AQA A-Level Chemistry: Acids and bases (7405)

This guide teaches section 3.1.12 Acids and bases of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for exams from June 2016 onwards: 3.1.12.1 to 3.1.12.6 and Required practical 9. The whole section is A-level only. The specification places it on A-level Paper 1, and Paper 3 can test any content.

For quick recall, use the acids and bases revision notes. To test yourself, use the acids and bases practice questions. Kw and Ka are equilibrium constants, so revise equilibria and Kc first if Kc is shaky. Titration arithmetic uses the mole revision notes. The course hub is AQA A-Level Chemistry, the printable checklist lists every specification point, and the free 10-minute diagnostics show where to start.

What this unit covers

Spec point	What you must be able to do	Status
3.1.12.1	Define acids as proton donors and bases as proton acceptors; show proton transfer in acid–base equilibria	A-level only
3.1.12.2	Use pH = −log₁₀[H⁺]; convert [H⁺] to pH and back; find the pH of a strong acid	A-level only
3.1.12.3	Derive Kw = [H⁺][OH⁻]; explain that Kw varies with temperature; find the pH of a strong base	A-level only
3.1.12.4	Write Ka; relate pH, acid concentration and Ka; convert Ka and pKa	A-level only
3.1.12.5	Titration calculations; sketch and explain pH curves; choose an indicator	A-level only
Required practical 9	Follow pH as a weak acid reacts with a strong base, and a strong acid with a weak base	A-level only
3.1.12.6	Explain acidic and basic buffers; give applications; calculate the pH of acidic buffers	A-level only

3.1.12.1 Brønsted–Lowry acids and bases

• An acid is a proton donor.
• A base is a proton acceptor.
• Acid–base equilibria involve the transfer of protons (H⁺ ions).

HNO₃ + H₂O → H₃O⁺ + NO₃⁻
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

In the first equation HNO₃ donates a proton (acid) and water accepts it (base). In the second, water donates a proton to NH₃, so water is the acid and NH₃ the base.

In the reverse of the ammonia reaction, NH₄⁺ donates a proton to OH⁻, so NH₄⁺ is the acid and OH⁻ the base. Look at both directions of an equilibrium.

3.1.12.2 pH and how to find it

[H⁺] in aqueous solution covers a very wide range, so a logarithmic scale is used:

pH = −log₁₀[H⁺]          [H⁺] = 10^(−pH)

A fall of one pH unit means [H⁺] is ten times larger.

A strong acid such as HCl or HNO₃ is fully dissociated in water. For a monoprotic strong acid, [H⁺] equals the acid concentration.

Worked example 1

(a) [H⁺] = 3.6 × 10⁻³ mol dm⁻³. pH = −log₁₀(3.6 × 10⁻³) = 2.44

(b) pH = 11.20. [H⁺] = 10^(−11.20) = 6.31 × 10⁻¹² mol dm⁻³

(c) 0.0250 mol dm⁻³ HNO₃: [H⁺] = 0.0250 mol dm⁻³, so pH = −log₁₀(0.0250) = 1.60

Give pH to 2 decimal places unless told otherwise.

3.1.12.3 The ionic product of water, Kw

Water is slightly dissociated:

H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)        Kc = [H⁺][OH⁻] / [H₂O]

So little water dissociates that [H₂O] is effectively constant, so it is combined with Kc:

Kw = Kc × [H₂O] = [H⁺][OH⁻]        units: mol² dm⁻⁶

At 298 K, Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶.

Kw varies with temperature. It rises with temperature, so the dissociation is endothermic. In pure water [H⁺] = [OH⁻], so [H⁺] = √Kw. When Kw is larger, pure water has pH below 7 but is still neutral, because [H⁺] still equals [OH⁻].

Strong bases such as NaOH and KOH are fully dissociated, so [OH⁻] equals the base concentration. Then [H⁺] = Kw / [OH⁻].

Worked example 2

(a) pH of 0.0400 mol dm⁻³ NaOH at 298 K.

[OH⁻] = 0.0400 mol dm⁻³
[H⁺] = 1.00 × 10⁻¹⁴ / 0.0400 = 2.50 × 10⁻¹³ mol dm⁻³
pH = −log₁₀(2.50 × 10⁻¹³) = 12.60

(b) At a higher temperature, Kw = 2.40 × 10⁻¹⁴ mol² dm⁻⁶. In pure water, [H⁺] = √(2.40 × 10⁻¹⁴) = 1.55 × 10⁻⁷ mol dm⁻³, so pH = 6.81. The water is neutral.

