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A pH curve (or titration curve) is a plot of pH against the volume of titrant added during a titration. It reveals the chemistry of the acid–base reaction far more completely than a single end-point reading: the starting pH tells us about the strength of the acid (or base) being titrated, the steepness of the rise tells us how cleanly the equivalence point can be located, the position of the equivalence point on the pH axis reflects the strength of the conjugate species, and a "flat" region around half-equivalence reveals buffering by a weak acid and its salt. There are four classic curves at A-Level — strong acid + strong base, strong acid + weak base, weak acid + strong base, and weak acid + weak base — and an indicator is only useful if the pH range over which it changes colour sits inside the steep portion of the curve. This lesson covers all four shapes, the equivalence point and half-equivalence point, indicator choice, and the limits of the indicator method.
Spec mapping (AQA 7405): This lesson maps to §3.1.12 (acids and bases — pH curves, titrations and indicators), building on lesson 3 of this course (§3.1.12, pH of weak acids and weak bases — pre-equivalence calculations) and lesson 4 (§3.1.12, buffer action, where pH = pKa at half-equivalence). It anchors Required Practical 9 (lesson 7), which delivers the experimental measurement of a titration curve with a pH meter. The stoichiometric basis of every titration calculation was developed in atomic-structure lesson 8 (titration calculations from §3.1.2.5). Refer to the official AQA specification document for the exact wording.
Assessment objectives: AO1 (recall): the shape of each of the four standard curves and the pH ranges of the common indicators (methyl orange, methyl red, bromothymol blue, phenolphthalein, thymolphthalein). AO2 (apply): sketch a pH curve given the strengths of the acid and base, identify the equivalence and half-equivalence points, and read pKa off a weak-acid curve at half-equivalence. AO3 (analyse and evaluate): justify the choice of indicator for a specific titration, explain why a weak-acid–weak-base titration cannot be detected with a colour indicator, and reason about the position of the equivalence point on the pH axis from the strengths (Ka, Kb) of the species involved.
A pH curve is recorded by adding the titrant (the solution in the burette) in small increments to a fixed volume of analyte (the solution in the conical flask, usually 25.0 cm³), measuring the pH after each addition with a calibrated pH meter, and plotting pH (y-axis) against volume of titrant (x-axis). The curve has three diagnostic regions:
The volume of titrant required to reach equivalence is fixed by stoichiometry and is independent of acid/base strength. It is the pH at the equivalence point that depends on strength. That distinction matters for indicator choice.
Example: 25.0 cm³ of 0.100 mol dm⁻³ HCl titrated with 0.100 mol dm⁻³ NaOH.
Reaction: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
| Volume NaOH added (cm³) | Approximate pH |
|---|---|
| 0.0 | 1.0 |
| 5.0 | 1.18 |
| 12.5 | 1.48 |
| 24.0 | 2.79 |
| 24.9 | 3.70 |
| 25.0 (equivalence) | 7.00 |
| 25.1 | 10.30 |
| 26.0 | 11.21 |
| 50.0 | 12.52 |
Shape: The curve starts at pH ≈ 1, rises only slowly until approaching equivalence, then jumps almost vertically from about pH 3.5 to pH 10.5 over the addition of roughly 0.2 cm³ of titrant either side of equivalence, then flattens at pH ≈ 13 as excess base accumulates.
Why pH = 7 at equivalence? The salt formed is NaCl. Na⁺ is the conjugate of the strong base NaOH (and is essentially neutral in water — it neither donates nor accepts protons); Cl⁻ is the conjugate of the strong acid HCl (also essentially neutral). Neither ion hydrolyses water, so the equivalence-point solution is just water + spectator ions, with pH = 7 at 298 K.
Indicator choice: The steep portion spans roughly pH 4 to pH 10. Any indicator whose colour-change range lies entirely within that window is acceptable — methyl orange (3.1–4.4), methyl red (4.2–6.3), bromothymol blue (6.0–7.6), and phenolphthalein (8.2–10.0) all work for a strong–strong titration. This is the only one of the four classic titrations for which all four indicators are usable.
Example: 25.0 cm³ of 0.100 mol dm⁻³ HCl titrated with 0.100 mol dm⁻³ NH₃(aq).
