Edexcel A-Level Chemistry: Acids, Bases and Buffers — Complete Revision Guide (9CH0)
Edexcel A-Level Chemistry: Acids, Bases and Buffers — Complete Revision Guide (9CH0)
Acids, bases and buffers is one of the most calculation-heavy topics on Edexcel 9CH0, and one of the most rewarding to master. The whole section reduces to a small set of equilibrium expressions — Ka, Kw and Kb — combined with a logarithmic conversion to pH. Once you can move between concentrations and pH fluently, every standard question (strong acid pH, weak acid pH, buffer pH, titration curve interpretation, indicator choice) becomes a clean substitution.
This topic is also one of the most synoptic. The equilibrium constants here are special cases of the Kc framework from kinetics and equilibrium. The acid-base behaviour of amino acids in organic advanced is directly described by Ka. The titration arithmetic builds directly on the mole calculations from amounts and redox. Time spent here returns across the rest of the course.
This guide walks through the acids, bases and buffers content in 9CH0 topic by topic. It covers the Brønsted-Lowry theory; conjugate acid-base pairs; Ka, Kb and Kw; pH calculations for strong and weak acids and bases; buffer composition and the Henderson-Hasselbalch equation; how buffers resist pH change; titration curves for the four combinations of acid and base; and the choice of indicators. For each topic you will find the core ideas, common pitfalls, a worked example and a link into the LearningBro Acids, Bases and Buffers course.
What the Edexcel 9CH0 Specification Covers
Edexcel A-Level Chemistry (9CH0) is examined through Paper 1 (Inorganic and Physical, 1h45, 90 marks), Paper 2 (Organic and Physical, 1h45, 90 marks) and Paper 3 (General and Practical, 2h30, 120 marks). Acids and buffers is examined heavily on Paper 1 and on Paper 3 through practical-based titration questions.
| Sub-topic | Spec area | Typical paper weight |
|---|---|---|
| Brønsted-Lowry theory and conjugate pairs | Topic 12 | 2-4 marks |
| Ka and pKa | Topic 12 | 3-5 marks |
| Strong acid pH calculations | Topic 12 | 3-5 marks |
| Weak acid pH calculations | Topic 12 | 4-6 marks |
| Strong and weak base pH | Topic 12 | 3-5 marks |
| Buffer composition and Henderson-Hasselbalch | Topic 12 | 4-8 marks |
| How buffers resist pH change | Topic 12 | 3-5 marks |
| Titration curves (4 combinations) | Topic 12 | 4-6 marks |
| Indicator choice | Topic 12 | 2-4 marks |
These weights are estimates, modelled on the Edexcel 9CH0 paper format. What is reliable is that an extended-response buffer or pH-after-titration question — typically eight to ten marks — appears on essentially every Paper 1.
Brønsted-Lowry Theory and Conjugate Pairs
The Brønsted-Lowry theory defines an acid as a proton donor and a base as a proton acceptor. This is broader than the simpler Arrhenius definition (acids release H+ in water, bases release OH-) and applies in non-aqueous solvents and to species without explicit OH groups.
Every Brønsted acid has a corresponding conjugate base — what is left after the proton has been donated. Every Brønsted base has a conjugate acid. These come in pairs: HCl/Cl-, CH3COOH/CH3COO-, H2O/OH- (water can be a base) or H2O/H3O+ (water can be an acid). Water is amphoteric because it can act as either.
Worked example. Identify the conjugate acid-base pairs in NH3 + H2O ⇌ NH4+ + OH-. NH3 (base) and NH4+ (its conjugate acid) form one pair. H2O (acid) and OH- (its conjugate base) form the other.
A common pitfall is to identify the conjugate base as the original acid plus an extra proton (it should be minus one proton). Another is to call something a "Brønsted-Lowry acid" without specifying proton donor.
See the Brønsted-Lowry lesson.
Ka and pKa
For a weak acid HA dissociating in water, HA ⇌ H+ + A-, the acid dissociation constant Ka = [H+][A-] / [HA]. The smaller Ka, the weaker the acid. pKa = -log10 Ka, so the larger pKa, the weaker the acid.
| Acid | Ka | pKa |
|---|---|---|
| HCl (strong) | ~10^7 | ~-7 |
| H2SO4 (first dissociation) | very large | very negative |
| HNO3 (strong) | ~24 | ~-1.4 |
| CCl3COOH | 0.20 | 0.7 |
| HF | 6.6 × 10^-4 | 3.2 |
| CH3COOH | 1.8 × 10^-5 | 4.76 |
| HCN | 4.9 × 10^-10 | 9.3 |
| H2O (autoprotolysis) | 1.0 × 10^-14 | 14 |
Strong acids essentially fully dissociate in water (Ka very large). Weak acids only partially dissociate; the equilibrium lies well to the left.
A common pitfall is to forget that pKa is the negative log of Ka — students sometimes write pKa = log Ka. Another is to use Ka concentrations in g dm^-3 instead of mol dm^-3.
