UK Edexcel A-Level Chemistry 9CH0 · Year 13 · Topic 12

Part 1: Proton transfer, acid strength and the pH scale

Reviewed 9 October 2026.

Connect acid–base definitions to particles, then calculate pH for strong acids and bases without confusing strength with concentration.

Acids and bases are roles in proton transfer

A Brønsted–Lowry acid donates a proton, H⁺; a Brønsted–Lowry base accepts a proton. In HCl + H₂O → H₃O⁺ + Cl⁻, HCl is the donor and water is the acceptor. In water, a proton is hydrated: H₃O⁺ is a useful explicit representation, while H⁺(aq) is the conventional abbreviation in calculations.

After HA loses H⁺ it becomes A⁻, its conjugate base. After B gains H⁺ it becomes BH⁺, its conjugate acid. A conjugate pair differs by exactly one proton, not one electron. In NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, the pairs are NH₄⁺/NH₃ and H₂O/OH⁻. NH₃ and H₂O react with each other but are not a conjugate pair.

Water can donate or accept a proton depending on its partner and is amphiprotic. For HSO₄⁻ + H₂O ⇌ SO₄²⁻ + H₃O⁺, HSO₄⁻ acts as an acid; for HSO₄⁻ + H₃O⁺ ⇌ H₂SO₄ + H₂O it acts as a base. Follow the transferred H and the charge rather than classifying every negatively charged ion as only a base.

HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)
B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)

Strong and concentrated describe different things

A strong acid is essentially completely dissociated in dilute aqueous solution. A weak acid establishes an equilibrium in which only a proportion of its molecules has donated a proton. Concentration instead measures the total amount of solute per volume. A dilute strong acid can have a higher pH than a concentrated weak acid.

For equal analytical concentrations of monobasic acids, a strong acid normally gives a larger [H⁺] than a weak acid. For equal amounts of monobasic acid, however, the same amount of hydroxide is needed for complete neutralisation: removing H⁺ makes more weak acid dissociate. Weak does not mean that some acid can never react.

A strong soluble hydroxide supplies its hydroxide ions essentially completely; a weak base such as ammonia reacts only partly with water. Strong/weak describes extent of reaction with water, whereas soluble/insoluble describes how much material enters solution. Do not use these terms interchangeably.

The pH scale is logarithmic

At this level, pH = −log₁₀[H⁺], using the numerical concentration in mol dm⁻³. The inverse is [H⁺] = 10⁻ᵖᴴ mol dm⁻³. A fall of one pH unit means ten times the hydrogen-ion concentration; pH itself is not a concentration and has no units.

For 0.0300 mol dm⁻³ HNO₃, complete dissociation gives [H⁺] = 0.0300 mol dm⁻³ and pH = −log₁₀(0.0300) = 1.52. Conversely pH 3.40 gives [H⁺] = 10⁻³·⁴⁰ = 3.98 × 10⁻⁴ mol dm⁻³. Use base-ten logarithms here, not the natural logarithm used with Arrhenius or Gibbs energy.

If 20.0 cm³ of 0.150 mol dm⁻³ HCl is diluted to a total volume of 100.0 cm³, n(H⁺) = 0.150 × 0.0200 = 0.00300 mol and [H⁺] = 0.00300/0.1000 = 0.0300 mol dm⁻³, giving pH 1.52. “Add 100 cm³ water” would make a different total volume.

For a strong monoprotic acid, [H⁺] ≈ acid concentration provided its contribution dominates that from water. Do not blindly multiply every diprotic acid concentration by two: sulfuric acid has different successive dissociation behaviour and the assumptions must be stated or supplied. At very high dilution, water prevents an acid solution from becoming alkaline simply because a naive −log c exceeds 7 at 25 °C.

Kw connects hydrogen and hydroxide ions

Water self-ionises according to 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq). Its ionic product is Kw = [H⁺][OH⁻], with units mol² dm⁻⁶ in the concentration convention. At 25 °C, Kw is approximately 1.00 × 10⁻¹⁴ mol² dm⁻⁶. Kw changes with temperature.

A neutral solution has [H⁺] = [OH⁻], so [H⁺] = √Kw and neutral pH = ½pKw. Define pKw = −log₁₀Kw and pKa = −log₁₀Ka using the quoted numerical constants. Neutral pH is 7.00 only when pKw = 14.00; warm pure water can have pH below 7 while remaining neutral.

For 0.0200 mol dm⁻³ NaOH at 25 °C, [OH⁻] = 0.0200, [H⁺] = 1.00 × 10⁻¹⁴/0.0200 = 5.00 × 10⁻¹³ mol dm⁻³ and pH = 12.30. For a fully dissociated M(OH)₂ solution of stated concentration 0.00500 mol dm⁻³, [OH⁻] = 0.0100 mol dm⁻³ and pH = 12.00. Account for the formula before applying Kw.

Equivalently pH + pOH = pKw, where pOH = −log₁₀[OH⁻]. The frequently used 14 − pOH shortcut assumes 25 °C; use the supplied Kw at other temperatures.

Interpret measured pH, including salts

For equimolar solutions at a fixed temperature, compare the extent of ion formation. HCl gives more H⁺ than ethanoic acid; NaOH gives more OH⁻ than ammonia. A pH measurement alone cannot establish strength unless the analytical concentration and conditions are known.

NaCl is approximately neutral because neither ion reacts appreciably with water. Sodium ethanoate is alkaline: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. Ammonium chloride is acidic: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺. The spectator counter-ion balances charge but need not determine pH.

Calibrate a pH probe using appropriate known buffers, rinse between samples and keep temperature comparable. Universal indicator can show broad differences, but a colour category is too coarse for a reliable Ka calculation. A lower recorded pH might reflect a higher concentration, an acidic impurity or calibration error, as well as greater acid strength.

Quick checks

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

Q1. Identify the conjugate pairs in HCO₃⁻ + H₂O ⇌ CO₃²⁻ + H₃O⁺.Show answer

HCO₃⁻/CO₃²⁻ differ by one H⁺, as do H₃O⁺/H₂O. HCO₃⁻ donates the proton in the forward direction.

Q2. Calculate the pH of 0.00400 mol dm⁻³ HCl.Show answer

Assume complete dissociation and negligible water contribution. [H⁺] = 0.00400 mol dm⁻³; pH = −log₁₀(0.00400) = 2.40.

Q3. A solution has pH 4.25. Find [H⁺] and compare it with pH 6.25.Show answer

[H⁺] = 10⁻⁴·²⁵ = 5.62 × 10⁻⁵ mol dm⁻³. It is 100 times the hydrogen-ion concentration at pH 6.25, not twice as large.

Q4. Kw is 4.00 × 10⁻¹⁴ mol² dm⁻⁶ at a stated temperature. Find the pH of neutral water.Show answer

[H⁺] = √Kw = 2.00 × 10⁻⁷ mol dm⁻³. pH = 6.70. It is neutral because [H⁺] equals [OH⁻].

Q5. Explain why 0.100 mol dm⁻³ NH₄Cl is acidic, whereas NaCl is approximately neutral.Show answer

NH₄⁺ can donate a proton to water, forming H₃O⁺. Na⁺ and Cl⁻ do not appreciably produce H₃O⁺ or OH⁻ under these conditions. Being a salt does not guarantee pH 7.

Sources

Sources and examiner guidance (reviewed 9 October 2026)

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