Titration Curves

What Is a Titration Curve?

A titration curve is a graph of pH (y-axis) against volume of titrant added (x-axis). It shows how the pH of the solution in the conical flask changes as the titrant is added from the burette. The shape of the curve depends on the strengths of the acid and base involved.

Understanding titration curves is essential for choosing the correct indicator and for interpreting experimental data.

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Strong Acid + Strong Base

Consider titrating 25.0 cm³ of 0.10 mol dm⁻³ HCl with 0.10 mol dm⁻³ NaOH.

Key features of the curve:

• Starting pH: Low (around 1.0 for this concentration)
• Before equivalence: pH rises slowly as H⁺ is gradually neutralised
• Near equivalence: A steep vertical section — the pH jumps dramatically (roughly pH 3 to pH 11) over a tiny volume change
• Equivalence point: pH = 7.00 (neutral salt NaCl formed)
• After equivalence: pH continues to rise but levels off, approaching pH 13

The steep section spans a wide pH range (approximately 3 to 11), which means many indicators will work for this titration.

Why is the equivalence point exactly pH 7? Because the salt formed (NaCl) is from a strong acid and strong base. Neither Na⁺ nor Cl⁻ hydrolyse in water, so the solution is neutral.

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Calculating pH at Different Points (SA + SB)

Before equivalence (e.g. 20.0 cm³ of NaOH added):

Moles H⁺ = 0.10 × 0.0250 = 2.50 × 10⁻³ mol Moles OH⁻ = 0.10 × 0.0200 = 2.00 × 10⁻³ mol Excess H⁺ = 0.50 × 10⁻³ mol Total volume = 45.0 cm³ = 0.0450 dm³ [H⁺] = 0.50 × 10⁻³ / 0.0450 = 0.0111 mol dm⁻³ pH = -log(0.0111) = 1.95

After equivalence (e.g. 30.0 cm³ of NaOH added):

Moles OH⁻ excess = 0.10 × 0.0300 - 2.50 × 10⁻³ = 0.50 × 10⁻³ mol Total volume = 55.0 cm³ = 0.0550 dm³ [OH⁻] = 0.50 × 10⁻³ / 0.0550 = 9.09 × 10⁻³ mol dm⁻³ [H⁺] = Kw / [OH⁻] = 1.00 × 10⁻¹⁴ / 9.09 × 10⁻³ = 1.10 × 10⁻¹² pH = -log(1.10 × 10⁻¹²) = 11.96

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Strong Acid + Weak Base

Consider titrating 25.0 cm³ of 0.10 mol dm⁻³ HCl with 0.10 mol dm⁻³ NH₃.

Key features:

• Starting pH: Low (around 1.0)
• Before equivalence: pH rises slowly
• Equivalence point: pH < 7 (approximately 5.3). This is because the salt formed (NH₄Cl) contains the weak conjugate acid NH₄⁺, which partially dissociates and makes the solution slightly acidic
• Steep section: Smaller range, roughly pH 3 to pH 7
• After equivalence: pH rises more gradually and levels off at a lower value (around 10–11)

The steep section is in the acidic range, so methyl orange is suitable but phenolphthalein is not.

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Weak Acid + Strong Base

Consider titrating 25.0 cm³ of 0.10 mol dm⁻³ CH₃COOH with 0.10 mol dm⁻³ NaOH.

Key features:

• Starting pH: Higher than a strong acid of the same concentration (around 2.9) because the weak acid only partially dissociates
• Buffer region: As NaOH is added, a buffer forms (weak acid + its conjugate base). The pH rises gradually and smoothly in this region
• Half-equivalence point: The point where exactly half the acid has been neutralised. Here, [HA] = [A⁻], so pH = pKa. This is a key feature for determining Ka experimentally
• Equivalence point: pH > 7 (approximately 8.7). The salt CH₃COONa contains the weak conjugate base CH₃COO⁻, which reacts with water to produce OH⁻
• Steep section: Smaller range, roughly pH 7 to pH 11
• After equivalence: pH rises and levels off

The steep section is in the alkaline range, so phenolphthalein is suitable but methyl orange is not.

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Weak Acid + Weak Base

Consider titrating CH₃COOH with NH₃.

