The Common-Ion Effect

Core Concept

When a solution already contains one of the ions that an equilibrium produces, that equilibrium shifts away from producing more of it — Le Châtelier's principle applied to a shared ion. Adding acetate (CH₃COO⁻) to acetic acid suppresses the acid's ionization, lowering [H₃O⁺] and raising the pH. Adding NH₄⁺ to ammonia suppresses its ionization. The same principle lowers the solubility of a sparingly soluble salt when one of its ions is already present — and it is the foundational reason buffers work.

Key Ideas

• Definition: the shift of an equilibrium caused by adding an ion already involved in it.
• Effect on weak acids: adding the conjugate base A⁻ drives HA ⇌ H⁺ + A⁻ to the left, so [H₃O⁺] drops and pH rises.
• Effect on weak bases: adding BH⁺ drives B + H₂O ⇌ BH⁺ + OH⁻ to the left, so [OH⁻] drops and pH falls.
• Effect on solubility: adding a common ion to a saturated salt lowers its molar solubility.
• Quantitative consequence: the ionization is no longer x²/C — now one of the products starts at a nonzero concentration.

Equations and Variables

• Weak acid + common ion: HA ⇌ H₃O⁺ + A⁻, with [A⁻]₀ supplied by the salt.
• ICE with a nonzero product start: Ka = x([A⁻]₀ + x) / ([HA]₀ − x)
• Approximation: Ka ≈ x[A⁻]₀ / [HA]₀, so x = [H₃O⁺] = Ka × [HA]₀/[A⁻]₀
• Solubility analog: AgCl(s) ⇌ Ag⁺ + Cl⁻; adding NaCl shifts left and lowers [Ag⁺].

How It Works

1. Write the ionization equilibrium and identify the shared ion.
2. Add the salt that supplies that ion; its concentration is now the initial concentration of that product.
3. Rebuild the ICE table with a nonzero starting amount of the common product.
4. Solve for x (the newly produced H₃O⁺ or OH⁻), which is now smaller than in the pure acid/base because the product side already starts populated.
5. Use the approximation x = Ka([HA]₀/[A⁻]₀), which is simply a rearranged form of the Henderson–Hasselbalch equation.

Worked Example

Find the pH of a solution that is 0.10 M in acetic acid and 0.10 M in sodium acetate (Ka = 1.8 × 10⁻⁵). Compare to 0.10 M acetic acid alone (pH 2.87).

CH₃COOH ⇌ H₃O⁺ + CH₃COO⁻, with [CH₃COO⁻]₀ = 0.10 M from the salt.

ICE: [CH₃COOH] = 0.10 − x; [H₃O⁺] = x; [CH₃COO⁻] = 0.10 + x.

Ka = x(0.10 + x)/(0.10 − x) ≈ x(0.10)/(0.10) = x

x = [H₃O⁺] = 1.8 × 10⁻⁵ M → pH = 4.74

The common acetate ion has suppressed [H₃O⁺] from 1.34 × 10⁻³ M down to 1.8 × 10⁻⁵ M — a ~75-fold drop — raising the pH from 2.87 to 4.74.

How it works

1. Write the ionization equilibrium and identify the shared ion.
2. Add the salt that supplies that ion; its concentration is now the initial concentration of that product.
3. Rebuild the ICE table with a nonzero starting amount of the common product.
4. Solve for x (the newly produced H₃O⁺ or OH⁻), which is now smaller than in the pure acid/base because the product side already starts populated.
5. Use the approximation x = Ka([HA]₀/[A⁻]₀), which is simply a rearranged form of the Henderson–Hasselbalch equation.

Common confusions

• "The common ion must be H₃O⁺ or OH⁻." — Any shared ion counts; the acetate or ammonium ion is the common ion in buffer-style problems.
• "The salt fully neutralizes the acid." — No; the acid still ionizes, just less. Some H₃O⁺ remains (here 1.8 × 10⁻⁵ M, not zero).
• "Adding a common ion increases ionization." — It decreases it; you are adding a product, so the reaction shifts toward reactants.
• "The common-ion effect only applies to acids." — It applies to any equilibrium: weak bases and solubility equilibria too.
• "Forgetting the salt contributes the product at the start." — The whole point is that [A⁻]₀ is not zero in the ICE table.

Quick review

• Common ion = an ion already in the equilibrium.
• Shifts the equilibrium to consume the added ion (Le Châtelier).
• Weak acid + its salt: [H₃O⁺] = Ka × [HA]/[A⁻]; pH rises.
• Weak base + its salt: [OH⁻] falls; pH falls.
• Also lowers solubility of sparingly soluble salts.