Weak-Acid pH from an Equilibrium Table

Introduction

A weak acid ionises only partly in water, so its pH cannot be found simply from its concentration. Instead you set up an equilibrium table, write the Ka expression and solve for the hydrogen-ion concentration. Very often a shortcut makes the algebra easy, but the shortcut is only valid when ionisation is small. This page shows how to use the shortcut, how to test it, and what to do when it breaks down.

Core explanation

The table. For a weak acid HA with initial concentration c:

HA ⇌ H⁺ + A⁻

- Initial: c, 0, 0 - Change: −x, +x, +x - Equilibrium: c − x, x, x

Here x is [H⁺] (ignoring the tiny contribution from water), so Ka = x² / (c − x).

The approximation. If the acid is only slightly ionised, x is much smaller than c and c − x ≈ c. Then Ka ≈ x² / c and x ≈ √(Ka c).

Testing it. After solving, calculate x / c × 100. If this is below about 5%, the approximation introduces an error in x of only a few per cent and the pH is reliable to about ±0.01. If it is above 5%, the approximation should be abandoned.

Example that passes. Ethanoic acid has Ka = 1.8 × 10⁻⁵ mol dm⁻³ (pKa = 4.76). For c = 0.10 mol dm⁻³, x ≈ √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol dm⁻³, so pH = 2.87. The degree of ionisation is 1.34 × 10⁻³ / 0.10 = 1.3%, well under 5%. The exact quadratic gives x = 1.33 × 10⁻³ and the same pH to two decimal places.

The exact route. Rearranging Ka = x² / (c − x) gives x² + Ka x − Ka c = 0, so

x = (−Ka + √(Ka² + 4Ka c)) / 2

Only the positive root is physical; the negative root would imply a negative [H⁺].

When the test fails. The approximation weakens when the acid is relatively strong (large Ka) or very dilute (small c). A convenient pre-check is the ratio c / Ka: if it is greater than about 400, the 5% test will almost always be passed.

Formulae

Ka = [H⁺][A⁻] / [HA]; approximate [H⁺] = √(Ka c); exact [H⁺] = (−Ka + √(Ka² + 4Ka c)) / 2; pH = −log₁₀[H⁺]; degree of ionisation = x / c.

Step-by-step reasoning

1. Write the ionisation equation and the Ka expression. 2. Fill in the equilibrium table with c and x. 3. Solve with the approximation x ≈ √(Ka c). 4. Test x / c against 5%. 5. If the test fails, solve the quadratic and take the positive root. 6. Convert [H⁺] to pH and give two decimal places.

Visual explanation

Sketch a bar for the initial acid concentration c. Shade the thin slice that ionises. If the slice is a sliver less than one twentieth of the bar, removing it hardly changes the bar length, so c − x ≈ c. If the slice is a large chunk, the bar is visibly shorter and the approximation is no longer acceptable.

Real-world analogy

If you take £1 from a £100 note, you still have "about £100". If you take £30 from £100, saying you still have "about £100" would be misleading. The 5% test is simply a check on whether the amount removed is small enough to ignore.

Real-world example

Vinegar is roughly 0.8 mol dm⁻³ ethanoic acid. With Ka = 1.8 × 10⁻⁵ the approximation gives [H⁺] ≈ 3.8 × 10⁻³ mol dm⁻³ and a pH of about 2.4, with less than 1% ionisation. This is why vinegar tastes sharp yet is far less corrosive than a strong acid of the same concentration.

Why?

Why does dilution increase the degree of ionisation? In the approximate form, x / c = √(Ka / c). Lowering c raises this fraction. Physically, dilution lowers the chance of H⁺ and A⁻ meeting to recombine, so the equilibrium shifts towards ions.

Common misconception

"The pH of a weak acid equals −log c." That is only true for a strong acid. For a weak acid, [H⁺] is much smaller than c, so using −log c overestimates acidity by more than a whole pH unit in typical cases.

Worked example

Question: Chloroethanoic acid has Ka = 1.4 × 10⁻³ mol dm⁻³. Find the pH of a 0.010 mol dm⁻³ solution.

Reasoning: Approximation: x ≈ √(1.4 × 10⁻⁵) = 3.74 × 10⁻³, which is 37% of c, so the test fails. Exact: x² + 1.4 × 10⁻³ x − 1.4 × 10⁻⁵ = 0. The discriminant is 1.96 × 10⁻⁶ + 5.6 × 10⁻⁵ = 5.80 × 10⁻⁵, with square root 7.61 × 10⁻³. So x = (7.61 × 10⁻³ − 1.4 × 10⁻³) / 2 = 3.11 × 10⁻³ mol dm⁻³.

Answer: pH = 2.51 (the invalid approximation would give 2.43).

Quick check

1. A weak acid calculation gives x = 2.0 × 10⁻³ mol dm⁻³ for c = 0.020 mol dm⁻³. Is the approximation acceptable? Answer: No; x / c = 10%, which exceeds 5%, so the quadratic should be solved.

Exam focus

Show the equilibrium table, state the approximation explicitly, and quote the percentage ionisation to justify it. Examiners often ask you to "state one assumption": say that ionisation is small enough for [HA] ≈ c, and that H⁺ from water is negligible.

Advanced insight

For extremely dilute or extremely weak acids, water's own ionisation (Kw = 1.0 × 10⁻¹⁴ at 25 °C) is no longer negligible. A charge balance, [H⁺] = [A⁻] + [OH⁻], must then be combined with Ka and Kw, giving a cubic in [H⁺]. At higher concentrations, activity coefficients below 1 make the calculated pH slightly too low.

Summary

For a weak acid, Ka = x² / (c − x). If ionisation is small, x ≈ √(Ka c); check that x is under 5% of c. If it is not, solve x² + Ka x − Ka c = 0 and take the positive root. Dilution raises the percentage ionisation even as it raises the pH.

Practice questions

1. Calculate the pH of 0.10 mol dm⁻³ methanoic acid, Ka = 1.8 × 10⁻⁴ mol dm⁻³, and test the approximation. Answer: x ≈ √(1.8 × 10⁻⁵) = 4.2 × 10⁻³, so pH = 2.37; ionisation is 4.2%, just under 5%, so the approximation is acceptable. 2. A 0.050 mol dm⁻³ weak acid has pH 3.00. Calculate Ka. Answer: x = 1.0 × 10⁻³; Ka = (1.0 × 10⁻³)² / (0.050 − 0.001) = 2.0 × 10⁻⁵ mol dm⁻³. 3. What is the percentage ionisation of 0.0010 mol dm⁻³ ethanoic acid using the approximation, and is it valid? Answer: x ≈ √(1.8 × 10⁻⁸) = 1.34 × 10⁻⁴, which is 13% of c, so the approximation is not valid. 4. Why is the negative root of the quadratic rejected? Answer: It would correspond to a negative hydrogen-ion concentration, which is physically impossible.