Mini-Lesson

Acids and Bases

This mini-lesson covers AQA 3.1.12 Acids and bases: Brønsted–Lowry theory, the pH scale, K_w, K_a and pK_a for weak acids, calculating the pH of strong acids, weak acids, strong bases and buffers, and reading titration curves to choose an indicator.

Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Everything here is A-level standard — the maths is done properly, not skipped. Press Start when you're ready.

Acids and bases · Brønsted–Lowry

Brønsted–Lowry acids, bases and the pH scale

A Brønsted–Lowry acid is a proton (H⁺) donor; a base is a proton acceptor. Every acid has a conjugate base — what is left after it has donated its proton.

HA + H₂O ⇌ H₃O⁺ + A⁻HA/A⁻ is one conjugate pair; H₂O/H₃O⁺ is the other. We usually write H⁺ for H₃O⁺ for brevity

Hydrogen ion concentrations span many orders of magnitude, so we use a log scale:

pH = −log₁₀[H⁺] and [H⁺] = 10^−pHa pH change of 1 unit means a TEN-fold change in [H⁺]

Strong ≠ concentrated. Strong means fully dissociated (HCl, HNO₃, H₂SO₄). Concentrated means a lot of moles per dm³. A concentrated weak acid can easily have a higher pH than a dilute strong acid.

Calculate

Your turn

1Calculate the pH of 0.0100 mol dm⁻³ hydrochloric acid. (Give your answer to 2 decimal places.)

Hint: HCl is strong, so [H⁺] = 0.0100 mol dm⁻³. pH = −log(0.0100).

Acids and bases · weak acids

K_a, pK_a and the pH of a weak acid

A weak acid only partially dissociates, so an equilibrium is set up:

HA ⇌ H⁺ + A⁻ · K_a = [H⁺][A⁻] ÷ [HA]units: mol dm⁻³ · a LARGER K_a means a stronger acid

pK_a = −log₁₀K_a · K_a = 10^−pK_aa SMALLER pK_a means a stronger acid — the logs invert the comparison

Two approximations make the weak-acid pH calculation possible:

The dissociation of water is negligible, so [H⁺] = [A⁻].
Dissociation is so slight that [HA] at equilibrium ≈ the initial concentration.

K_a = [H⁺]² ÷ c → [H⁺] = √(K_a × c)

Worked example

0.100 mol dm⁻³ ethanoic acid, K_a = 1.74 × 10⁻⁵ mol dm⁻³

[H⁺] = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³

pH = −log(1.32 × 10⁻³) = 2.88

Calculate

Your turn

2Calculate the pH of 0.100 mol dm⁻³ ethanoic acid (K_a = 1.74 × 10⁻⁵ mol dm⁻³). Give your answer to 2 decimal places.

Hint: [H⁺] = √(K_a × c) = √(1.74 × 10⁻⁵ × 0.100) = 1.32 × 10⁻³; then pH = −log[H⁺].

Quick check

?Ethanoic acid has pK_a = 4.76 and chloroethanoic acid has pK_a = 2.86. Which is the stronger acid, and why?

Sort it

Sort each substance

Tap a substance, then tap whether it is a strong acid, a weak acid, or a base.

🟩 Strong acid

🟪 Weak acid

🟦 Base

Acids and bases · K_w

K_w and the pH of a strong base

Water itself ionises very slightly — it is amphoteric, acting as both acid and base:

H₂O ⇌ H⁺ + OH⁻ · K_w = [H⁺][OH⁻]at 298 K, K_w = 1.0 × 10⁻¹⁴ mol² dm⁻⁶

In pure water [H⁺] = [OH⁻] = √(1.0 × 10⁻¹⁴) = 1.0 × 10⁻⁷ mol dm⁻³, giving pH = 7.00. That is why neutral is pH 7 — but only at 298 K. Ionisation of water is endothermic, so heating water increases K_w, increases [H⁺] and lowers the pH of neutral water below 7 — the water is still neutral, because [H⁺] still equals [OH⁻].

To find the pH of a strong base: [OH⁻] = the concentration of the base; then

[H⁺] = K_w ÷ [OH⁻]then pH = −log[H⁺] as usual

Calculate

Your turn

3Calculate the pH of 0.0500 mol dm⁻³ NaOH. (K_w = 1.0 × 10⁻¹⁴ mol² dm⁻⁶.) Give your answer to 2 decimal places.

Hint: [OH⁻] = 0.0500. [H⁺] = 1.0 × 10⁻¹⁴ ÷ 0.0500 = 2.0 × 10⁻¹³. pH = −log(2.0 × 10⁻¹³).

Quick check

?Water is heated from 298 K to 323 K. The pH of the pure water falls to 6.6. Is the water now acidic?

