The Common Ion Effect in Acid–Base Systems

Take a solution of ethanoic acid. Measure its pH: about 2.9 at 0.10 mol/dm³. Now dissolve some sodium ethanoate in it, a salt that on its own makes water slightly alkaline. The pH jumps to around 4.8, even though you’ve added no base in the usual sense. The ethanoic acid has become much less ionised.

This is the common ion effect: adding an ion that’s already part of an equilibrium pushes that equilibrium backwards. It’s Le Chatelier’s principle applied to ions, and it’s the reason buffers work.

The principle

A weak acid sets up an equilibrium:

HA ⇌ H⁺ + A⁻

If you add extra A⁻ (from a soluble salt like NaA), you increase the concentration of a product. By Le Chatelier’s principle, the equilibrium shifts to the left to use up some of the added A⁻. That combines H⁺ with A⁻ to make HA, so:

• [H⁺] falls
• the pH rises
• the percentage of acid ionised falls

The added ion is called the common ion because it’s shared by the acid and the salt.

The same happens if you add extra H⁺ (from a strong acid) to a weak acid solution: the weak acid’s ionisation is suppressed even further.

Why it happens: the Ka expression

The equilibrium constant doesn’t change:

Ka = [H⁺][A⁻] ÷ [HA]

If [A⁻] is raised by adding salt, then for Ka to stay the same, [H⁺] must fall (and [HA] rises slightly). The equilibrium constant is fixed; the individual concentrations adjust.

Worked example 1: ethanoic acid with sodium ethanoate

Compare the pH and percent ionisation of 0.10 mol/dm³ ethanoic acid (Ka = 1.8 × 10⁻⁵) alone, and with 0.10 mol/dm³ sodium ethanoate added.

Acid alone:

• [H⁺] ≈ √(1.8 × 10⁻⁵ × 0.10) = 1.34 × 10⁻³ mol/dm³
• pH = 2.87
• % ionised = 1.34 × 10⁻³ ÷ 0.10 × 100 = 1.3%

With 0.10 mol/dm³ ethanoate added:

	CH₃COOH	H⁺	CH₃COO⁻
Initial	0.10	0	0.10
Change	−x	+x	+x
Equilibrium	0.10 − x	x	0.10 + x

Ka = x(0.10 + x) ÷ (0.10 − x)

Because x is now tiny, 0.10 + x ≈ 0.10 and 0.10 − x ≈ 0.10:

1.8 × 10⁻⁵ = x × 0.10 ÷ 0.10, so x = 1.8 × 10⁻⁵ mol/dm³

• pH = 4.74
• % ionised = 1.8 × 10⁻⁵ ÷ 0.10 × 100 = 0.018%

Adding the common ion cut the ionisation by a factor of about 75 and raised the pH by nearly two units. That new pH equals the pKa, which is exactly what the Henderson–Hasselbalch equation predicts for equal amounts of acid and conjugate base.

Worked example 2: a weak acid in a strong acid

What fraction of 0.10 mol/dm³ ethanoic acid is ionised in a solution that also contains 0.010 mol/dm³ HCl?

The HCl supplies [H⁺] = 0.010 mol/dm³, which swamps the small amount from the weak acid.

Ka = [H⁺][A⁻] ÷ [HA] → 1.8 × 10⁻⁵ = (0.010)(x) ÷ 0.10

x = [CH₃COO⁻] = 1.8 × 10⁻⁴ mol/dm³

% ionised = 1.8 × 10⁻⁴ ÷ 0.10 × 100 = 0.18%

Compared with 1.3% in pure water, the strong acid’s H⁺ has suppressed ethanoic acid’s ionisation about sevenfold. The pH is set almost entirely by the HCl: pH ≈ 2.00.

This matters in practice. When you mix a strong acid and a weak acid, you can usually ignore the weak acid’s contribution to [H⁺].

Worked example 3: a weak base with its salt

Find the pH of 0.20 mol/dm³ ammonia containing 0.10 mol/dm³ ammonium chloride. Kb(NH₃) = 1.8 × 10⁻⁵.

