Acids, bases and buffers — OCR A-Level Chemistry

This criterion covers the Brønsted–Lowry definition: an acid is a proton (H⁺) donor and a base is a proton acceptor. A reaction between them involves proton transfer, producing a conjugate acid and a conjugate base that differ by one H⁺. For example, in CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺, ethanoic acid donates a proton to water, so CH₃COOH is the acid, H₂O the base, CH₃COO⁻ the conjugate base and H₃O⁺ the conjugate acid. Recognise that water is amphiprotic, acting as acid or base, and that the definition does not require aqueous conditions.

The Brønsted–Lowry definition classifies a species by what it does with a proton, H⁺. An acid donates a proton to another species; a base accepts a proton from another species. The definition is about behaviour in a reaction, not about a permanent label, so the same species can act as an acid in one reaction and as a base in another. For example, in NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, water donates a proton to ammonia, so water is the acid and ammonia is the base. In HCl + H₂O → H₃O⁺ + Cl⁻, hydrogen chloride donates a proton to water, so HCl is the acid and water is the base. Notice that water changes role between the two reactions. A proton here means the H⁺ ion, a hydrogen atom that has lost its electron.

A conjugate acid–base pair consists of two species that differ by one proton, H⁺. The acid member of the pair has one more proton than the base member. When an acid donates a proton it becomes its conjugate base; when a base accepts a proton it becomes its conjugate acid. In NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ there are two conjugate pairs: NH₄⁺ and NH₃ differ by one proton, and H₂O and OH⁻ differ by one proton. Label each pair by matching the species that gains or loses the single proton, and remember that the pair is written with the acid first by convention. The term lets you describe both directions of a reversible proton transfer without inventing new vocabulary.

Basicity counts the number of moles of H⁺ that one mole of an acid can donate in an acid–base reaction, so it is a property of the acid's formula and structure rather than of its concentration. Hydrochloric acid, HCl, is monobasic because each molecule supplies one H⁺. Sulfuric acid, H₂SO₄, is dibasic because it can release two H⁺, giving HSO₄⁻ and then SO₄²⁻. Phosphoric acid, H₃PO₄, is tribasic because it can release three H⁺, giving H₂PO₄⁻, HPO₄²⁻ and PO₄³⁻. Ethanoic acid, CH₃COOH, is monobasic: only the carboxyl hydrogen is acidic. Compare this with concentration, which is about how much acid is dissolved in a given volume; a dilute dibasic acid can still donate two moles of H⁺ per mole of acid. When neutralising, a dibasic acid needs twice as many moles of hydroxide as a monobasic acid for the same number of moles of acid.

In aqueous solution an acid supplies H⁺(aq), and this ion is the species that reacts in each case. With a reactive metal such as magnesium, the H⁺ ions are reduced to hydrogen gas while the metal is oxidised to its cation: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g). With a metal oxide such as copper(II) oxide, the oxide is a base and the H⁺ neutralises it to form water and the metal cation: CuO(s) + 2H⁺(aq) → Cu²⁺(aq) + H₂O(l). With a carbonate such as calcium carbonate, the H⁺ produces carbon dioxide and water: CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + CO₂(g) + H₂O(l). With an alkali such as sodium hydroxide, the H⁺ is neutralised by OH⁻ to give water: H⁺(aq) + OH⁻(aq) → H₂O(l). In every case the ionic equation shows H⁺ as the reacting acid species, with spectator ions such as chloride omitted.

For a weak acid HA dissociating in water, HA ⇌ H⁺ + A⁻, the acid dissociation constant is Ka = [H⁺][A⁻]/[HA], where concentrations are equilibrium values. A larger Ka means the acid dissociates to a greater extent, so Ka measures acid strength. Because Ka can span many orders of magnitude, it is often expressed as pKa = −log₁₀Ka, so a smaller pKa corresponds to a stronger acid. The value of Ka depends on temperature. Learners should be able to write the Ka expression for a given weak acid, use it to compare the extent of dissociation of different acids, and convert between Ka and pKa.

pKa is a logarithmic scale that makes the wide range of Ka values easier to handle. The relationship is pKa = −log₁₀Ka, and rearranging gives Ka = 10^(−pKa). Because of the negative sign, a larger Ka gives a smaller pKa, so a stronger acid has the lower pKa. For example, if Ka = 1.8 × 10⁻⁵ mol dm⁻³ then pKa = −log₁₀(1.8 × 10⁻⁵) ≈ 4.74. In an MCQ, convert between the two forms carefully, keep the negative sign, and use the inverse relationship to rank acid strength: lowest pKa means strongest acid.

