Percent Ionisation of Weak Acids

Here’s a result that surprises most students. Take a bottle of vinegar and dilute it with water. The solution obviously becomes less acidic: its pH goes up. But at the same time, a larger share of the ethanoic acid molecules in it have given up their protons. Diluting a weak acid makes it less acidic overall, yet more ionised.

Percent ionisation is the quantity that captures this, and it’s one of the clearest ways to see what “weak acid” really means.

Definition

Percent ionisation (also called percent dissociation, or degree of dissociation × 100) is the fraction of acid molecules that have ionised at equilibrium, expressed as a percentage:

% ionisation = ([H⁺] at equilibrium ÷ initial acid concentration) × 100

For HA ⇌ H⁺ + A⁻, each molecule that ionises produces one H⁺, so [H⁺] measures how many have ionised.

Strong acids are essentially 100% ionised. Weak acids at typical lab concentrations are usually between about 0.01% and 10% ionised.

A useful shortcut

From the weak acid approximation [H⁺] ≈ √(Ka × c):

% ionisation ≈ (√(Ka × c) ÷ c) × 100 = √(Ka ÷ c) × 100

This shows immediately how the percentage depends on the two variables:

• Larger Ka → more ionisation. Stronger acids ionise more.
• Smaller c → more ionisation. Dilute solutions ionise more.

The shortcut is only reliable while the answer stays below about 5%. Beyond that, solve the full equation, as shown in pH of weak acids.

Worked example 1: from Ka

Calculate the percent ionisation of 0.10 mol/dm³ ethanoic acid (Ka = 1.8 × 10⁻⁵).

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

So in a solution of this concentration, only about 1 molecule in 75 has given away its proton at any moment.

Worked example 2: from a measured pH

A 0.050 mol/dm³ solution of a weak acid has a pH of 2.85. What percentage is ionised?

• [H⁺] = 10⁻²·⁸⁵ = 1.41 × 10⁻³ mol/dm³
• % ionisation = 1.41 × 10⁻³ ÷ 0.050 × 100 = 2.8%

This route needs no Ka at all, which makes it a common exam question: measure pH, and you know how much of the acid has ionised.

Worked example 3: the effect of dilution

Calculate the percent ionisation of ethanoic acid at 1.0, 0.10, 0.010 and 0.0010 mol/dm³.

Concentration (mol/dm³)	[H⁺] (mol/dm³)	pH	% ionised
1.0	4.2 × 10⁻³	2.37	0.42%
0.10	1.3 × 10⁻³	2.87	1.3%
0.010	4.2 × 10⁻⁴	3.37	4.2%
0.0010	1.25 × 10⁻⁴	3.90	12.5%

The first three rows use the shortcut. For the last row the shortcut would predict 13.4%, which fails the 5% check, so the value shown comes from solving the quadratic x² + (1.8 × 10⁻⁵)x − 1.8 × 10⁻⁸ = 0.

Two trends run in opposite directions:

• As the solution gets more dilute, pH rises (less acidic).
• As the solution gets more dilute, % ionisation rises (more of the acid has reacted).

Both are true at once. There’s less acid in total, so there’s less H⁺ in total, even though a bigger proportion of what’s there has ionised.

Why dilution increases ionisation

Think about Le Chatelier’s principle. The equilibrium is:

HA ⇌ H⁺ + A⁻

There’s one particle on the left and two on the right. When you add water, every concentration drops. The system responds by shifting towards the side with more dissolved particles, partly offsetting the dilution. That’s the right-hand side, so more HA ionises.

You can see the same thing mathematically. After dilution by a factor of 10, if nothing shifted, the reaction quotient Q = [H⁺][A⁻] ÷ [HA] would fall to one tenth of Ka (two concentrations on top fall by 10 each, one on the bottom falls by 10). Q is now less than Ka, so the reaction moves forward until Q = Ka again.