Hold on — by convention the curve is drawn with the burette holding the titrant. The classical setup is the acid in the burette, base in the flask (or vice versa). Here we take 25.0 cm³ of NH₃(aq) in the flask and titrate with HCl from the burette — pH starts high and falls — or, equivalently, 25.0 cm³ HCl in the flask and NH₃ from the burette. The shape is the mirror image but the equivalence-point pH is the same. We will describe the latter: NH₃ added to HCl.
Reaction: HCl(aq) + NH₃(aq) → NH₄Cl(aq) (i.e. NH₄⁺(aq) + Cl⁻(aq))
| Volume NH₃ added (cm³) | Approximate pH |
|---|---|
| 0.0 | 1.0 |
| 12.5 | 1.48 |
| 24.0 | 2.79 |
| 24.9 | 3.70 |
| 25.0 (equivalence) | ≈ 5.3 |
| 25.1 | 6.5 |
| 26.0 | 8.5 |
| 50.0 | 11.10 |
Shape: The curve starts at pH ≈ 1, rises in the normal way through the pre-equivalence region, but the steep rise at equivalence is shorter — only from about pH 3.5 to pH 7. After equivalence the curve flattens at pH ≈ 11 (the pH of excess 0.05 mol dm⁻³ NH₃), not pH 13 as for a strong base.
Why is the equivalence-point pH ≈ 5.3? The salt formed is NH₄Cl. The chloride ion is a spectator. The ammonium ion NH₄⁺ is the conjugate acid of the weak base NH₃ and does donate protons to water:
NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq)
with Ka(NH₄⁺) = Kw / Kb(NH₃) = 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰, giving pKa ≈ 9.25. For 0.0500 mol dm⁻³ NH₄⁺ (concentration at equivalence after dilution by mixing equal volumes), [H⁺] ≈ √(Ka × c) = √(5.6 × 10⁻¹⁰ × 0.0500) = 5.3 × 10⁻⁶ mol dm⁻³, giving pH ≈ 5.28. The equivalence-point solution is acidic because the salt's cation is a weak acid.
Indicator choice: The steep portion spans about pH 3.5 to pH 7. Methyl orange (3.1–4.4) is the standard choice; methyl red (4.2–6.3) also works. Phenolphthalein (8.2–10.0) is unsuitable — its colour change occurs well above the steep portion and would record an end-point much later than the true equivalence.
Example: 25.0 cm³ of 0.100 mol dm⁻³ CH₃COOH (ethanoic acid, pKa = 4.76) titrated with 0.100 mol dm⁻³ NaOH.
Reaction: CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l)
| Volume NaOH added (cm³) | Approximate pH |
|---|---|
| 0.0 | 2.87 |
| 5.0 | 4.14 |
| 12.5 (half-equivalence) | 4.76 = pKa |
| 20.0 | 5.36 |
| 24.0 | 6.14 |
| 24.9 | 7.16 |
| 25.0 (equivalence) | ≈ 8.72 |
| 25.1 | 10.30 |
| 50.0 | 12.52 |
Shape: The curve starts at pH ≈ 2.9 (not pH 1 — CH₃COOH is only partially dissociated). It rises rapidly at first (the unbuffered region before significant A⁻ builds up), then crosses a long flat buffer region centred on the half-equivalence volume (12.5 cm³) where pH = pKa = 4.76. After half-equivalence the curve rises again, climbs steeply through equivalence from about pH 7 to pH 10.5, and levels off at pH ≈ 13.
The half-equivalence point. At half-equivalence, exactly half of the CH₃COOH has been converted to CH₃COO⁻. Therefore [HA] = [A⁻]. Substituting into the Henderson–Hasselbalch equation:
pH = pKa + log₁₀([A⁻]/[HA]) = pKa + log₁₀(1) = pKa + 0 = pKa
This is the diagnostic feature of a weak-acid curve: drop a vertical line from the volume axis at half-equivalence, project across to the pH axis, and you read pKa directly. The technique is used routinely to measure pKa values of newly-synthesised acids.