This topic links the hydrogen ion concentration of an aqueous solution to its pH through the logarithmic relationship pH = –log₁₀[H⁺(aq)]. Because the scale is logarithmic, a tenfold change in [H⁺(aq)] shifts pH by one unit: lowering [H⁺(aq)] from 1.0 × 10⁻² to 1.0 × 10⁻³ mol dm⁻³ raises pH from 2.00 to 3.00. Work in mol dm⁻³, take log₁₀ of the concentration, then apply the negative sign. For a solution with [H⁺(aq)] = 1.0 × 10⁻³ mol dm⁻³, log₁₀(1.0 × 10⁻³) = –3, so pH = –(–3) = 3.00. Reverse the process with [H⁺(aq)] = 10^(–pH). Recognise that acidic solutions have pH below 7 at 298 K and that a smaller [H⁺(aq)] gives a larger pH. Keep significant figures sensible: pH quoted to two decimal places corresponds to concentration to two significant figures.

This criterion requires confident use of both forms of the pH expression: pH = –log₁₀[H⁺(aq)] and its rearrangement [H⁺(aq)] = 10^(–pH). The logarithm is base 10, and the negative sign converts the negative logarithm of a dilute acid into a positive pH. For [H⁺(aq)] = 1.0 × 10⁻² mol dm⁻³, log₁₀(1.0 × 10⁻²) = –2, so pH = –(–2) = 2.00. Reversing, a pH of 2.00 gives [H⁺(aq)] = 10^(–2.00) = 1.0 × 10⁻² mol dm⁻³. Note that for concentrations below 1 mol dm⁻³ the logarithm is negative, so omitting the minus sign would report a negative pH for an ordinary acid. Practise both directions, keep concentrations in mol dm⁻³, and quote pH to two decimal places.

Kw is the equilibrium constant for the self-ionisation of water, H₂O ⇌ H⁺ + OH⁻, written Kw = [H⁺][OH⁻]. At 298 K its value is 1.00 × 10⁻¹⁴ mol² dm⁻⁶, so in pure water [H⁺] = [OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³ and pH = 7.00. Because the self-ionisation is endothermic, Kw increases as temperature rises, so neutral water at a higher temperature has a lower pH than 7 even though [H⁺] = [OH⁻]. Use Kw to find [H⁺] from a known [OH⁻] in a strong base, or to find [OH⁻] from a known [H⁺] in a strong acid, then convert to pH. Remember that Kw is constant at a fixed temperature and that the units are mol² dm⁻⁶.

pH is defined as pH = −log₁₀[H⁺], where [H⁺] is in mol dm⁻³, and [H⁺] = 10^(−pH). For a strong monoprotic acid such as HCl the acid is fully dissociated, so [H⁺] equals the acid concentration and pH follows directly. For a strong base, first find [OH⁻] from the base concentration, then use Kw = [H⁺][OH⁻] to obtain [H⁺] before taking the logarithm. For a weak acid, use Ka with the approximation [H⁺] ≈ √(Ka × [acid]) when dissociation is small, or solve the equilibrium expression more fully. Related quantities include [H⁺], [OH⁻], pOH and Ka, so choose the correct relationship for the species given.

A strong monobasic acid releases exactly one proton per molecule and ionises essentially completely in water, so the acid is treated as the sole source of H⁺ and the equilibrium lies far to the right. For a solution of concentration c, [H⁺] ≈ c, giving pH = −log₁₀[H⁺]; for example 0.010 mol dm⁻³ HCl gives [H⁺] = 0.010 mol dm⁻³ and pH = 2.00. The label monobasic matters: one mole of acid supplies one mole of H⁺, so no factor of two appears. Learners must distinguish strong (full ionisation) from concentrated (many moles per dm³), and recognise that dilution by a factor of ten raises pH by one unit. Weak monobasic acids such as ethanoic acid ionise only partially, so [H⁺] is far below c and needs Ka; the strong-acid shortcut does not apply to them.