3.1.12.4 Weak acids, Ka and pKa

Weak acids and weak bases dissociate only slightly in aqueous solution. Ethanoic acid and ammonia are examples. For a weak acid HA:

HA ⇌ H⁺ + A⁻          Ka = [H⁺][A⁻] / [HA]          units: mol dm⁻³
pKa = −log₁₀ Ka        Ka = 10^(−pKa)

Ka is the acid dissociation constant. A larger Ka (smaller pKa) means a stronger acid.

To relate pH, concentration and Ka, make two approximations:

1. [H⁺] = [A⁻], because each HA that dissociates gives one of each (H⁺ from water is ignored).
2. [HA] at equilibrium ≈ the starting concentration, because so little dissociates.

Then Ka = [H⁺]² / [HA], so [H⁺] = √(Ka × [HA]).

Worked example 3: pH from Ka

Ethanoic acid has Ka = 1.74 × 10⁻⁵ mol dm⁻³ (pKa 4.76). Find the pH of a 0.150 mol dm⁻³ solution.

[H⁺]² = 1.74 × 10⁻⁵ × 0.150 = 2.61 × 10⁻⁶
[H⁺] = 1.62 × 10⁻³ mol dm⁻³
pH = 2.79

Worked example 4: Ka from pH

A 0.0800 mol dm⁻³ solution of a weak acid HA has pH 3.05. Find Ka and pKa.

[H⁺] = 10^(−3.05) = 8.91 × 10⁻⁴ mol dm⁻³
Ka = (8.91 × 10⁻⁴)² / 0.0800 = 9.93 × 10⁻⁶ mol dm⁻³
pKa = −log₁₀(9.93 × 10⁻⁶) = 5.00

3.1.12.5 Titrations, pH curves and indicators

Calculations from titration results

Use the mean of concordant titres, moles = concentration × volume (dm³), then the equation ratio.

Worked example 5. 25.0 cm³ of NaOH needs a mean titre of 21.30 cm³ of 0.120 mol dm⁻³ HCl.

n(HCl) = 0.120 × 21.30/1000 = 2.556 × 10⁻³ mol = n(NaOH)   (1 : 1)
[NaOH] = 2.556 × 10⁻³ / 0.0250 = 0.102 mol dm⁻³

Worked example 6: pH after mixing a strong acid and a strong base. 20.0 cm³ of 0.150 mol dm⁻³ HCl is added to 25.0 cm³ of 0.100 mol dm⁻³ NaOH.

n(H⁺) = 0.150 × 0.0200 = 3.00 × 10⁻³ mol
n(OH⁻) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol
excess H⁺ = 5.0 × 10⁻⁴ mol in 45.0 cm³
[H⁺] = 5.0 × 10⁻⁴ / 0.0450 = 0.0111 mol dm⁻³      pH = 1.95

If OH⁻ is in excess instead, find [OH⁻] in the total volume, then use Kw.

Shapes of pH curves

A pH curve plots pH against volume added. Each curve has a steep section around the equivalence point, except weak acid with weak base.

Acid + base	Start (acid in flask)	Steep section	pH at equivalence
Strong + strong	Low (about 1)	Long, spanning acidic to alkaline	About 7
Weak + strong	Higher, rises then levels into a buffer region	Shorter, in the alkaline range	Above 7
Strong + weak	Low (about 1)	Shorter, in the acidic range	Below 7
Weak + weak	Higher	No sharp vertical section	Hard to read

Explanations to learn:

• A weak acid starts at higher pH because it is only slightly dissociated.
• After the first additions of a strong base to a weak acid, the flask contains HA and A⁻. This is a buffer, so pH changes slowly.
• At half-neutralisation, [HA] = [A⁻], so Ka = [H⁺] and pH = pKa. This is how you can find Ka experimentally.
• At equivalence, the salt of a weak acid and a strong base gives a solution above pH 7. The salt of a strong acid and a weak base gives a solution below pH 7.

Choosing an indicator

An indicator is suitable when its whole colour-change range lies within the steep section of the curve.