Key features:

• Starting pH: Moderately low (around 2.9)
• Equivalence point: Approximately pH 7 (but depends on the specific Ka and Kb)
• Steep section: There is no steep vertical section. The pH change near the equivalence point is very gradual
• No sharp colour change: Because there is no steep section, no indicator gives a sharp end point. This titration cannot be reliably monitored with an indicator

This is why weak acid/weak base titrations are typically monitored using a pH meter instead.

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Summary of All Four Titration Curve Shapes

flowchart TD
    A[Identify acid and base strengths] --> B{Both strong?}
    B -->|Yes| C["SA + SB: Equiv pH = 7, steep pH 3-11"]
    B -->|No| D{Acid strong, base weak?}
    D -->|Yes| E["SA + WB: Equiv pH < 7, steep pH 3-7"]
    D -->|No| F{Acid weak, base strong?}
    F -->|Yes| G["WA + SB: Equiv pH > 7, steep pH 7-11"]
    F -->|No| H["WA + WB: Equiv pH ~ 7, NO steep section"]

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Identifying Key Points on a Curve

Feature	How to identify	Significance
Starting pH	The y-intercept	Tells you about the strength and concentration of the acid
Buffer region	The gently sloping section before equivalence (weak acid curves)	pH changes slowly because a buffer is present
Half-equivalence point	Halfway along the buffer region (half the equivalence volume)	pH = pKa at this point
Equivalence point	The midpoint of the steep section	All acid has reacted with base
Steep section	The near-vertical portion	The pH range where indicator must change colour

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Determining Ka from a Titration Curve

For a weak acid titrated with a strong base:

1. Find the equivalence point (midpoint of the steep section)
2. Read the volume at equivalence (Veq)
3. Find the half-equivalence point: volume = Veq/2
4. Read the pH at this point
5. pH at half-equivalence = pKa, so Ka = 10⁻ᵖᴴ

This is an elegant and practical method for determining Ka experimentally.

Worked Example: A titration curve for a weak acid shows the equivalence point at 24.0 cm³ of NaOH. At 12.0 cm³ (half-equivalence), the pH is 4.76. Therefore pKa = 4.76 and Ka = 10⁻⁴·⁷⁶ = 1.74 × 10⁻⁵ mol dm⁻³ — this is ethanoic acid.

Worked Example: Another curve shows equivalence at 30.0 cm³. At 15.0 cm³, pH = 3.75. pKa = 3.75, Ka = 10⁻³·⁷⁵ = 1.78 × 10⁻⁴ mol dm⁻³ — this is methanoic acid.

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Sketching Titration Curves

When sketching, pay attention to:

1. The correct starting pH (strong acid lower than weak acid of same concentration)
2. Whether there is a buffer region (only for weak acids)
3. The pH at the equivalence point (= 7 for SA/SB, < 7 for SA/WB, > 7 for WA/SB)
4. The size of the steep section (large for SA/SB, smaller for mismatched pairs, absent for WA/WB)
5. The final pH the curve approaches

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Summary Table

Titration	Equiv. pH	Steep Range	Suitable Indicator	Buffer region?
SA + SB	7	~3 to ~11	Either methyl orange or phenolphthalein	No
SA + WB	< 7	~3 to ~7	Methyl orange	After equivalence
WA + SB	> 7	~7 to ~11	Phenolphthalein	Before equivalence
WA + WB	≈ 7	None	No suitable indicator	Before and after

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Common Exam Mistakes with Titration Curves

Mistake	Correction
Assuming equivalence point is always pH 7	Only true for SA + SB; WA + SB gives pH > 7, SA + WB gives pH < 7
Confusing the half-equivalence point with the equivalence point	Half-equivalence is at half the equivalence volume; pH = pKa here
Drawing a buffer region for a strong acid	Buffer regions only appear for weak acids being titrated
Saying WA + WB has a steep section	There is no steep section — the pH changes very gradually
Reading Ka as the pH at equivalence	Ka comes from pH at the half-equivalence point
Drawing the starting pH too low for a weak acid	Weak acids start higher than strong acids of the same concentration

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A-Level Deep Dive: Titration Curves

Spec mapping

Edexcel 9CH0 specification, Topic 12: Acid-base equilibria, sub-strand 12.5 requires you to sketch and interpret pH vs titrant volume curves for the four canonical combinations: strong acid + strong base; weak acid + strong base; strong acid + weak base; weak acid + weak base; identify equivalence point and half-equivalence point on each curve; read pKa from the half-equivalence pH on a weak-acid-vs-strong-base curve; recognise the buffering region around the half-equivalence point (refer to the official specification document for exact wording). Examined in Paper 1 (curve interpretation) and Paper 3 (CP13 directly). Linked synoptically to indicator selection (Lesson 9).