For a strong base such as NaOH or KOH, the hydroxide ion concentration equals the concentration of the base because each formula unit dissociates fully. Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K. Rearranging gives [H⁺] = Kw ÷ [OH⁻], and pH = −log₁₀[H⁺]. For 0.0100 mol dm⁻³ NaOH, [OH⁻] = 0.0100 mol dm⁻³, so [H⁺] = 1.00 × 10⁻¹⁴ ÷ 0.0100 = 1.00 × 10⁻¹² mol dm⁻³ and pH = 12.00. For a diprotic strong base such as Ba(OH)₂, [OH⁻] is twice the base concentration. Remember that Kw changes with temperature, so use the value quoted for the stated temperature.

For a weak monobasic acid HA, the acid-dissociation constant is Ka = [H⁺][A⁻]/[HA]. The approximation treats the undissociated concentration as essentially the initial concentration, so [HA] ≈ [HA]₀, and assumes [H⁺] ≈ [A⁻] because the acid is the main source of protons and autoionisation of water is negligible. Combining these gives [H⁺] ≈ √(Ka × [HA]₀), so pH = −log₁₀[H⁺] and pKa = −log₁₀Ka. For example, with Ka = 1.8 × 10⁻⁵ mol dm⁻³ and [HA]₀ = 0.10 mol dm⁻³, [H⁺] ≈ √(1.8 × 10⁻⁶) ≈ 1.34 × 10⁻³ mol dm⁻³ and pH ≈ 2.87. Rearranged forms let you find Ka from pH or find the concentration needed for a target pH. Always keep units consistent and check that the answer is physically sensible for a weak acid.

The approximations [HA] ≈ [HA]₀ and [H⁺] ≈ [A⁻] work well only when the acid is weak enough that a small fraction dissociates. For a ‘stronger’ weak acid, with a larger Ka and a greater degree of dissociation, the amount of HA that has dissociated becomes a significant fraction of the initial concentration, so [HA] is noticeably less than [HA]₀. The assumption that [H⁺] equals [A⁻] also weakens because protons from water autoionisation and other sources matter more at higher dilution. Using [H⁺] ≈ √(Ka × [HA]₀) then overestimates [H⁺] and underestimates pH. The exact treatment solves the quadratic from Ka = x²/([HA]₀ − x), where x = [H⁺]. Recognise when the approximation is unsafe and say why.

A buffer is a solution that resists pH change when small amounts of acid or base are added. An acidic buffer is made from a weak acid and its conjugate base, for example ethanoic acid and sodium ethanoate; a basic buffer uses a weak base and its conjugate acid, such as ammonia and ammonium chloride. The weak acid supplies a reservoir of undissociated molecules and the salt supplies the conjugate base. Added H⁺ ions are removed by the conjugate base, and added OH⁻ ions are removed by the weak acid, so [H⁺] stays nearly constant. Calculations use Ka = [H⁺][A⁻]/[HA], rearranged to [H⁺] = Ka × [HA]/[A⁻], then pH = −log₁₀[H⁺]. Buffers are used in blood, in shampoos and in biochemical work.

A buffer solution is defined as a system that minimises pH changes when small amounts of an acid or a base are added. The key idea is resistance, not complete prevention: the pH shifts only slightly because the solution contains reservoirs that remove added H⁺ or OH⁻ ions. In an acidic buffer, the weak acid removes added OH⁻ ions and the conjugate base removes added H⁺ ions. In a basic buffer, the weak base removes added H⁺ ions and its conjugate acid removes added OH⁻ ions. The buffer works until one reservoir is exhausted, after which the pH changes sharply. Recognise that the definition applies to both acidic and basic buffers and to small additions only.

