Part 2: Kw, strong bases and neutralisation calculations

Use the ionic product of water and calculate what remains after an acid and alkali react.

Neutral means equal ion concentrations

Water undergoes a small amount of self-ionisation. Its nearly constant solvent contribution is incorporated into Kw. In the A-level concentration convention, Kw = [H⁺][OH⁻] with units mol² dm⁻⁶. It applies to acidic, neutral and alkaline aqueous solutions at the stated temperature.

At 298 K, use Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ unless another value is supplied. Neutrality means [H⁺] = [OH⁻], so each equals √Kw. pH 7.00 is neutral at 298 K, but not at every temperature. Water ionisation is endothermic, so warming increases Kw and lowers the neutral pH while the two ion concentrations remain equal.

Find hydroxide first, hydrogen ions second

A dissolved strong hydroxide provides hydroxide ions according to its formula. For NaOH, [OH⁻] ≈ c; for fully dissociated dissolved Ba(OH)₂, [OH⁻] ≈ 2c. Use the dissolved concentration, not the mass of undissolved solid.

For 0.0150 mol dm⁻³ NaOH at 298 K, [H⁺] = (1.00 × 10⁻¹⁴)/0.0150 = 6.67 × 10⁻¹³ mol dm⁻³. pH = 12.18. For 0.00400 mol dm⁻³ dissolved Ba(OH)₂, [OH⁻] = 0.00800 mol dm⁻³ and pH = 11.90.

The shortcut pH + pOH = pKw follows from Kw. It equals 14.00 only when the supplied Kw makes pKw = 14.00.

Neutralisation is a mole calculation first

Calculate the amounts of H⁺ and OH⁻ supplied, use the balanced reaction to find the excess, then divide excess moles by the combined solution volume. Only then calculate pH. Do not subtract pH values, average them or subtract concentrations when volumes differ.

If alkali remains, use Kw after the mole balance

Mix 25.0 cm³ of 0.100 mol dm⁻³ HCl with 40.0 cm³ of 0.100 mol dm⁻³ NaOH. Excess OH⁻ = 0.00400 − 0.00250 = 0.00150 mol. In 0.0650 dm³, [OH⁻] = 0.0230769… mol dm⁻³. At 298 K, [H⁺] = 4.333… × 10⁻¹³ and pH = 12.36.

At exact strong-acid/strong-base equivalence, neither has a stoichiometric excess. Water sets the neutral pH if the salt ions do not undergo appreciable hydrolysis. Do not set [H⁺] = 0 and attempt log(0). A weak-acid or weak-base salt changes the equivalence-point pH and needs different reasoning.

Worked neutral water at another temperature

If Kw = 4.00 × 10⁻¹⁴ mol² dm⁻⁶ at a stated higher temperature, neutral [H⁺] = 2.00 × 10⁻⁷ mol dm⁻³ and pH = 6.70. The sample is still neutral because [OH⁻] has increased by the same amount.

For a base calculation at that temperature, use that Kw. A solution with [OH⁻] = 0.0100 mol dm⁻³ then has [H⁺] = 4.00 × 10⁻¹² and pH = 11.40, not 12.00.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.

Sources