Lesson 2 of 6

Calculate pH

Choose the strong or weak approximation and check its assumptions.

A-Level 9476 (2026-2027)

I can define and calculate with acid-base equilibrium quantities.Syllabus
Syllabus 9476, 10.1(b). Define and calculate with acid-base equilibrium quantities.
I can calculate hydrogen-ion concentration and pH.Syllabus
Syllabus 9476, 10.1(c). Calculate hydrogen-ion concentration and pH.

Choose the chemical model before taking a logarithm

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

Useful dilute-solution models
Solution	First find	Then 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(K_a 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(K_b c), under corresponding assumptions.	pOH, then pH using pKw.

The weak-acid shortcut comes from K_a = x²/(c - x), where x = [H⁺] = [A^-]. If x is much smaller than c, you can replace c - x with c. Check by working out the fraction ionised: if it is small, the shortcut is fine. If it is not small, the shortcut is not justified. Syllabus questions will not ask you to solve a quadratic.

Worked exampleCompare 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 K_a = 1.80 × 10^-5 mol dm^-3; and 0.100 mol dm^-3 NH₃ with K_b = 1.80 × 10^-5 mol dm^-3.

Try it first. Then show the worked stepsHide the worked steps

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

Try it step by step

Work it out in steps: [H⁺] = sqrt(Ka x c), then pH = -log[H⁺].

Problem 1. Find the pH of 0.0500 mol dm^-3 ethanoic acid, Ka = 1.80 x 10^-5 mol dm^-3, at 298 K. Give [H⁺] to 3 significant figures and pH to 2 decimal places.
Problem 2. Find the pH of 0.200 mol dm^-3 of a weak acid HB with Ka = 1.60 x 10^-4 mol dm^-3 at 298 K. Give [H⁺] to 3 significant figures and pH to 2 decimal places.
Problem 3. Find the pH of 0.0100 mol dm^-3 of a weak acid HA with Ka = 4.00 x 10^-6 mol dm^-3 at 298 K. Give [H⁺], in standard form or as a decimal, to 3 significant figures and pH to 2 decimal places.

Show the answer

Problem 1.

[H⁺] = sqrt(Ka x c): [H⁺] = sqrt(1.80 x 10^-5 x 0.0500) = sqrt(9.00 x 10^-7) = 9.49 x 10^-4 mol dm^-3. Only about 2% of the acid ionises, so the approximation holds.

pH = -log[H⁺]: pH = -log(9.49 x 10^-4) = 3.02.

Problem 2.

[H⁺] = sqrt(Ka x c): [H⁺] = sqrt(1.60 x 10^-4 x 0.200) = sqrt(3.20 x 10^-5) = 5.66 x 10^-3 mol dm^-3.

pH = -log[H⁺]: pH = -log(5.66 x 10^-3) = 2.25.

Problem 3.

[H⁺] = sqrt(Ka x c): [H⁺] = sqrt(4.00 x 10^-6 x 0.0100) = sqrt(4.00 x 10^-8) = 2.00 x 10^-4 = 0.000200 mol dm^-3.

pH = -log[H⁺]: pH = -log(2.00 x 10^-4) = 3.70.

Watch out for this

A student says: "To find the pH after mixing an acid and an alkali, calculate the pH of each and take the average." What is wrong with this?

Show the answer

Close to 10^-7 mol dm^-3, the ions from water itself matter at 298 K, so you cannot ignore them. When an acid and a base are mixed, first let them neutralise each other in their mole ratio. Then divide any excess by the total volume. Only then calculate the pH.