Part 1: Proton transfer, strong acids and strong bases

Connect acid–base equations to hydrogen-ion concentration and calculate pH only after finding the species left in solution.

Follow the proton and identify both conjugate pairs

A Brønsted–Lowry acid donates H⁺; a base accepts H⁺. In NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, ammonia is the base and water the acid. NH₄⁺ is the conjugate acid of NH₃, and OH⁻ is the conjugate base of H₂O. A conjugate pair differs by exactly one proton, so its charges also differ by one.

Water can act as an acid or as a base depending on its partner. In HCl + H₂O → H₃O⁺ + Cl⁻, water accepts the proton. The shorthand H⁺(aq) represents hydrated hydrogen ions in aqueous calculations; it does not imply isolated bare protons drifting in water.

Monobasic, dibasic and tribasic acids can donate one, two and three protons respectively per molecule in complete neutralisation. H₃PO₄ is tribasic, but this does not mean every one of its three protons dissociates completely in an ordinary aqueous solution. Count ionisable acid protons, not every H in the molecular formula.

Connect acid reactions to the role of H⁺

Metals such as magnesium reduce hydrogen ions to hydrogen while the metal is oxidised. Oxide, hydroxide and carbonate bases accept protons. The ionic equations identify the reacting particles and leave out spectator ions.

For a weak acid, its dissolved molecules replenish some H⁺ as it reacts, so an overall equation may use HA explicitly. Weak does not mean unable to neutralise a base; strength describes the extent of ionisation, whereas total neutralising capacity depends on amount and stoichiometry.

pH is logarithmic, not a direct concentration scale

Use pH = −log₁₀[H⁺] with concentration in mol dm⁻³ in the A-Level convention. A tenfold increase in [H⁺] lowers pH by one. A change from pH 4 to pH 2 represents a hundredfold increase, not a doubling. pH has no unit.

For an illustrative 0.00400 mol dm⁻³ strong monobasic acid, complete dissociation gives [H⁺] = 0.00400 mol dm⁻³ and pH = 2.40. Conversely a pH of 3.25 gives [H⁺] = 10⁻³·²⁵ = 5.62 × 10⁻⁴ mol dm⁻³. Do not use e⁻pH: the logarithm here is base ten.

Strong means essentially complete ionisation in the usual aqueous model; concentrated means a large amount per volume. A dilute strong acid can have a higher pH than a concentrated weak acid. Compare both concentration and strength. For extremely dilute solutions the contribution from water cannot be neglected; do not extrapolate [H⁺] = acid concentration beyond that approximation.

Use Kw to connect hydroxide and hydrogen ions

Kw = [H⁺][OH⁻] at a stated temperature. At 25 °C, use 1.00 × 10⁻¹⁴ mol² dm⁻⁶ unless different data are supplied. For 0.0150 mol dm⁻³ fully dissociated Ba(OH)₂, [OH⁻] = 0.0300 mol dm⁻³ because each formula unit supplies two hydroxide ions. Thus [H⁺] = Kw/[OH⁻] = 3.33 × 10⁻¹³ mol dm⁻³ and pH = 12.48.

Neutral means [H⁺] = [OH⁻], so [H⁺] = √Kw. If an illustrative higher-temperature Kw is 4.00 × 10⁻¹⁴, neutral [H⁺] is 2.00 × 10⁻⁷ and neutral pH is 6.70. The solution remains neutral: pH 7 is not the universal definition of neutrality.

Worked example: neutralise first, then calculate pH

Mix 25.0 cm³ of 0.100 mol dm⁻³ HCl with 15.0 cm³ of 0.100 mol dm⁻³ NaOH. Initial amounts are 0.00250 mol H⁺ and 0.00150 mol OH⁻. Neutralisation removes equal amounts, leaving 0.00100 mol H⁺.

The total volume is 40.0 cm³ = 0.0400 dm³, so [H⁺] = 0.00100/0.0400 = 0.0250 mol dm⁻³ and pH = 1.60. Averaging the initial pH values would have no chemical meaning. If base were in excess, calculate its remaining [OH⁻] first and then use Kw.

Quick checks

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

Sources