pH of Weak Acids: ICE Tables and Approximations

For a strong acid, pH is a one-line calculation: [H⁺] equals the acid concentration. Weak acids are different. Only a small fraction of their molecules give up a proton, and that fraction depends on both the acid’s strength and its concentration. To find the pH you have to treat the ionisation as an equilibrium.

That sounds heavy, but the method is always the same, and in most exam questions a simple approximation turns it into a two-line calculation. This guide shows the full method, the shortcut, and how to tell when the shortcut is allowed.

The setup

A weak acid HA ionises reversibly in water:

HA ⇌ H⁺ + A⁻

The acid dissociation constant is:

Ka = [H⁺][A⁻] ÷ [HA]

The bigger Ka is, the more the acid ionises and the stronger it is. (If Ka and pKa are new to you, read Ka and pKa explained first.)

The ICE table method

ICE stands for Initial, Change, Equilibrium. It’s a way of keeping track of concentrations as the equilibrium sets up. Let the starting concentration of acid be c, and let x be the amount that ionises.

	HA	H⁺	A⁻
Initial	c	0	0
Change	−x	+x	+x
Equilibrium	c − x	x	x

(We ignore the tiny amount of H⁺ from water, which is fine unless the acid is extremely weak or extremely dilute.)

Substitute the equilibrium row into the Ka expression:

Ka = x² ÷ (c − x)

Solve for x. Since x = [H⁺], the pH is −log x.

The approximation

If the acid is weak enough that only a small fraction ionises, then x is tiny compared with c, and c − x ≈ c. The equation becomes:

Ka ≈ x² ÷ c, so x ≈ √(Ka × c)

This is the formula most students use:

[H⁺] ≈ √(Ka × c)

When is it safe?

Check after you’ve calculated x. If x is less than 5% of c, the approximation is fine. If it’s more than 5%, solve the full equation (a quadratic) instead.

A quicker test before you start: if c ÷ Ka is greater than about 400, the approximation will pass the 5% check.

Worked example 1: methanoic acid

Calculate the pH of 0.20 mol/dm³ methanoic acid, HCOOH. Ka = 1.8 × 10⁻⁴ mol/dm³.

• [H⁺] ≈ √(1.8 × 10⁻⁴ × 0.20) = √(3.6 × 10⁻⁵) = 6.0 × 10⁻³ mol/dm³
• pH = −log(6.0 × 10⁻³) = 2.22
• Check: 6.0 × 10⁻³ ÷ 0.20 = 3.0%. Below 5%, so the approximation is valid.

Worked example 2: hypochlorous acid

Calculate the pH of 0.15 mol/dm³ hypochlorous acid, HOCl. Ka = 3.0 × 10⁻⁸ mol/dm³.

• [H⁺] ≈ √(3.0 × 10⁻⁸ × 0.15) = √(4.5 × 10⁻⁹) = 6.7 × 10⁻⁵ mol/dm³
• pH = 4.17
• Check: 6.7 × 10⁻⁵ is about 0.04% of 0.15. Easily valid.

Hypochlorous acid is the active disinfectant formed when chlorine dissolves in swimming pool water. Its weakness matters: at the pool’s pH of about 7.4, a large share of it has turned into the hypochlorite ion, which is a much less effective disinfectant. That’s one reason pool operators control pH so carefully.

Worked example 3: when the shortcut fails (hydrofluoric acid)

Calculate the pH of 0.050 mol/dm³ HF. Ka = 6.8 × 10⁻⁴ mol/dm³.

Try the shortcut first:

• [H⁺] ≈ √(6.8 × 10⁻⁴ × 0.050) = √(3.4 × 10⁻⁵) = 5.83 × 10⁻³ mol/dm³
• Check: 5.83 × 10⁻³ ÷ 0.050 = 11.7%. Too big. The approximation isn’t valid.

Solve the full equation:

6.8 × 10⁻⁴ = x² ÷ (0.050 − x)

Rearrange: x² + (6.8 × 10⁻⁴)x − 3.4 × 10⁻⁵ = 0

Use the quadratic formula, taking the positive root:

x = [−6.8 × 10⁻⁴ + √((6.8 × 10⁻⁴)² + 4 × 3.4 × 10⁻⁵)] ÷ 2 = 5.50 × 10⁻³ mol/dm³

pH = −log(5.50 × 10⁻³) = 2.26

The shortcut would have given 2.23. The difference looks small, but in an exam that asks for two decimal places, it’s the difference between the mark and no mark.

