Acids — OCR A-Level Chemistry

This section introduces the Brønsted–Lowry model: an acid is a proton donor and a base is a proton acceptor. An alkali is a base that dissolves in water to release hydroxide ions, so all alkalis are bases but not all bases are alkalis. Neutralisation occurs when an acid reacts with a base to form a salt and water, and in terms of ions the hydrogen ions from the acid combine with hydroxide ions to form water. Read the specification statements alongside your textbook and class notes, and check that you can write balanced equations for reactions such as hydrochloric acid with sodium hydroxide, and identify the acid, base and salt in each case. Use the reading to build a clear distinction between the general term base and the more specific term alkali, and to practise writing ionic equations for neutralisation.

Acids and alkalis are identified by their formulae and by the ions they release in water. Learn the four common acids: hydrochloric acid HCl, sulfuric acid H₂SO₄, nitric acid HNO₃ and ethanoic acid CH₃COOH. Learn the three common alkalis: sodium hydroxide NaOH, potassium hydroxide KOH and ammonia NH₃. In aqueous solution an acid releases H⁺ ions, so HCl gives H⁺ and Cl⁻, and H₂SO₄ gives 2H⁺ and SO₄²⁻. An alkali releases OH⁻ ions in aqueous solution: NaOH gives Na⁺ and OH⁻, KOH gives K⁺ and OH⁻. Ammonia is the exception to watch: NH₃ accepts H⁺ from water to form NH₄⁺ and OH⁻, so the OH⁻ ions appear indirectly. A multiple-choice question may ask which species is released, which formula is an alkali, or which ion is present in excess.

A strong acid dissociates almost completely in aqueous solution, so nearly every acid molecule releases its H⁺ ions. Hydrochloric acid is a strong acid: HCl → H⁺ + Cl⁻, and the equilibrium lies far to the right. A weak acid dissociates only partially, so most molecules remain undissociated and only a small fraction release H⁺ ions. Ethanoic acid is a weak acid: CH₃COOH ⇌ CH₃COO⁻ + H⁺, with the equilibrium lying to the left. The qualitative explanation compares the extent of dissociation, not the concentration of the solution. A dilute strong acid can have a lower H⁺ concentration than a concentrated weak acid, so keep the two ideas separate. A multiple-choice question may ask which acid is strong, which equilibrium position applies, or how the relative numbers of ions compare.

This row is a heading that introduces the sub-statement (i), so read it as the frame for the whole neutralisation idea rather than as a standalone fact. In OCR A Level Chemistry (H432) section 2.1.4 Acids, neutralisation is defined by what reacts: an acid supplies H⁺ ions and a base supplies OH⁻ ions, and those ions combine. The heading tells you the specification defines neutralisation by the reacting species, not by a colour change or a pH value. When you revise, build the ionic equation H⁺ + OH⁻ → H₂O and be ready to apply it to strong and weak acids and to alkalis such as NaOH and NH₃. Treat the sub-statement (i) as the assessed content and use this heading to organise your notes around the ions involved.

This sub-statement requires you to know that in neutralisation, hydrogen ions from the acid react with hydroxide ions from the base to form water. The ionic equation is H⁺ + OH⁻ → H₂O, and it is the same for any strong acid reacting with any strong alkali because the other ions are spectators. For example, hydrochloric acid and sodium hydroxide give Na⁺ and Cl⁻ as spectator ions, so the ionic equation is H⁺ + OH⁻ → H₂O. A multiple-choice question may ask you to pick the correct ionic equation, identify the spectator ions, or recognise that the reaction is H⁺ donating to OH⁻. Check the charges balance: one 1⁺ ion plus one 1⁻ ion gives a neutral product.

An acid reacts with a base to give a salt and, in most cases, water. A base is a proton acceptor; an alkali is a base that dissolves in water, so all alkalis are bases but not all bases are alkalis. With a metal oxide such as copper(II) oxide, the acid is neutralised: 2HCl(aq) + CuO(s) → CuCl₂(aq) + H₂O(l). With an alkali such as sodium hydroxide, the ionic equation is H⁺(aq) + OH⁻(aq) → H₂O(l). With a carbonate such as calcium carbonate, the acid also releases carbon dioxide: 2HCl(aq) + CaCO₃(s) → CaCl₂(aq) + H₂O(l) + CO₂(g). Full equations must balance in atoms and charge, and the salt formula follows from the ions involved, for example Al³⁺ with SO₄²⁻ gives Al₂(SO₄)₃.

An acid–base titration uses a burette to add a solution of known concentration to a measured volume of the other solution until the reaction is just complete, marked by an indicator colour change or a pH meter. A suitable indicator changes colour close to the equivalence point; phenolphthalein suits a strong acid with a strong base, while methyl orange suits a strong acid with a weak base. The result allows calculation of an unknown concentration: moles = concentration × volume in dm³, and the balanced equation gives the reacting ratio. For example, if 25.0 cm³ of 0.100 mol dm⁻³ NaOH needs 20.0 cm³ of HCl, then moles of NaOH = 0.100 × 0.0250 = 2.50 × 10⁻³ mol, and the 1:1 equation gives the acid concentration as 2.50 × 10⁻³ ÷ 0.0200 = 0.125 mol dm⁻³.

This point requires practical technique, not just description. For a standard solution, weigh the solid accurately, dissolve it in a small volume of distilled water in a beaker, transfer the solution and rinsings through a funnel into a volumetric flask, then add distilled water to the graduation mark and invert to mix. For a titration, rinse and fill the burette with the titrant, pipette a measured volume of the analyte into a conical flask, and add the indicator before you begin. Run titrant in, swirling continuously, and add it dropwise as the end point approaches. Record the volume at the first permanent colour change and repeat for concordant results.

Titration calculations start from the mean titre in cm³, converted to dm³ by dividing by 1000, and the known concentration of the standard solution, giving amount in moles by n = c × V. The balanced equation gives the mole ratio between the two reactants, so the amount of the unknown is found. Dividing by its volume in dm³ gives its concentration; multiplying by its molar mass gives mass concentration if needed. Structured questions lead you through these steps, while non-structured questions require you to choose the route yourself. The method works for familiar acids and bases such as hydrochloric acid and sodium hydroxide and for unfamiliar ones, because only the balanced equation and the data change.