Why is the equivalence-point pH ≈ 8.7? The salt is sodium ethanoate, CH₃COONa. Na⁺ is a spectator; the ethanoate ion CH₃COO⁻ is the conjugate base of a weak acid and hydrolyses water:
CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq)
with Kb = Kw / Ka = 10⁻¹⁴ / 1.74 × 10⁻⁵ = 5.7 × 10⁻¹⁰. For 0.0500 mol dm⁻³ CH₃COO⁻ at equivalence, [OH⁻] ≈ √(Kb × c) = √(5.7 × 10⁻¹⁰ × 0.0500) = 5.4 × 10⁻⁶ mol dm⁻³, giving pOH = 5.27 and pH = 14 − 5.27 = 8.73. The equivalence-point solution is basic because the salt's anion is a weak base.
Indicator choice: The steep portion spans about pH 7 to pH 10.5. Phenolphthalein (8.2–10.0) is the standard choice. Methyl orange (3.1–4.4) would change colour very early — in the middle of the buffer region, long before equivalence — and is unsuitable. Bromothymol blue (6.0–7.6) is borderline; it changes colour just before the steep rise begins and is normally avoided.
Example: 25.0 cm³ of 0.100 mol dm⁻³ CH₃COOH titrated with 0.100 mol dm⁻³ NH₃.
Reaction: CH₃COOH(aq) + NH₃(aq) ⇌ CH₃COO⁻(aq) + NH₄⁺(aq)
Shape: The curve starts at pH ≈ 2.9, rises through a buffer region around pH = pKa(CH₃COOH) = 4.76, climbs through an equivalence-point pH of about 7 (because pKa(CH₃COOH) ≈ pKa(NH₄⁺), so the conjugate-base and conjugate-acid hydrolyses approximately cancel), then continues to rise gently toward pH ≈ 11 with no sharp inflection. There is no steep portion at all. The transition from "before equivalence" to "after equivalence" plays out over several cm³ of titrant rather than a fraction of a cm³.
Indicator detection fails. Because no indicator has a colour-change range narrow enough to fit inside the (non-existent) steep portion, the colour change of any indicator added to the flask drifts gradually as titrant is added — the end point cannot be located within ±0.5 cm³, and the titration is useless for quantitative work using indicators.
The correct method. Use a pH meter (or a conductivity meter) to record the curve numerically, then locate the equivalence point by finding the point of inflection of the curve (the point where the second derivative changes sign, or, in practice, where the first derivative dpH/dV is at a maximum). This is the standard approach in Required Practical 9 for any titration where indicator detection is unreliable.
These two terms sound similar but mean different things:
If the indicator is chosen well, end point and equivalence point coincide to within experimental precision (typically ±0.05 cm³). If the indicator's pH range does not overlap the steep portion of the curve, the end point can be systematically off from equivalence by several cm³ — this is called indicator error or titration error, and it is the main reason for choosing the indicator carefully.
The half-equivalence point is the most diagnostic single feature of a weak-acid curve. It is defined by:
V(half-equivalence) = ½ × V(equivalence)
At this volume, exactly half of the weak acid HA has reacted with the strong base to form A⁻. Therefore moles HA remaining = moles A⁻ formed, and since both are in the same flask, [HA] = [A⁻]. Substituting into Ka:
Ka = [H⁺][A⁻] / [HA] = [H⁺] × 1 = [H⁺]
Taking −log₁₀ of both sides: pH = pKa.
This is the experimental basis of pKa determination by titration. To find pKa of an unknown weak acid:
The method is also used for weak bases titrated against strong acids, with pH = pKa(BH⁺) at half-equivalence, where pKa(BH⁺) is the pKa of the conjugate acid of the base.
The flat-ish region around half-equivalence is the buffer region. In this region the flask contains comparable concentrations of HA and A⁻ — exactly the composition required for buffering — and small additions of strong base shift the [HA]/[A⁻] ratio only slightly, with correspondingly small changes in pH (lesson 4 derives the Henderson–Hasselbalch equation that quantifies this).
Buffering is most effective when [HA] and [A⁻] are equal (at half-equivalence), and the buffer "works" — keeping pH within about pKa ± 1 — across the volume range from roughly 10% to 90% of the equivalence volume. Outside that range the buffering breaks down: at very low volumes there is too little A⁻; near the equivalence volume there is too little HA.
