OCR A Level Chemistry A: Acids, bases and buffers (H432)

This guide teaches section 5.1.3, Acids, bases and buffers, from the H432 specification for OCR A Level Chemistry A (Version 3.1, May 2026; first assessment 2017). Its home is Module 5 (Physical chemistry and transition elements); this page works through each lettered outcome, 5.1.3(a) to 5.1.3(o), of the full H432 A level. The assessment overview places Module 5 in Component 01, and Component 03 draws on every module.

It builds on 2.1.4 Acids and Amount of substance. Afterwards, use the revision notes and the practice questions. The course hub lists every topic, the OCR chemistry checklist is printable, and a free diagnostic finds your weakest areas.

Coverage table

Specification statement	Skill
5.1.3(a)	Brønsted–Lowry acids and bases; conjugate pairs; basicity
5.1.3(b)	Ionic equations for H⁺ with metals and bases
5.1.3(c)	Ka expressions; Ka and pKa
5.1.3(d)	pH = –log[H⁺] and [H⁺] = 10⁻ᵖᴴ
5.1.3(e)	The ionic product of water, Kw
5.1.3(f)	pH of strong monobasic acids and strong bases
5.1.3(g)	pH or Ka of a weak monobasic acid, using approximations
5.1.3(h)	Limits of the approximations for “stronger” weak acids
5.1.3(i)–(k)	What a buffer is, how to make one, how it works
5.1.3(l)	Buffer pH and related quantities
5.1.3(m)	Control of blood pH
5.1.3(n)	pH titration curves and indicators
5.1.3(o)	Using a pH meter (PAG11)

Brønsted–Lowry acids and bases: 5.1.3(a)

A Brønsted–Lowry acid is a species that donates a proton (H⁺). A Brønsted–Lowry base is a species that accepts a proton. The earlier model from 2.1.4 linked acids to H⁺ ions and alkalis to OH⁻ ions. The proton-transfer model is wider, covering bases such as NH₃ that contain no OH⁻: a model developed over time.

When an acid loses H⁺, what remains is its conjugate base; the two form a conjugate acid–base pair and differ by exactly one H⁺.

Worked example. Identify the conjugate pairs in this equilibrium.

NH₄⁺ + CO₃²⁻ ⇌ NH₃ + HCO₃⁻

• NH₄⁺ donates H⁺ and becomes NH₃. Pair 1: acid NH₄⁺, base NH₃.
• CO₃²⁻ accepts H⁺ and becomes HCO₃⁻. Pair 2: acid HCO₃⁻, base CO₃²⁻.

Basicity is the number of protons one molecule of the acid can donate:

• monobasic: HCl, HNO₃, CH₃COOH (only the H in COOH is donated, not the three C–H hydrogens)
• dibasic: H₂SO₄
• tribasic: H₃PO₄

The role of H⁺ in acid reactions: 5.1.3(b)

The reacting particle is always H⁺(aq); ionic equations leave out spectator ions.

• Metal: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g). This is also redox.
• Carbonate: CO₃²⁻(aq) + 2H⁺(aq) → CO₂(g) + H₂O(l). An insoluble carbonate stays as a solid formula: MgCO₃(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂O(l) + CO₂(g).
• Metal oxide: CuO(s) + 2H⁺(aq) → Cu²⁺(aq) + H₂O(l)
• Alkali: H⁺(aq) + OH⁻(aq) → H₂O(l)

Ka and pKa: 5.1.3(c)

A weak acid HA only partly dissociates: HA(aq) ⇌ H⁺(aq) + A⁻(aq). The equilibrium constant, the acid dissociation constant, measures the extent of dissociation:

Ka = [H⁺(aq)][A⁻(aq)] / [HA(aq)] (units mol dm⁻³)

Bigger Ka means a stronger acid. Small Ka values are often converted to pKa:

pKa = –log Ka and Ka = 10⁻ᵖᴷᵃ

A smaller pKa means a stronger acid.

Worked example. Propanoic acid has Ka = 1.35 × 10⁻⁵ mol dm⁻³. pKa = –log(1.35 × 10⁻⁵) = 4.87. A second acid has pKa = 4.56, so Ka = 10⁻⁴·⁵⁶ = 2.75 × 10⁻⁵ mol dm⁻³. It is the stronger acid.

pH, Kw and strong acids and bases: 5.1.3(d)–(f)

pH is a convenient way to report [H⁺]:

pH = –log[H⁺] and [H⁺] = 10⁻ᵖᴴ

A tiny fraction of water molecules split into ions, H₂O(l) ⇌ H⁺(aq) + OH⁻(aq), and the product of the two ion concentrations is a constant called Kw:

Kw = [H⁺(aq)][OH⁻(aq)] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K

Kw varies with temperature, but pure water always has [H⁺] = [OH⁻], so it is neutral even when its pH is not 7.

