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Chemistry of Aqueous Solutions

Topic 2 of 6

Calculate pH

Choose the strong or weak approximation and check its assumptions.

A-Level 9476 (2026-2027)

Choose the chemical model before taking a logarithm

Strong substances use stoichiometry; weak substances use an equilibrium approximation.

Useful dilute-solution models
SolutionFirst findThen calculate
Strong monoprotic acid at concentration c[H+] approximately c.pH = -log10[H+].
Strong base[OH-] from complete dissociation and the formula stoichiometry.pOH = -log10[OH-]; pH = pKw - pOH.
Weak monoprotic acid HA at concentration c[H+] approximately sqrt(Ka c), if ionisation is small and water contribution negligible.pH from [H+]; check [H+]/c is small.
Weak monoacidic base B at concentration c[OH-] approximately sqrt(Kb c), under corresponding assumptions.pOH, then pH using pKw.

The weak-acid approximation comes from Ka = x2/(c - x), with x = [H+] = [A-]. If x is small compared with c, replace c - x with c. A small computed ionised fraction supports the approximation. If it is not small, the approximation is not justified; the required syllabus problems will not demand solving a quadratic.

Worked example

Compare four solutions at 298 K

Find pH for 0.0200 mol dm-3 HNO3; 0.00500 mol dm-3 dissolved Ca(OH)2; 0.100 mol dm-3 HA with Ka = 1.80 × 10-5; and 0.100 mol dm-3 NH3 with Kb = 1.80 × 10-5.

  1. HNO3: [H+] = 0.0200; pH = 1.70.
  2. Ca(OH)2: [OH-] = 2(0.00500) = 0.0100; pOH = 2.00 and pH = 12.00.
  3. HA: [H+] = sqrt(1.80 × 10-5 × 0.100) = 1.34 × 10-3; pH = 2.87. The ionised fraction is 1.34%, supporting the approximation.
  4. NH3: [OH-] = 1.34 × 10-3; pOH = 2.87, so pH = 11.13.
Answer

The pH values are 1.70, 12.00, 2.87 and 11.13 respectively. The calcium hydroxide calculation concerns the stated dissolved concentration, not an unlimited solubility assumption.