Part 1: Proton transfer, strong acids and pH

Identify conjugate pairs and calculate hydrogen-ion concentration, pH and the effect of dilution.

Follow the proton

A Brønsted–Lowry acid donates a proton, H⁺; a base accepts one. A conjugate acid–base pair differs by exactly one proton, including a one-unit change in charge. Water can act as either partner depending on what it reacts with.

In HCl + H₂O → H₃O⁺ + Cl⁻, HCl donates and water accepts a proton. The conjugate pairs are HCl/Cl⁻ and H₃O⁺/H₂O. In the ammonia equilibrium below, NH₃ accepts a proton from water. Its conjugate acid is NH₄⁺; water’s conjugate base is OH⁻.

H⁺(aq) is the usual A-level shorthand for hydrated hydrogen ions; isolated bare protons do not float independently in water. H₃O⁺ notation can make the proton transfer clearer.

Strength is not concentration

A strong acid is effectively fully ionised in dilute aqueous solution; a weak acid ionises only partly and establishes an equilibrium. Concentration is the amount of acid per unit solution volume. A concentrated weak acid and a dilute strong acid are both possible.

At equal analytical concentration, a strong monoprotic acid normally has a lower pH than a weak monoprotic acid. Across different concentrations, pH alone does not identify acid strength. Solubility is also a separate property: a sparingly soluble base can still dissociate fully in the portion that dissolves.

pH is a logarithmic measure

For the concentration-based A-level treatment, pH = −log₁₀[H⁺], with concentration expressed numerically in mol dm⁻³. The inverse is [H⁺] = 10⁻ᵖᴴ. A fall of one pH unit means ten times the hydrogen-ion concentration, not a fixed extra amount. pH itself has no units.

For [H⁺] = 3.20 × 10⁻³ mol dm⁻³, pH = 2.49. Conversely pH 2.80 gives [H⁺] = 1.58 × 10⁻³ mol dm⁻³. Keep unrounded concentration values until the last step. Give the decimal places requested; two decimal places are commonly required in these calculations.

The familiar 0–14 range is not an absolute mathematical limit. Very concentrated solutions can fall outside it and need more careful activity-based treatment. These notes use the dilute-solution model expected in the supplied A-level examples.

Count the released protons

For a dilute strong monoprotic acid such as HCl, [H⁺] is approximately its analytical concentration, provided the contribution from water is negligible. Thus 0.0250 mol dm⁻³ HCl has pH = −log₁₀(0.0250) = 1.60.

Do not multiply concentration by the number of H atoms in a formula. Only acidic protons matter, and successive dissociations may not be complete. For a polyprotic acid, use the dissociation information supplied rather than automatically assuming all protons are released fully.

Dilute concentration before taking the logarithm

Dilution conserves solute amount if no reaction occurs. Use the final total volume, not merely the volume of water added. For 20.0 cm³ of 0.150 mol dm⁻³ HCl diluted to 250 cm³, cfinal = 0.150 × 20.0/250 = 0.0120 mol dm⁻³ and pH = 1.92.

Diluting an ordinary strong monoprotic acid tenfold raises pH by approximately one unit. At extremely low acid concentration, water’s ionisation is no longer negligible: an acid does not become alkaline simply because −log₁₀ of its analytical concentration exceeds seven at 298 K.

Quick checks

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

Sources