Strong monobasic acid: [H⁺] = [acid]. For 0.0320 mol dm⁻³ HNO₃: pH = –log(0.0320) = 1.49.

Strong base: work out [OH⁻] first, then use Kw. For 0.0560 mol dm⁻³ KOH: [OH⁻] = 0.0560 mol dm⁻³ [H⁺] = Kw / [OH⁻] = 1.00 × 10⁻¹⁴ / 0.0560 = 1.79 × 10⁻¹³ mol dm⁻³ pH = –log(1.79 × 10⁻¹³) = 12.75

For Ba(OH)₂, double the concentration to get [OH⁻].

Weak acid calculations: 5.1.3(g)

The specification gives two approximations:

1. [H⁺]equilibrium ≈ [A⁻]equilibrium: the dissociation of water adds a negligible amount of H⁺.
2. [HA]equilibrium ≈ [HA]undissociated: so little HA dissociates that its concentration hardly changes.

With both in place the expression simplifies to Ka = [H⁺]² / [HA], which rearranges to [H⁺] = √(Ka × [HA]). Quadratic equations are not required.

Worked example: pH from Ka. Find the pH of 0.150 mol dm⁻³ butanoic acid, Ka = 1.48 × 10⁻⁵ mol dm⁻³. [H⁺] = √(1.48 × 10⁻⁵ × 0.150) = √(2.22 × 10⁻⁶) = 1.49 × 10⁻³ mol dm⁻³ pH = –log(1.49 × 10⁻³) = 2.83

Worked example: Ka from pH. A weak monobasic acid HT at 0.0650 mol dm⁻³ has pH 3.02. [H⁺] = 10⁻³·⁰² = 9.55 × 10⁻⁴ mol dm⁻³ Ka = [H⁺]² / [HT] = (9.55 × 10⁻⁴)² / 0.0650 = 1.40 × 10⁻⁵ mol dm⁻³

When the approximations fail: 5.1.3(h)

[HA]equilibrium ≈ [HA]undissociated only works while the fraction dissociated is small. A “stronger” weak acid (larger Ka), especially when dilute, dissociates enough for [HA]equilibrium to fall clearly below the starting value.

Worked example. Chloroethanoic acid, Ka = 1.36 × 10⁻³ mol dm⁻³, at 0.0100 mol dm⁻³. Approximate [H⁺] = √(1.36 × 10⁻³ × 0.0100) = 3.69 × 10⁻³ mol dm⁻³, pH 2.43. That answer implies about 37% of the acid has dissociated, which contradicts the assumption that hardly any has. The real [HA] is smaller, so the real [H⁺] is smaller too (about 3.07 × 10⁻³ mol dm⁻³, pH 2.51, found by a method you will not be asked to use). The approximation overestimates [H⁺] and underestimates pH.

The other approximation, [H⁺] = [A⁻], breaks down when the acid is so weak or so dilute that H⁺ from water is no longer negligible.

Buffers: 5.1.3(i)–(m)

A buffer solution is a system that minimises pH changes on addition of small amounts of an acid or a base.

Making an acid buffer (j)

• (j)(i) Mix a weak acid with a salt of that acid, e.g. CH₃COOH with CH₃COONa. The salt fully dissociates and supplies a large [A⁻].
• (j)(ii) Add a strong alkali to an excess of weak acid, e.g. excess CH₃COOH with NaOH. Part of the acid becomes salt; the rest stays as HA.

How the conjugate pair controls pH (k)

For CH₃COOH/CH₃COO⁻ the equilibrium is CH₃COOH ⇌ H⁺ + CH₃COO⁻, and the buffer holds a large reservoir of each partner.

• Small amount of acid added: [H⁺] rises. The extra H⁺ combines with the conjugate base, CH₃COO⁻, so the equilibrium moves to the left and most of the added H⁺ is used up.
• Small amount of alkali added: OH⁻ reacts with H⁺ to form water. More CH₃COOH dissociates, moving the equilibrium to the right and replacing the lost H⁺.

Because both reservoirs are large, a small addition barely alters the [HA] : [A⁻] ratio, so [H⁺] and pH hardly move.

Buffer calculations (l)

Rearrange Ka: [H⁺] = Ka × [HA] / [A⁻], then pH = –log[H⁺]. When [HA] = [A⁻], pH = pKa.