A buffer resists pH change when small amounts of acid or alkali are added, because it contains a reservoir of both a weak acid and its conjugate base. The specification introduces the formation of buffers from two routes, and this statement is the umbrella heading for both. Route one mixes a weak acid with a salt of that weak acid, for example CH₃COOH with CH₃COONa. Route two makes the conjugate base in situ by partial neutralisation: an excess of weak acid reacts with a limited amount of strong alkali, so some weak acid remains alongside the salt formed. In both routes the mixture supplies undissociated weak acid to remove added OH⁻ ions and conjugate base to remove added H⁺ ions, so the pH stays nearly constant.

This route forms a buffer by dissolving a weak acid together with a salt of that same weak acid. Ethanoic acid, CH₃COOH, is only partially dissociated, so the mixture contains a substantial reservoir of undissociated CH₃COOH molecules. Sodium ethanoate, CH₃COONa, is a soluble ionic salt that dissociates fully to give CH₃COO⁻ ions and Na⁺ ions. The result is a solution holding both the weak acid and its conjugate base, CH₃COO⁻, in appreciable amounts. Added H⁺ ions are removed by the CH₃COO⁻ ions, and added OH⁻ ions are removed by the CH₃COOH molecules, so the pH changes only slightly. The salt must contain the conjugate base of the same weak acid; a salt of a different acid does not give this buffer pair.

This item concerns a mixture in which a weak acid such as CH₃COOH is present in excess and a strong alkali such as NaOH is the limiting reagent. The strong alkali is fully ionised, so its OH⁻ ions are neutralised almost completely by the weak acid, which is only partially dissociated. Because unreacted weak acid remains in large amount, the resulting solution is a buffer: the undissociated CH₃COOH and its conjugate base CH₃COO⁻ form a conjugate acid–base pair that resists pH change. The pH lies in the acidic region, close to the pKa of the weak acid, not in the alkaline region, because the excess weak acid dominates. Learners should reason from amounts in moles, not from the labels 'acid' and 'alkali', and should recognise that the strong alkali is used up while the weak acid survives.

An acid buffer contains a weak acid and its conjugate base in comparable amounts, for example CH₃COOH and CH₃COO⁻ supplied as the sodium salt. The weak acid is only partially dissociated, so a reservoir of undissociated CH₃COOH exists alongside a reservoir of CH₃COO⁻. Added H⁺ ions are removed by the conjugate base: CH₃COO⁻ + H⁺ → CH₃COOH. Added OH⁻ ions are removed by the weak acid: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. In each case the added ion is consumed by one member of the pair and replaced by the other, so the ratio of the two reservoirs changes only slightly and the pH stays nearly constant. A strong acid or strong alkali has no such reservoir and its pH changes sharply.

A buffer resists pH change because a weak acid HA and its conjugate base A⁻ are both present in significant amount. The acid dissociation constant Ka = [H⁺][A⁻]/[HA] rearranges to [H⁺] = Ka × [HA]/[A⁻]. Taking negative logarithms gives pH = pKa + log([A⁻]/[HA]), where pKa = −log Ka. In a typical buffer made from a weak acid partly neutralised by strong base, [HA] and [A⁻] come from the amounts present after reaction, divided by the total volume. Because both concentrations share the same volume, the ratio [A⁻]/[HA] equals the ratio of amounts in moles, so volume often cancels. Related quantities include [H⁺], pKa, Ka, the ratio of acid to base, and the pH after adding small amounts of acid or base. Worked example: a buffer with [HA] = 0.10 mol dm⁻³ and [A⁻] = 0.20 mol dm⁻³ and Ka = 1.8 × 10⁻⁵ mol dm⁻³ gives [H⁺] = 1.8 × 10⁻⁵ × 0.10/0.20 = 9.0 × 10⁻⁶ mol dm⁻³, so pH = 5.05.