See the Ka and pKa lesson.
Strong Acid pH
For a strong acid that fully dissociates, [H+] equals the initial acid concentration. pH = -log10 [H+].
Worked example. Calculate the pH of 0.10 mol dm^-3 HCl. [H+] = 0.10 mol dm^-3. pH = -log10 (0.10) = 1.0.
For a diprotic strong acid like H2SO4 in dilute solution, [H+] is twice the initial acid concentration (both protons fully dissociate at moderate concentrations). For 0.05 mol dm^-3 H2SO4, [H+] = 0.10 mol dm^-3 and pH = 1.0.
A common pitfall is to forget the doubling for diprotic strong acids. Another is to give pH values like 1.001 — three significant figures of pH means three significant figures of [H+], not three decimals of pH. See the strong acid lesson.
Weak Acid pH
For a weak acid HA at initial concentration c, only a small fraction dissociates. Two assumptions usually apply: [H+] = [A-] (each dissociation gives one of each) and [HA] ≈ c (the small loss to dissociation is negligible). Then Ka = [H+]^2 / c, so [H+] = √(Ka × c) and pH = ½ (pKa - log10 c).
Worked example. Calculate the pH of 0.10 mol dm^-3 ethanoic acid (Ka = 1.8 × 10^-5). [H+] = √(1.8 × 10^-5 × 0.10) = √(1.8 × 10^-6) = 1.34 × 10^-3 mol dm^-3. pH = -log10 (1.34 × 10^-3) = 2.87.
The "[HA] ≈ c" assumption is valid when less than ~5 percent of the acid dissociates. For very dilute weak acids this can fail; the full quadratic equation is then needed. Edexcel rarely sets such cases.
A common pitfall is to forget the square root and write [H+] = Ka × c. Another is to use the initial acid concentration as [H+] (which is only true for strong acids). See the weak acid lesson.
Strong and Weak Base pH
Kw is the ionic product of water: Kw = [H+][OH-] = 1.0 × 10^-14 mol^2 dm^-6 at 298 K. So in any aqueous solution at 298 K, [H+] × [OH-] = 10^-14, pH + pOH = 14.
For a strong base that fully dissociates, [OH-] equals the initial base concentration; [H+] = Kw / [OH-]; pH follows.
Worked example. Calculate the pH of 0.10 mol dm^-3 NaOH. [OH-] = 0.10. [H+] = 10^-14 / 0.10 = 10^-13. pH = -log10 (10^-13) = 13.0.
For a weak base B + H2O ⇌ BH+ + OH-, Kb = [BH+][OH-] / [B], analogous to Ka. Kb × Ka = Kw for any conjugate acid-base pair, so pKa + pKb = 14.
A common pitfall is to confuse base concentration with [OH-] for weak bases — they are not equal because dissociation is partial. See the bases lesson.
Buffer Composition and Henderson-Hasselbalch
A buffer is a solution that resists changes in pH when small amounts of acid or base are added. Buffers contain comparable concentrations of a weak acid and its conjugate base (an "acidic buffer", e.g. ethanoic acid + sodium ethanoate), or a weak base and its conjugate acid (a "basic buffer", e.g. ammonia + ammonium chloride).
The pH of a buffer is calculated from the Henderson-Hasselbalch equation: pH = pKa + log10 ([A-] / [HA]). When [A-] = [HA], pH = pKa, so the buffer's effective pH is set by the choice of weak acid.
Worked example. Calculate the pH of a buffer made from 0.20 mol dm^-3 ethanoic acid (pKa 4.76) and 0.10 mol dm^-3 sodium ethanoate. pH = 4.76 + log10 (0.10 / 0.20) = 4.76 + log10 (0.5) = 4.76 - 0.30 = 4.46.
The buffer capacity — the amount of acid or base that can be added before pH changes significantly — is largest when [HA] = [A-] (pH = pKa) and decreases as the ratio moves away from 1.
A common pitfall is to confuse the ratio direction in the Henderson-Hasselbalch equation. The conjugate base [A-] is on top; the acid [HA] is on the bottom. See the Henderson-Hasselbalch lesson.
How Buffers Resist pH Change
When a small amount of acid is added to a buffer, the conjugate base A- reacts with the added H+ to form more HA: A- + H+ → HA. Both [A-] decreases and [HA] increases, but only slightly, so the ratio [A-]/[HA] changes little, and pH changes little.
When a small amount of base is added, HA reacts with OH-: HA + OH- → A- + H2O. [HA] decreases and [A-] increases slightly; the ratio changes only modestly.
Worked example. The buffer above (0.20 ethanoic acid, 0.10 ethanoate, 1 dm^3) has 0.005 mol HCl added. New [HA] = 0.205, [A-] = 0.095. New pH = 4.76 + log10 (0.095/0.205) = 4.76 - 0.33 = 4.43, a change of only 0.03 pH units. Adding 0.005 mol HCl to pure water at pH 7 would drop pH below 3 — buffering is dramatic.