Worked example (j)(i). A buffer is 0.180 mol dm⁻³ in propanoic acid and 0.240 mol dm⁻³ in sodium propanoate. Ka = 1.35 × 10⁻⁵ mol dm⁻³. [H⁺] = 1.35 × 10⁻⁵ × 0.180 / 0.240 = 1.01 × 10⁻⁵ mol dm⁻³ pH = 4.99

Worked example (j)(ii). 40.0 cm³ of 0.250 mol dm⁻³ butanoic acid (Ka = 1.48 × 10⁻⁵ mol dm⁻³) is mixed with 25.0 cm³ of 0.160 mol dm⁻³ NaOH.

• n(acid) = 0.0400 × 0.250 = 0.0100 mol; n(OH⁻) = 0.0250 × 0.160 = 0.00400 mol
• After reaction: n(HA) = 0.00600 mol, n(A⁻) = 0.00400 mol
• Both are in the same 65.0 cm³, so the mole ratio can replace the concentration ratio: [H⁺] = 1.48 × 10⁻⁵ × 0.00600 / 0.00400 = 2.22 × 10⁻⁵ mol dm⁻³
• pH = 4.65

“Related quantities” include the ratio, amount or mass of salt needed for a target pH. Work backwards: find [H⁺] = 10⁻ᵖᴴ, then [A⁻] = Ka × [HA] / [H⁺].

Blood pH (m)

In blood plasma, a carbonic acid–hydrogencarbonate buffer keeps the pH between 7.35 and 7.45:

H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq)

• Extra acid: H⁺ combines with HCO₃⁻, shifting the equilibrium left and forming H₂CO₃. The body converts H₂CO₃ to CO₂ and H₂O, and the CO₂ is breathed out.
• Extra alkali: OH⁻ removes H⁺; H₂CO₃ dissociates, shifting the equilibrium right to replace it.

pH titration curves and indicators: 5.1.3(n)

A pH titration curve plots pH against volume added. To sketch one, mark the starting pH, the equivalence volume and the steep part, then level off towards the pH of the excess reagent. For base added to acid, both at roughly 0.1 mol dm⁻³:

Acid in flask, base from burette	Starting pH	Steep section spans	pH where the moles match
strong acid with strong base	near 1	about pH 3 up to 11	7
weak acid with strong base	near 3	about pH 7 up to 11	above 7
strong acid with weak base	near 1	about pH 3 up to 7	below 7
weak acid with weak base	near 3	no sharp steep section	depends on the relative strengths

A weak acid–strong base curve rises gently through a buffer region before equivalence. Halfway to equivalence [HA] = [A⁻], so pH = pKa.

Choosing an indicator

Pick an indicator whose entire colour-change range falls inside the steep section of the curve. Commonly quoted ranges are about 3.1–4.4 for methyl orange and about 8.3–10.0 for phenolphthalein, but always use the ranges a question supplies.

• strong acid–strong base: either works
• weak acid–strong base: phenolphthalein
• strong acid–weak base: methyl orange
• weak acid–weak base: no indicator is suitable, because there is no sharp steep section

Why an indicator changes colour

Treat the indicator as a weak acid, HIn, where HIn and In⁻ have different colours:

HIn(aq) ⇌ H⁺(aq) + In⁻(aq)

• Adding acid raises [H⁺]; equilibrium shifts left, showing the colour of HIn.
• Adding alkali removes H⁺ (OH⁻ + H⁺ → H₂O); equilibrium shifts right, showing the colour of In⁻.
• At the end point, [HIn] = [In⁻] and you see a mixture of the two colours.

Measuring pH with a pH meter: 5.1.3(o)

PAG11 covers pH measurement, for example identifying unknown solutions. Good technique:

1. Calibrate the meter with buffer solutions of known pH (for example pH 4, 7 and 10).
2. Rinse the electrode with deionised water between solutions and blot dry, to avoid carry-over.
3. Immerse the bulb fully, stir gently and wait until the reading is steady before recording it.
4. Keep the temperature constant, since pH depends on temperature.
5. Store the electrode wet, in its storage solution.

Common errors

• Writing pH = –log[OH⁻] for a strong base instead of using Kw.
• In a (j)(ii) buffer, using the starting moles of acid, not the moles left.
• Saying a buffer keeps the pH constant: it only minimises the change.

Official syllabus

Source document: OCR A Level Chemistry A (code H432), specification Version 3.1 dated May 2026; first assessment 2017. This page follows section 5.1.3 Acids, bases and buffers, in Module 5: Physical chemistry and transition elements.