Blood pH is held close to 7.4 by a buffer containing carbonic acid, H₂CO₃, and hydrogencarbonate ions, HCO₃⁻. Carbonic acid is a weak acid that dissociates only partially: H₂CO₃ ⇌ H⁺ + HCO₃⁻. The mixture of the weak acid and its conjugate base resists pH change. When extra H⁺ enters the blood, for example from exercise or metabolism, the H⁺ reacts with HCO₃⁻ to form more H₂CO₃, removing the added acid and shifting the equilibrium to the left. When OH⁻ or a base enters, it reacts with H₂CO₃, which donates H⁺ and forms more HCO₃⁻, shifting the equilibrium to the right. Because both components are present in significant amounts, the ratio [HCO₃⁻]/[H₂CO₃] changes only slightly, so [H⁺] and hence pH stay nearly constant. The lungs remove CO₂ and the kidneys regulate HCO₃⁻, linking the buffer to respiration and excretion.

Neutralisation is the reaction of an acid with a base to form a salt and water; in ionic terms H⁺ from the acid reacts with OH⁻ from the base to form H₂O. For a strong acid and strong base the reaction goes essentially to completion. When one partner is weak, the strong partner still reacts essentially completely: the weak acid or weak base equilibrium shifts to replace the species being removed, so neutralisation is not less complete. For example, ethanoic acid with sodium hydroxide forms sodium ethanoate and water, and the ethanoic acid equilibrium shifts as H⁺ is removed. The enthalpy change of neutralisation for strong acid with strong base is about −57.1 kJ mol⁻¹ because the reaction is essentially H⁺ + OH⁻ → H₂O.

A pH titration curve plots pH against the volume of titrant added during a titration. The shape depends on whether the acid and base are strong or weak. Strong acid with strong base gives a sharp, near-vertical jump centred near pH 7. Strong acid with weak base starts low and the jump is shorter and lower, ending below pH 7. Weak acid with strong base starts higher, shows a buffer region and an equivalence point above pH 7. Weak acid with weak base gives a small, gradual change with no sharp jump. Learners should recognise the starting pH, the equivalence point and the steep section for each combination, and link the curve to the relative strengths and the salt formed.

In this buffer context you sketch and then interpret titration curves. A sketch plots pH on the vertical axis against volume of added acid or base on the horizontal axis, marking the start pH, the buffer region, the equivalence point and the final plateau. Interpretation means reading those features back to the chemistry: a weak acid–strong base titration begins above pH 1, shows a shallow buffer region where HA and A⁻ coexist, and has an equivalence pH above 7 because the salt hydrolyses. A strong acid–strong base curve starts near pH 1 and has equivalence at pH 7. You justify each feature using Ka, the buffer equation and the dominant species, and you label axes with quantity and unit.

An indicator is a weak acid whose two coloured forms change ratio over a narrow pH range, its end point range. To choose a suitable indicator you compare that range with the steep vertical section of the titration curve, which contains the equivalence point. If the indicator range falls within the steep section, the colour change coincides with the equivalence point and the titration is accurate. For a strong acid–strong base titration the steep section spans roughly pH 3 to 11, so phenolphthalein (about pH 8.2–10.0) or methyl orange (about pH 3.1–4.4) both work. For a weak acid–strong base titration the equivalence pH is above 7 and the steep section is higher, so phenolphthalein suits but methyl orange does not.

An acid–base indicator is itself a weak acid, written HA, that dissociates in water: HA ⇌ H⁺ + A⁻. The two forms have different colours, so the colour you see depends on the ratio [HA]/[A⁻]. Adding acid raises [H⁺]; by Le Chatelier's principle the position of equilibrium shifts to the left, towards HA, so the colour of the HA form dominates. Adding alkali removes H⁺ (neutralising it), so equilibrium shifts to the right, towards A⁻, and the colour of the A⁻ form dominates. The indicator changes colour over a pH range roughly centred on its pKa, because the ratio changes rapidly there. For a titration, choose an indicator whose end-point range lies within the steep section of the titration curve.

A pH meter measures [H⁺] and displays pH. Correct procedure: calibrate the meter with at least two buffer solutions of known pH that bracket the expected value, rinsing the electrode with distilled water between buffers and blotting gently. Rinse the electrode with distilled water and blot it before each measurement. Place the electrode in the sample, stir gently and wait for a stable reading before recording the pH to the precision the meter allows. Rinse the electrode with distilled water after use and store it in the recommended solution. Keep the bulb wet and handle it carefully. Record the temperature if the meter requires it, since pH depends on temperature.