A common pitfall is to forget that the buffer capacity can be exceeded if too much acid or base is added. Once one component is exhausted, the solution behaves like an unbuffered acid or base. See the buffer mechanism lesson.
Titration Curves: Four Combinations
A titration curve plots the pH of the analyte solution against the volume of titrant added. The shape depends on whether the acid and base are strong or weak.
| Combination | Initial pH | Equivalence pH | Steep portion | Suitable indicator |
|---|---|---|---|---|
| Strong acid + strong base | Low (~1) | 7 | pH 3-11 (very steep) | Methyl orange or phenolphthalein |
| Strong acid + weak base | Low (~1) | <7 (acidic salt) | pH 3-7 | Methyl orange |
| Weak acid + strong base | Higher (~3) | >7 (basic salt) | pH 7-11 | Phenolphthalein |
| Weak acid + weak base | Higher | ~7 | No steep portion | Indicator unsuitable; use pH meter |
The equivalence point is where moles of acid equal moles of base. For strong-strong, equivalence is at pH 7. For strong acid + weak base, the equivalence solution contains the conjugate acid of the weak base, which is acidic. For weak acid + strong base, the equivalence solution contains the conjugate base of the weak acid, which is basic. For weak acid + weak base, the curve has no steep section, so titration is impractical.
The half-equivalence point in a weak-acid/strong-base titration is where half the acid has been neutralised. At that point, [HA] = [A-], so pH = pKa. This is a useful experimental method for determining pKa.
A common pitfall is to expect equivalence pH of 7 for every titration. Another is to forget that the steep section determines indicator choice — the indicator's pKa must lie within the steep region. See the titration curves lesson.
Indicators
An indicator is itself a weak acid (HIn) whose conjugate base (In-) has a different colour. The colour change occurs over a pH range of approximately pKa(In) ± 1. The indicator's pKa must lie within the steep portion of the titration curve to give a sharp end point.
| Indicator | pKa | Colour low pH → high pH | Useful for |
|---|---|---|---|
| Methyl orange | 3.7 | Red → yellow | Strong acid + weak/strong base |
| Bromothymol blue | 7.0 | Yellow → blue | Strong acid + strong base |
| Phenolphthalein | 9.3 | Colourless → pink | Weak acid + strong base, strong-strong |
Worked example. Choose an indicator for the titration of 0.1 M ethanoic acid against 0.1 M NaOH. Equivalence pH is around 8.7 (basic, since the conjugate base of a weak acid). Steep portion runs from about pH 7 to pH 11. Phenolphthalein (pKa 9.3) lies within this range, so it gives a sharp end point. Methyl orange (pKa 3.7) lies below the steep section, so its colour changes early and the end point is poorly defined.
A common pitfall is to choose any indicator that "lies in pH 4-10" without checking the specific titration. Each combination has a specific suitable indicator. See the indicators lesson.
Common Mark-Loss Patterns
- Forgetting the negative sign in pH = -log10 [H+].
- Treating [H+] of a weak acid as equal to its concentration.
- Forgetting the doubling for diprotic strong acids like H2SO4.
- Mis-using the Henderson-Hasselbalch equation (acid on top instead of conjugate base).
- Confusing buffer mechanism (which reacts with which addition).
- Choosing an indicator that lies outside the steep portion of the curve.
- Forgetting that equivalence pH is not always 7.
- Mis-computing pKa from Ka (forgetting the negative).
- Using base concentration as [OH-] for weak bases.
- Failing to recognise the half-equivalence point as pH = pKa.
How to Revise This Topic
- Drill pH conversions both ways daily. Concentration to pH and pH to concentration. Twenty conversions a day for a fortnight.
- Practise weak-acid pH problems in batches of ten until √(Ka × c) is automatic.
- Build a titration-curve flashcard set with one card per combination, showing the curve shape, equivalence pH and suitable indicator.
- Drill buffer addition problems: given a buffer composition and a small addition of strong acid or base, calculate the new pH.
- Memorise the three standard indicators (methyl orange, bromothymol blue, phenolphthalein) by pKa and colour change.
- Use the LearningBro practice quizzes to test under timed conditions.
Linking to Other Topics
Acids and buffers builds on the kinetics and equilibrium framework — Ka, Kb and Kw are equilibrium constants. The titration calculations are direct extensions of the amounts and redox titration arithmetic. Amino acid pH behaviour (see organic advanced) is exactly the buffer chemistry of zwitterions. The inorganic chemistry tests for ions also use acid-base chemistry — the dissolution of Al(OH)3 in excess NaOH is a Brønsted reaction. Time spent here returns in many other topics.
Final Word
Acids and buffers is a calculation-heavy topic, but the calculations are formulaic once you know which formula to apply. Drill pH and pKa arithmetic, build automaticity with Henderson-Hasselbalch, and learn the four titration curves cold. The full LearningBro Acids, Bases and Buffers course walks through every sub-topic with worked examples and AI tutor feedback. Get this section fluent and a substantial fraction of the physical chemistry on every paper becomes routine.