant, so the other must be much scarcer to keep the product constant.

Quantitative treatment. Build an ICE table with the common ion's initial concentration already in place:

  • Initial: [common ion] = concentration from the added salt; [other ion] = 0.
  • Change: +s for each ion, scaled by coefficients.
  • Equilibrium: [common ion] = initial + s ≈ initial (since s is tiny), [other ion] = s.

Because Ksp values are small, the approximation (initial + s) ≈ initial is almost always valid — and it turns the problem into one line of algebra.

Where this appears elsewhere: the common-ion effect is exactly the mechanism behind buffer action (8.8) — adding acetate to acetic acid suppresses the acid's ionization the same way adding chloride suppresses AgCl's dissolution.

Key points

  • A common ion always decreases solubility.
  • Ksp is unchanged — only the ion distribution shifts.
  • Put the common ion’s concentration in the Initial row of the ICE table.
  • Because s is tiny, (initial + s) ≈ initial almost always holds.

Common mistakes

  • Thinking the common ion changes Ksp. Only temperature does.
  • Forgetting the ion source multiplies. 0.10 M CaCl₂ supplies 0.20 M Cl⁻.
  • Starting the ICE table with the common ion at zero.
  • Assuming a spectator ion has an effect. Adding NaNO₃ to a AgCl solution does essentially nothing.

Worked example

Ksp for AgCl is 1.8 × 10⁻¹⁰. Calculate its molar solubility (a) in pure water and (b) in 0.10 M NaCl. (c) Explain the difference.

(a) In pure water.
AgCl(s) ⇌ Ag⁺ + Cl⁻; [Ag⁺] = [Cl⁻] = s
Ksp = s² = 1.8 × 10⁻¹⁰
s = 1.3 × 10⁻⁵ M

(b) In 0.10 M NaCl. NaCl is soluble and supplies 0.10 M Cl⁻ before any AgCl dissolves.

Ag⁺Cl⁻
I00.10
C+s+s
Es0.10 + s ≈ 0.10

Ksp = (s)(0.10) = 1.8 × 10⁻¹⁰
s = 1.8 × 10⁻⁹ M

(Check the approximation: 1.8 × 10⁻⁹ is utterly negligible next to 0.10 ✓)

(c) Explanation. Solubility fell by a factor of about 7000. The chloride ion already in solution drives Qsp above Ksp, so the equilibrium shifts toward the solid until Qsp = Ksp again. Since the product [Ag⁺][Cl⁻] is pinned at 1.8 × 10⁻¹⁰ and [Cl⁻] is now roughly 7000 times larger than it was in pure water, [Ag⁺] — and therefore the amount of AgCl dissolved — must be smaller by the same factor. Ksp itself is unchanged.

Full notes for topic 7.12 →