Worked example 4: finding Ka from pH

A 0.10 mol/dm³ solution of a weak acid HA has a pH of 3.00. Calculate Ka.

• [H⁺] = 10⁻³·⁰⁰ = 1.0 × 10⁻³ mol/dm³, and [A⁻] = [H⁺]
• [HA] at equilibrium = 0.10 − 0.0010 = 0.099 mol/dm³
• Ka = (1.0 × 10⁻³)² ÷ 0.099 = 1.0 × 10⁻⁵ mol/dm³
• pKa = 5.00

Many exams let you use 0.10 instead of 0.099 here. It makes almost no difference, but using the exact value never loses marks.

Worked example 5: how dilution changes ionisation

Calculate the pH and percentage ionisation of ethanoic acid (Ka = 1.8 × 10⁻⁵) at 0.10 and at 0.010 mol/dm³.

At 0.10 mol/dm³:

• [H⁺] = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ → pH = 2.87
• Percentage ionised = 1.34 × 10⁻³ ÷ 0.10 × 100 = 1.3%

At 0.010 mol/dm³:

• [H⁺] = √(1.8 × 10⁻⁷) = 4.24 × 10⁻⁴ → pH = 3.37
• Percentage ionised = 4.24 × 10⁻⁴ ÷ 0.010 × 100 = 4.2%

Two things to notice. First, diluting ten times raised the pH by only 0.5, not by 1 as it would for a strong acid. Second, the fraction ionised went up on dilution. That’s Le Chatelier’s principle in action: adding water spreads the particles out and favours the side with more dissolved particles (H⁺ + A⁻). Read more in percent ionisation of weak acids.

Why dilution raises pH by 0.5 for a weak acid

From [H⁺] ≈ √(Ka × c), reducing c by a factor of 10 reduces [H⁺] by √10 ≈ 3.16. Since log(3.16) = 0.5, the pH goes up by 0.5 units per tenfold dilution. For a strong acid, [H⁺] drops by the full factor of 10, so pH goes up by 1. This is a neat way of telling a strong acid from a weak one experimentally.

Summary of the method

1. Write the equilibrium: HA ⇌ H⁺ + A⁻.
2. Set up the ICE table with starting concentration c.
3. Write Ka = x² ÷ (c − x).
4. Try the approximation x ≈ √(Ka × c).
5. Check x is below 5% of c. If not, solve the quadratic.
6. pH = −log x.

Practice questions

1. Calculate the pH of 0.25 mol/dm³ benzoic acid (Ka = 6.3 × 10⁻⁵).
2. A 0.020 mol/dm³ solution of a weak acid has pH 3.40. Calculate Ka and pKa.
3. Calculate the pH of 0.010 mol/dm³ HNO₂ (Ka = 5.0 × 10⁻⁴). Is the approximation valid?

Answers:

1. [H⁺] = √(1.58 × 10⁻⁵) = 3.97 × 10⁻³ (1.6% of c, valid); pH = 2.40.
2. [H⁺] = 3.98 × 10⁻⁴; Ka = (3.98 × 10⁻⁴)² ÷ (0.020 − 0.000398) = 8.1 × 10⁻⁶; pKa = 5.09.
3. Shortcut gives √(5.0 × 10⁻⁶) = 2.24 × 10⁻³, which is 22% of c, so it’s not valid. Solving x² + (5.0 × 10⁻⁴)x − 5.0 × 10⁻⁶ = 0 gives x = 2.0 × 10⁻³, so pH = 2.70.

Common mistakes

• Using [H⁺] = c for a weak acid. That’s only true for strong acids.
• Forgetting to square-root. [H⁺] = Ka × c gives a ridiculous answer; it’s the square root.
• Skipping the 5% check. Relatively strong “weak” acids like HF, HNO₂ and chloroethanoic acids often fail it.
• Using pKa in place of Ka. Convert first: Ka = 10⁻ᵖᴷᵃ.
• Applying the method to polyprotic acids without thought. Usually only the first ionisation matters for pH, but check your course’s convention.

Key takeaways

• Weak acids only partly ionise, so pH needs Ka and an equilibrium calculation.
• The ICE table gives Ka = x² ÷ (c − x), and usually [H⁺] ≈ √(Ka × c).
• Check that x is below 5% of c; if not, use the quadratic.
• Dilution increases the fraction ionised but still raises the pH, by about 0.5 per tenfold dilution.
• Check your answers with the pH calculator, which solves the exact equation including water’s own ionisation.