This mini-lesson covers the whole of Edexcel Topic 12: Brønsted–Lowry acids, bases and conjugate pairs; pH, Ka, pKa and Kw; the pH of strong acids, strong bases and weak acids; buffers and the Henderson–Hasselbalch equation; and titration curves and indicator choice.
Work through each screen, answer the questions as you go (several are calculations) and collect ⭐ stars. Every calculation is worked through for you first. Press Start when you're ready.
Definitions
Brønsted–Lowry acids and bases
A Brønsted–Lowry acid is a proton (H⁺) donor. A Brønsted–Lowry base is a proton acceptor. Acid–base reactions are simply proton transfers — and they are equilibria.
HA + H₂O ⇌ H₃O⁺ + A⁻HA donates a proton (acid); H₂O accepts it (base)
A "free" H⁺ in water is really a hydroxonium (oxonium) ion, H₃O⁺. You may write H⁺(aq) as shorthand — Edexcel accepts both.
Amphoteric / amphiprotic: water can act as an acid or a base. With HCl it accepts a proton; with NH₃ it donates one (NH₃ + H₂O ⇌ NH₄⁺ + OH⁻).
Only species with an available lone pair can accept a proton — that is why NH₃, OH⁻ and H₂O are bases.
Exam habit: never say "acids release H⁺". Say proton donor, and always show the equilibrium arrow ⇌ for a weak acid.
Conjugate pairs
Conjugate acid–base pairs
When an acid loses its proton, what is left is its conjugate base. When a base gains a proton, it becomes its conjugate acid. A conjugate pair differs by exactly one H⁺ — one proton, one unit of charge.
Every proton-transfer equilibrium contains two conjugate pairs — an acid on the left with its base on the right, and a base on the left with its acid on the right.
Worked example — NH₃ in water
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
NH₃ accepts a proton → its conjugate acid is NH₄⁺. H₂O donates a proton → its conjugate base is OH⁻.
Rule of thumb: the stronger the acid, the weaker its conjugate base. Cl⁻ (from the strong acid HCl) is a hopeless base; CH₃COO⁻ (from a weak acid) is a decent one.
Quick check
Spot the conjugate base
1In the equilibrium HNO₂ + H₂O ⇌ H₃O⁺ + NO₂⁻, which species is the conjugate base of H₃O⁺?
Match it
Match the acid to its conjugate base
Tap an acid on the left, then the species left behind when it donates one proton.
Acid
Conjugate base
The pH scale
pH = −log₁₀[H⁺]
Hydrogen ion concentrations span many powers of ten, so we compress them onto a logarithmic scale.
pH = −log₁₀[H⁺] · [H⁺] = 10−pH[H⁺] is in mol dm⁻³ · a change of 1 pH unit = a factor of 10 in [H⁺]
Because HCl, HNO₃ and H₂SO₄ are strong acids, they are assumed to be fully dissociated: for a monoprotic strong acid (HCl, HNO₃), [H⁺] = concentration of the acid. H₂SO₄ is diprotic — 1 mol releases up to 2 mol of H⁺, so double the concentration before you take the log.
Worked example — pH of 0.150 mol dm⁻³ HNO₃
HNO₃ is strong and monoprotic, so [H⁺] = 0.150 mol dm⁻³.
pH = −log₁₀(0.150) = 0.82 (2 d.p.)
Significant figures: quote pH to 2 decimal places. Only the digits after the point count as significant, so 2 d.p. matches 2 s.f. data.
Calculate
Your turn — pH of a strong acid
2Calculate the pH of 0.0500 mol dm⁻³ hydrochloric acid, giving your answer to 2 decimal places.
pH
Hint: HCl is strong and monoprotic, so [H⁺] = 0.0500 mol dm⁻³. pH = −log₁₀(0.0500).
Calculate
Your turn — working backwards from pH
3A solution has pH = 2.40. Calculate [H⁺], giving your answer in units of 10⁻³ mol dm⁻³ to 3 significant figures. (For example, if [H⁺] = 5.00 × 10⁻³ mol dm⁻³ you would type 5.00.)
× 10⁻³ mol dm⁻³
Hint: [H⁺] = 10−pH = 10−2.40. That is 100.60 × 10−3.