Acids and bases · buffers

Buffer solutions

A buffer resists a change in pH when small amounts of acid or alkali are added, or when the solution is diluted. An acidic buffer is a weak acid plus its conjugate base (e.g. ethanoic acid + sodium ethanoate) — made either by mixing them, or by partially neutralising a weak acid with alkali.

How it works — both species are present in large amounts:

Add acid (H⁺): the conjugate base mops it up: CH₃COO⁻ + H⁺ → CH₃COOH. [H⁺] barely rises.
Add alkali (OH⁻): the weak acid mops it up: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. The acid dissociates further to replace the H⁺ removed.

[H⁺] = K_a × [HA] ÷ [A⁻]rearranged from K_a — the ratio is what matters, so diluting a buffer does not change its pH

Blood is buffered near pH 7.4 by the carbonic acid / hydrogencarbonate system: H₂CO₃ ⇌ H⁺ + HCO₃⁻. Excess H⁺ shifts it left; excess OH⁻ removes H⁺ and shifts it right.

Calculate

Your turn

4A buffer contains 0.200 mol dm⁻³ ethanoic acid and 0.100 mol dm⁻³ sodium ethanoate. K_a = 1.74 × 10⁻⁵ mol dm⁻³. Calculate its pH (2 d.p.).

Hint: [H⁺] = K_a × [HA] ÷ [A⁻] = 1.74 × 10⁻⁵ × (0.200 ÷ 0.100) = 3.48 × 10⁻⁵; pH = −log[H⁺].

Acids and bases · titration curves

Titration curves and choosing an indicator

A titration curve plots pH against volume of titrant added. The key feature is the vertical section at the equivalence point — where a tiny addition causes a huge pH jump.

Strong acid + strong base — a long vertical section, from about pH 3 to 11; equivalence at pH 7. Almost any indicator works.
Weak acid + strong base — equivalence above pH 7 (the salt is basic); vertical section roughly pH 7–11 → use phenolphthalein (range 8.3–10.0).
Strong acid + weak base — equivalence below pH 7; vertical section roughly pH 3–7 → use methyl orange (range 3.1–4.4).
Weak acid + weak base — there is no sharp vertical section, so no indicator is suitable.

The rule: an indicator is suitable only if its pH range lies entirely within the vertical section of the curve. Note also the half-equivalence point of a weak acid: there [HA] = [A⁻], so pH = pK_a — the standard way to measure pK_a from a curve.

Match it

Match the symbol to its definition

Tap a symbol on the left, then its definition on the right.

Symbol

Definition

Quick check

?Which indicator should be used for the titration of ethanoic acid with sodium hydroxide?

Quick check

?A small amount of NaOH is added to a buffer of ethanoic acid and sodium ethanoate. Which equation shows how the pH is kept nearly constant?

Acids and bases · exam traps

The five things to check before you press "="

Strong ≠ concentrated. Strong = fully dissociated. Concentrated = a lot of moles per dm³.
H₂SO₄ is diprotic. If the question says both protons dissociate, [H⁺] is twice the acid concentration.
For a strong base, you must go via K_w. Never take −log of the base concentration.
For a weak acid, [H⁺] = √(K_a × c) — never c itself.
pH is quoted to 2 decimal places. The digits before the point come from the power of ten and carry no significant figures.

Titration-curve facts worth memorising: at the half-equivalence point of a weak acid, [HA] = [A⁻], so pH = pK_a — the standard way to read pK_a off a curve. And a weak acid + weak base curve has no vertical section, so no indicator works.

Quick check

?What is the conjugate base of HSO₄⁻?

Calculate

Your turn

5Calculate the pH of 0.0500 mol dm⁻³ H₂SO₄, assuming both protons fully dissociate. (2 d.p.)

Hint: H₂SO₄ is diprotic, so [H⁺] = 2 × 0.0500 = 0.100 mol dm⁻³. pH = −log(0.100).

Recap

The big ideas to know

Brønsted–Lowry: an acid is a proton (H⁺) donor; a base is a proton acceptor

pH: pH = −log[H⁺] and [H⁺] = 10^−pH

Strong acid: fully dissociated, so [H⁺] = c (×2 for H₂SO₄ if both protons are released)

Weak acid: [H⁺] = √(K_a × c), assuming [H⁺] = [A⁻] and [HA] at equilibrium ≈ initial c

Strong base: find [OH⁻], then [H⁺] = K_w / [OH⁻]

Buffer: [H⁺] = K_a × [HA] ÷ [A⁻] — resists pH change on adding small amounts of acid or alkali

Indicators: choose one whose pH range lies inside the vertical section of the titration curve

That is the whole of AQA 3.1.12. Press Finish to see your score.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Acids and Bases for AQA A-level Chemistry (7405). 🎉

Your stars: 0 / 0

Next: test yourself in the Evaluate stage Confidence Quiz, then lock it in with Verify.

📣 Smashed it? Share your score

Challenge a mate to beat your stars, or show a parent how you got on.

→ Back to all subjects