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

The added NH₄⁺ is the common ion. With x small:

Kb = [NH₄⁺][OH⁻] ÷ [NH₃] → 1.8 × 10⁻⁵ = (0.10)(x) ÷ 0.20

x = [OH⁻] = 3.6 × 10⁻⁵ mol/dm³

pOH = 4.44, pH = 9.56

Without the ammonium chloride, the pH of 0.20 mol/dm³ ammonia is 11.28. The common ion lowered it by more than 1.7 units. See pH of weak bases.

The common ion effect and buffers

Every buffer is a common ion system. A buffer contains a weak acid and a salt of its conjugate base (or a weak base and its conjugate acid). Because the conjugate base is present in large amounts, the weak acid’s ionisation is heavily suppressed, and the equilibrium concentrations of HA and A⁻ are essentially the amounts you added. That’s what makes the Henderson–Hasselbalch approximation so accurate for buffers.

When a little strong acid is added to the buffer, the large reservoir of A⁻ absorbs it. When a little strong base is added, the large reservoir of HA absorbs it. See buffers explained and buffer calculations.

The common ion effect and solubility

The same principle applies to sparingly soluble salts. Silver chloride dissolves very slightly:

AgCl(s) ⇌ Ag⁺ + Cl⁻

Adding sodium chloride (extra Cl⁻) pushes the equilibrium left, so even less silver chloride dissolves. This is used in analysis to make precipitation more complete, and it’s covered in detail with solubility products (Ksp).

Real-world examples

Drug absorption. Many drugs are weak acids or bases. The pH and ions in the stomach and intestine control how much is in the neutral, membrane-crossing form. Formulating a drug as a particular salt, or with buffering agents, uses common ion principles to control how it dissolves and is absorbed.

Blood and body fluids. The large, relatively constant concentration of hydrogencarbonate in blood acts as a common ion reservoir for the carbonic acid equilibrium, part of what stabilises blood pH. See the blood buffer system.

Water treatment. Adding a common ion helps precipitate unwanted metal ions. For example, adding carbonate or hydroxide can reduce the solubility of calcium or heavy metal compounds so they settle out.

Analytical chemistry. Buffers based on the common ion effect keep pH constant during titrations with EDTA, colour tests and enzyme assays, where a small pH drift would change the result.

Controlling indicator colour. An indicator is a weak acid, so adding its conjugate base (or an acid) shifts its colour. That’s exactly how buffers of known pH are used to set up indicator colour standards.

Common misconceptions

• “Adding a salt of the acid neutralises it.” Sodium ethanoate doesn’t neutralise ethanoic acid; it suppresses its ionisation. The amount of acid is unchanged.
• “The equilibrium constant changes.” Ka stays the same at a given temperature; only the concentrations change.
• “The common ion effect only applies to acids.” It applies to any equilibrium involving ions, including weak bases and sparingly soluble salts.

Practice questions

1. Calculate the pH of a solution containing 0.050 mol/dm³ methanoic acid (Ka 1.8 × 10⁻⁴) and 0.10 mol/dm³ sodium methanoate.
2. What percentage of 0.20 mol/dm³ HF (Ka 6.8 × 10⁻⁴) is ionised in 0.10 mol/dm³ HCl?

Answers:

1. [H⁺] = Ka × [HA] ÷ [A⁻] = 1.8 × 10⁻⁴ × 0.050 ÷ 0.10 = 9.0 × 10⁻⁵ → pH = 4.05.
2. [F⁻] = Ka × [HF] ÷ [H⁺] = 6.8 × 10⁻⁴ × 0.20 ÷ 0.10 = 1.36 × 10⁻³; % = 1.36 × 10⁻³ ÷ 0.20 × 100 = 0.68%.

Key takeaways

• Adding an ion already present in an equilibrium shifts it back, suppressing ionisation.
• For a weak acid, adding its conjugate base (or a strong acid) lowers [H⁺] from the weak acid and reduces percent ionisation dramatically.
• Ka and Kb don’t change; the concentrations adjust to keep them constant.
• Buffers are common ion systems, which is why Henderson–Hasselbalch works so well for them.
• The same principle reduces the solubility of sparingly soluble salts.