Water & bases
Kw — the ionic product of water
Water self-ionises very slightly: 2H₂O ⇌ H₃O⁺ + OH⁻ (often written H₂O ⇌ H⁺ + OH⁻). Because [H₂O] is effectively constant, it is folded into the constant:
Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 Kin pure water [H⁺] = [OH⁻] = 1.0 × 10⁻⁷, so pH = 7.00
Kw lets you get the pH of a strong base: NaOH and KOH are fully dissociated, so [OH⁻] = concentration of the base (double it for Ba(OH)₂, which releases 2 OH⁻).
Temperature matters. Self-ionisation is endothermic, so raising T shifts the equilibrium right (Le Chatelier): Kw gets bigger and the pH of pure water falls below 7 — but the water is still neutral because [H⁺] still equals [OH⁻].
Calculate
Your turn — pH of a strong base
4Calculate the pH of 0.0200 mol dm⁻³ sodium hydroxide at 298 K (Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶). Give your answer to 2 decimal places.
pH
Hint: [OH⁻] = 0.0200. [H⁺] = 1.0 × 10⁻¹⁴ ÷ 0.0200 = 5.0 × 10⁻¹³. Now take −log₁₀.
Calculate
Your turn — Kw at a different temperature
5At 50 °C, Kw = 5.48 × 10⁻¹⁴ mol² dm⁻⁶. Calculate the pH of pure water at 50 °C, to 2 decimal places.
pH
Hint: in pure water [H⁺] = [OH⁻], so Kw = [H⁺]². Therefore [H⁺] = √(5.48 × 10⁻¹⁴) = 2.34 × 10⁻⁷ mol dm⁻³.
Quick check
Is that hot water acidic?
6Pure water at 50 °C has a pH of about 6.63. Which statement is correct?
Strong vs weak
Ka and pKa
A strong acid is fully dissociated in water. A weak acid is only partially dissociated — it sits at an equilibrium that lies well to the left. The position of that equilibrium is measured by the acid dissociation constant, Ka:
Ka = [H⁺][A⁻] ÷ [HA]units: mol dm⁻³ · pKa = −log₁₀Ka · Ka = 10−pKa
Larger Ka → stronger acid. Ethanoic acid Ka = 1.74 × 10⁻⁵; chloroethanoic acid Ka = 1.38 × 10⁻³ (the electron-withdrawing Cl stabilises the anion, so it is stronger).
Smaller pKa → stronger acid (the minus sign flips the order). pKa values are far easier to compare — most lie between 0 and 14.
Strong ≠ concentrated. "Strong/weak" is about the extent of dissociation; "concentrated/dilute" is about how much acid is in the solution. A very dilute strong acid can have a higher pH than a concentrated weak one.
Quick check
Strong or concentrated?
7Solution P is 0.0010 mol dm⁻³ HCl (strong). Solution Q is 1.00 mol dm⁻³ CH₃COOH (weak, Ka = 1.74 × 10⁻⁵). Which statement is correct?
Calculate
Your turn — Ka to pKa
8Ethanoic acid has Ka = 1.74 × 10⁻⁵ mol dm⁻³. Calculate its pKa to 2 decimal places.
When the approximation breaks: for a fairly strong weak acid (large Ka), an appreciable fraction dissociates, so [HA]eqm is noticeably less than c. The true [H⁺] is then smaller than √(Ka × c), so the pH you calculate from the approximation comes out slightly too low.
Calculate
Your turn — pH of a weak acid
9Calculate the pH of 0.100 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵ mol dm⁻³), to 2 decimal places.
10A 0.0500 mol dm⁻³ solution of a weak acid HA has a measured pH of 3.10. Calculate Ka, giving your answer in units of 10⁻⁵ mol dm⁻³ to 3 significant figures.
× 10⁻⁵ mol dm⁻³
Hint: [H⁺] = 10−3.10 = 7.94 × 10⁻⁴ mol dm⁻³. Then Ka = [H⁺]² ÷ c = (7.94 × 10⁻⁴)² ÷ 0.0500.
Buffers
What a buffer is and how it works
A buffer solution resists a change in pH when a small amount of acid or alkali is added, or when the solution is diluted. An acidic buffer is a mixture of a weak acid and a salt of that acid (its conjugate base) — for example CH₃COOH + CH₃COONa. You can also make one by partially neutralising a weak acid with a strong base.
CH₃COOH ⇌ H⁺ + CH₃COO⁻the salt supplies a large reservoir of CH₃COO⁻; the acid supplies a large reservoir of CH₃COOH
Add acid (H⁺): the added H⁺ reacts with the large reservoir of CH₃COO⁻ → CH₃COOH. The equilibrium shifts left, so most of the added H⁺ is removed and [H⁺] barely rises.