Ostwald’s dilution law

In 1888, Wilhelm Ostwald expressed this mathematically. If α is the fraction ionised (α = % ÷ 100), then:

Ka = α²c ÷ (1 − α)

For weak acids where α is small, 1 − α ≈ 1, giving α ≈ √(Ka ÷ c). This is exactly the shortcut above. Ostwald used electrical conductivity measurements to find α at different concentrations, and showed that Ka stayed constant, strong evidence for Arrhenius’s ionic theory. You can read more about that theory in Arrhenius acids and bases.

Worked example 4: comparing acids at the same concentration

Compare the percent ionisation of 0.10 mol/dm³ solutions of HF (Ka 6.8 × 10⁻⁴), methanoic acid (Ka 1.8 × 10⁻⁴), ethanoic acid (Ka 1.8 × 10⁻⁵) and hypochlorous acid (Ka 3.0 × 10⁻⁸).

Acid	Ka	% ionised at 0.10 mol/dm³
HF	6.8 × 10⁻⁴	7.9% (quadratic)
Methanoic acid	1.8 × 10⁻⁴	4.2%
Ethanoic acid	1.8 × 10⁻⁵	1.3%
Hypochlorous acid	3.0 × 10⁻⁸	0.055%

At the same concentration, percent ionisation ranks acids in exactly the same order as Ka. That’s why it’s a fair way to compare strength, as long as the concentrations match. Comparing % ionisation at different concentrations can be misleading.

Worked example 5: finding Ka from percent ionisation

A 0.20 mol/dm³ weak acid is 1.5% ionised. Find Ka.

• [H⁺] = [A⁻] = 0.015 × 0.20 = 3.0 × 10⁻³ mol/dm³
• [HA] = 0.20 − 0.0030 = 0.197 mol/dm³
• Ka = (3.0 × 10⁻³)² ÷ 0.197 = 4.6 × 10⁻⁵

Weak bases work the same way

For a weak base B, percent protonation is [OH⁻] ÷ c × 100, and it also rises on dilution. For 0.10 mol/dm³ ammonia (Kb 1.8 × 10⁻⁵), about 1.3% of molecules have picked up a proton at any moment, the same figure as ethanoic acid, because the two constants happen to be almost equal.

A classroom investigation

You can see percent ionisation change with your own eyes using a conductivity meter or a simple bulb-and-battery conductivity tester. Prepare 1.0, 0.10 and 0.010 mol/dm³ solutions of ethanoic acid and of hydrochloric acid, and measure each one.

The hydrochloric acid readings fall roughly in proportion to concentration: ten times less acid, about ten times less conductivity. The ethanoic acid readings fall much more slowly, because each dilution also increases the fraction of molecules that ionise. At the highest concentration, ethanoic acid conducts far worse than hydrochloric acid; at the lowest, the gap between them is noticeably smaller. Wear eye protection throughout, and dispose of the solutions by diluting them with plenty of water.

Why it matters

• Conductivity. A 0.1 mol/dm³ weak acid conducts electricity poorly because only a small percentage of it exists as ions.
• Reaction rates. Weak acids react more slowly with magnesium or carbonates than strong acids of the same concentration, because fewer H⁺ ions are available at any instant. As those ions are used up, more HA ionises to replace them, so the total amount of gas produced ends up the same.
• Titrations. Weak and strong acids of equal concentration and volume need exactly the same amount of alkali to neutralise, because the equilibrium keeps shifting until all the HA has reacted.

Key takeaways

• % ionisation = [H⁺] ÷ c × 100, and ≈ √(Ka ÷ c) × 100 for weak acids.
• Stronger acids (larger Ka) and more dilute solutions ionise to a greater extent.
• Dilution raises both pH and % ionisation at the same time.
• Ostwald’s dilution law, Ka = α²c ÷ (1 − α), connects α, c and Ka.
• Compare acid strengths using % ionisation only at equal concentrations; otherwise use Ka and pKa.