Indicator	Colour change range (pH)
Methyl orange	3.1–4.4
Phenolphthalein	8.3–10.0

• Strong acid + strong base: either works.
• Weak acid + strong base: phenolphthalein.
• Strong acid + weak base: methyl orange.
• Weak acid + weak base: no indicator gives a sharp end point. Use a pH meter.

Required practical 9

Investigate how pH changes when a weak acid reacts with a strong base, and when a strong acid reacts with a weak base.

1. Calibrate the pH meter with buffer solutions of known pH.
2. Put a measured volume of the acid (or base) in a beaker, using a pipette.
3. Add the other solution from a burette in small portions. Stir or swirl, then record the pH.
4. Use smaller portions near the end point, where pH changes fastest.
5. Plot pH against volume added. Read the equivalence volume from the middle of the steep section.
6. For the weak acid, read the pH at half the equivalence volume. That pH equals pKa.

3.1.12.6 Buffer action

A buffer solution keeps an approximately constant pH despite dilution or the addition of small amounts of acid or base.

• Acidic buffer: a weak acid and the salt of that weak acid, for example ethanoic acid and sodium ethanoate. It can be made by partly neutralising a weak acid with a strong base.
• Basic buffer: a weak base and the salt of that weak base, for example ammonia and ammonium chloride.

How an acidic buffer works

The buffer contains large amounts of both HA and A⁻.

HA ⇌ H⁺ + A⁻

• Add a little acid: the extra H⁺ reacts with A⁻ to form HA. The equilibrium shifts left and [H⁺] stays nearly the same.
• Add a little base: OH⁻ reacts with HA (OH⁻ + HA → A⁻ + H₂O), or with H⁺ which is then replaced as more HA dissociates.

Both [HA] and [A⁻] are large, so their ratio, and the pH, changes only a little. Dilution changes both by the same factor, so the ratio stays the same.

How a basic buffer works

• Add acid: NH₃ + H⁺ → NH₄⁺
• Add base: NH₄⁺ + OH⁻ → NH₃ + H₂O

Calculating the pH of an acidic buffer

Rearrange Ka:

[H⁺] = Ka × [HA] / [A⁻]

Take [HA] as the acid added and [A⁻] as the salt. The volume cancels, so moles can be used.

Worked example 7. 1.00 dm³ of buffer contains 0.200 mol ethanoic acid and 0.150 mol sodium ethanoate (Ka = 1.74 × 10⁻⁵ mol dm⁻³).

[H⁺] = 1.74 × 10⁻⁵ × 0.200/0.150 = 2.32 × 10⁻⁵      pH = 4.63

Add 0.0100 mol HCl. The H⁺ turns A⁻ into HA: HA = 0.210 mol, A⁻ = 0.140 mol.

[H⁺] = 1.74 × 10⁻⁵ × 0.210/0.140 = 2.61 × 10⁻⁵      pH = 4.58

The pH falls by only 0.05. The same acid in 1.00 dm³ of water would give pH 2.00.

Worked example 8: buffer by partial neutralisation. 15.0 cm³ of 0.100 mol dm⁻³ NaOH is added to 50.0 cm³ of 0.100 mol dm⁻³ ethanoic acid.

n(HA) start = 5.00 × 10⁻³ mol;  n(OH⁻) = 1.50 × 10⁻³ mol
HA left = 3.50 × 10⁻³ mol;  A⁻ formed = 1.50 × 10⁻³ mol
[H⁺] = 1.74 × 10⁻⁵ × 3.50/1.50 = 4.06 × 10⁻⁵      pH = 4.39

Applications of buffers

• Blood is kept between pH 7.35 and 7.45, mainly by a carbonic acid–hydrogencarbonate buffer.
• Fermentation and fabric dyeing need a set pH.
• pH meters are calibrated with buffer solutions of known pH.

Common errors

• Using √(Ka[HA]) for a buffer. In a buffer [H⁺] ≠ [A⁻]; use Ka × [HA]/[A⁻].
• Forgetting the total volume after mixing a strong acid and strong base.
• Saying hot pure water is acidic because its pH is below 7.
• Choosing an indicator whose range only overlaps the steep section.
• Not updating moles of HA and A⁻ after adding acid or base to a buffer.

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.1.12 Acids and bases (3.1.12.1 to 3.1.12.6, A-level only), with Required practical 9.