The buffer region is the shape signature of a weak acid being titrated. A strong acid produces no buffer region — the pre-equivalence curve rises monotonically with no plateau, because no weak conjugate base is present to buffer.
An indicator is itself a weak acid (HIn) whose acid form HIn and conjugate base In⁻ have different colours. The equilibrium is:
HIn(aq) ⇌ H⁺(aq) + In⁻(aq), Kln = [H⁺][In⁻] / [HIn]
with pKln the indicator's dissociation constant. By the same logic as Henderson–Hasselbalch, pH = pKln + log([In⁻]/[HIn]). The eye perceives the indicator colour as "mostly HIn" when [HIn] / [In⁻] > 10 (pH < pKln − 1) and as "mostly In⁻" when [In⁻] / [HIn] > 10 (pH > pKln + 1). The full colour transition therefore plays out across roughly pKln ± 1, a span of about 2 pH units.
Selection rule for titration indicators:
The pH-range over which the indicator changes colour (pKln ± 1) must lie entirely inside the steep portion of the titration curve. The pKln of the indicator should ideally match the pH at equivalence.
If the indicator range falls outside the steep portion, the indicator colour change occurs before or after the true equivalence, and the end-point reading is systematically wrong.
| Indicator | pKln | Colour change | pH range | Use for |
|---|---|---|---|---|
| Methyl orange | 3.7 | red → yellow | 3.1–4.4 | Strong acid + weak base (equivalence pH ≈ 5) |
| Methyl red | 5.1 | red → yellow | 4.2–6.3 | Strong acid + weak base (alternative to methyl orange) |
| Bromothymol blue | 7.0 | yellow → blue | 6.0–7.6 | Strong acid + strong base (pH 7) |
| Phenolphthalein | 9.4 | colourless → pink | 8.2–10.0 | Weak acid + strong base (equivalence pH ≈ 9) |
| Thymolphthalein | 10.0 | colourless → blue | 9.4–10.6 | Weak acid + strong base (alternative) |
A student plans to titrate 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid against 0.100 mol dm⁻³ sodium hydroxide. Which indicator should they use?
This is a weak acid + strong base titration. The equivalence-point pH is about 8.7 (calculated above). The steep portion of the curve spans approximately pH 7 to pH 10.5.
Answer: phenolphthalein (or thymolphthalein).
Polyprotic acids — those with more than one acidic proton, such as H₂SO₄ (diprotic, very strong for the first proton and moderately strong for the second), H₂CO₃ (diprotic, both weak), and H₃PO₄ (triprotic, all weak) — can in principle show multiple equivalence points, one for each successive deprotonation. The two (or three) equivalence points are clearly resolved on the curve only when successive pKa values are well separated — a useful rule of thumb is ΔpKa > 4.
For H₃PO₄: pKa₁ = 2.15, pKa₂ = 7.20, pKa₃ = 12.35. The differences are 5.05 and 5.15 — well above 4 — so all three equivalence points are clearly visible on the curve as three separate sharp inflections at pH ≈ 4.7, 9.8, and beyond about pH 13 (the third inflection is suppressed by the basicity of water itself). At each equivalence point an indicator could be chosen to detect that specific deprotonation.
For H₂SO₄: the first proton is fully dissociated (effectively pKa ≈ −3) and the second has pKa₂ ≈ 1.99 — ΔpKa is large but both deprotonations occur in the strong-acid region, so the curve looks essentially like a single strong-acid + strong-base titration with a slightly broadened pre-equivalence section.
Polyprotic curves are covered properly in lesson 6 of this course (polyprotic acids and salt hydrolysis); the present note exists only to flag that "the curve has one equivalence point" is a simplification true only for monoprotic acids.
| Method | Best for | Resolution | Limitations |
|---|---|---|---|
| Coloured indicator | Strong–strong, strong–weak, weak–strong (with correct indicator) | ±0.05 cm³ at end-point if matched | Useless for weak–weak; cannot give continuous curve |
| pH meter | Any titration including weak–weak; pKa determination | ±0.01 pH unit if calibrated | Requires calibration with buffers (pH 4, 7, 10) before use; electrode must be rinsed and not allowed to dry |
Common procedural errors:
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