Add alkali (OH⁻): OH⁻ reacts with the H⁺, but the large reservoir of CH₃COOH dissociates further (equilibrium shifts right) to replace it, so [H⁺] barely falls.
Basic buffers use a weak base and its salt, e.g. NH₃ / NH₄Cl. Blood is buffered near pH 7.40 by the H₂CO₃ / HCO₃⁻ system.
Quick check
Adding acid to a buffer
11A small volume of dilute HCl is added to a CH₃COOH / CH₃COO⁻ buffer. Which is the best explanation of what happens?
Buffer maths
The Henderson–Hasselbalch equation
Rearranging Ka = [H⁺][A⁻] ÷ [HA] gives [H⁺] = Ka × [HA] ÷ [A⁻], and taking −log₁₀ of both sides gives:
pH = pKa + log₁₀([A⁻] ÷ [HA])[A⁻] = the salt (conjugate base) · [HA] = the weak acid
When [A⁻] = [HA], log₁₀(1) = 0 and pH = pKa. This is the buffer's most effective point — and the half-equivalence point of a weak-acid titration.
Because only the ratio appears, you can use moles instead of concentrations (the volume cancels) — and diluting the buffer does not change its pH.
A buffer works best within about pKa ± 1. Choose the acid whose pKa is closest to your target pH.
Worked example
Buffer of 0.100 mol dm⁻³ HA (pKa = 4.20) and 0.0500 mol dm⁻³ NaA.
A titration curve plots pH against volume of base added. The near-vertical section is the equivalence point — where the acid and base have reacted in exactly the stoichiometric ratio. The pH at equivalence depends on the salt that is left behind:
Strong acid + strong base: long vertical section from about pH 3 to 11; equivalence at pH 7 (the salt is neutral).
Weak acid + strong base: equivalence above pH 7 (≈ 8–9) — the conjugate base of the weak acid hydrolyses. There is a flat buffer region before it, where pH = pKa at half-equivalence.
Strong acid + weak base: equivalence below pH 7 (≈ 5–6) — the conjugate acid of the weak base is acidic.
Weak acid + weak base:no sharp vertical section at all — no indicator can be used; you must use a pH meter.
Weak acid titrated with strong base: a buffer region, then a shorter vertical jump ending above pH 7.Quick check
pH at the equivalence point
1325.0 cm³ of 0.100 mol dm⁻³ ethanoic acid is titrated with 0.100 mol dm⁻³ NaOH. At the equivalence point the solution contains only sodium ethanoate and water. What is the pH there, and why?
Indicators
Choosing an indicator
An indicator is itself a weak acid whose acid form and conjugate base form are different colours:
HIn ⇌ H⁺ + In⁻colour A ⇌ colour B · the colour changes over roughly pKIn ± 1
An indicator is suitable only if its whole colour-change range lies inside the vertical section of the titration curve — that way one drop of titrant flips the colour completely.
Indicator
pH range
Colour change (acid → alkali)
Methyl orange
3.1 – 4.4
red → yellow
Phenolphthalein
8.3 – 10.0
colourless → pink
Strong + strong: the vertical section spans ~pH 3–11, so either indicator works.
Weak acid + strong base: the jump is in the alkaline region → phenolphthalein only.
Strong acid + weak base: the jump is in the acidic region → methyl orange only.
Weak + weak:neither — there is no vertical section.
Sort it
Which indicator for this titration?
Tap a titration, then tap the indicator you could use. (HCl, HNO₃ and H₂SO₄ are strong; NaOH and KOH are strong; CH₃COOH and HCOOH are weak acids; NH₃ is a weak base.)
🟠 Methyl orange only
🩷 Phenolphthalein only
🟪 Either works
Recap
The big ideas to know
Brønsted–Lowry: acid = proton donor, base = proton acceptor; conjugate pairs differ by one H⁺
pH: pH = −log₁₀[H⁺] and [H⁺] = 10−pH; for a strong monoprotic acid [H⁺] = c
Kw: [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K; use it to get the pH of strong bases; Kw rises with temperature
Ka and pKa: Ka = [H⁺][A⁻] ÷ [HA]; pKa = −log₁₀Ka; bigger Ka (smaller pKa) = stronger acid
Weak acids: [H⁺] = √(Ka × c), assuming [H⁺] = [A⁻] and [HA] ≈ c
Buffers: weak acid + its salt; pH = pKa + log₁₀([A⁻] ÷ [HA]); pH = pKa at half-equivalence