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Worked example with full mark scheme

Question (8 marks):

(a) Sketch the titration curve obtained when 25.0 cm³ of 0.10 mol dm⁻³ ethanoic acid (CH₃COOH, pKa = 4.76) is titrated with 0.10 mol dm⁻³ NaOH(aq), showing pH on the y-axis and volume of NaOH added on the x-axis. Mark the equivalence point and the half-equivalence point clearly. (4)

(b) State the volume of NaOH required to reach the equivalence point and explain why the equivalence pH is not 7. (2)

(c) State, with reasoning, which of methyl orange (pKa 3.7), bromothymol blue (pKa 7.1), and phenolphthalein (pKa 9.3) is most suitable for indicating the equivalence point. (2)

Solution with mark scheme:

(a) Curve features:

• Initial pH ≈ 2.88 (from Ka = 1.74 × 10⁻⁵ × 0.10 = 1.74 × 10⁻⁶, so pH = 2.88).
• Buffering region (between ≈ 5% and 95% titrant) where pH rises gradually from ≈ 3.5 to ≈ 6.0.
• Half-equivalence at 12.5 cm³ titrant: pH = pKa = 4.76.
• Equivalence at 25.0 cm³ titrant: sharp pH jump from ≈ 6 to ≈ 11.
• Equivalence pH ≈ 8.7 (from CH₃COO⁻ hydrolysis).
• Beyond equivalence, pH rises towards ≈ 12.5 as excess strong base accumulates.

M1 — overall sigmoidal shape with steep rise around equivalence. M1 — half-equivalence pH = pKa = 4.76 marked. M1 — equivalence point at 25.0 cm³ marked, pH > 7. A1 — final plateau pH approaching ≈ 12.5.

(b) V_eq = 25.0 cm³ (1:1 stoichiometry, equal concentrations).

Equivalence pH > 7 because the salt CH₃COONa ionises to give CH₃COO⁻, which is a weak base (conjugate base of a weak acid). It hydrolyses partially: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. The resulting [OH⁻] makes the equivalence solution slightly basic.

M1 — V_eq = 25.0 cm³. A1 — equivalence pH > 7 with hydrolysis equation/explanation.

(c) Phenolphthalein is most suitable because its pKa (9.3) is closest to the equivalence pH (8.7), and its colour-change range (8.2–10.0) encompasses the steep portion of the curve. Bromothymol blue (range ≈ 6.0–7.6) and methyl orange (range 3.1–4.4) change colour in the buffering region, before the equivalence — they would give inaccurate end points.

M1 — phenolphthalein selected. A1 — justified by pKa matched to equivalence pH and range within steep section.

Total: 8 marks (M5 A3, split as shown).

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Specimen question modelled on the Edexcel 9CH0 Paper 1 format

Question (6 marks): Compare the titration curves obtained when 25.0 cm³ samples of (a) HCl(aq) and (b) HCN(aq) (Ka = 6.2 × 10⁻¹⁰), each at 0.10 mol dm⁻³, are titrated against 0.10 mol dm⁻³ NaOH(aq). Identify three differences and explain each. (6)

Mark scheme decomposition by AO:

• B1 (AO2.1) — Initial pH: HCl gives pH 1.0 (full dissociation, [H⁺] = 0.10); HCN gives pH ≈ 5.1 (weak acid, [H⁺] = √(Ka × c) = √(6.2 × 10⁻¹¹) ≈ 7.9 × 10⁻⁶).
• B1 (AO2.1) — Half-equivalence: HCl has no buffering region (sharp jump at start); HCN has half-equivalence at 12.5 cm³ where pH = pKa = 9.21.
• B1 (AO2.1) — Equivalence pH: HCl + NaOH gives pH 7 (NaCl, neutral salt); HCN + NaOH gives pH ≈ 11.5 (NaCN, strongly basic salt because CN⁻ is a stronger conjugate base than CH₃COO⁻).
• B1 (AO2.1) — Steepness of jump: HCl curve has a wide vertical jump (pH 4 to 10); HCN curve has a much narrower jump (pH 9.5 to 11.5).
• B1 (AO3.1) — Indicator constraints: HCl + NaOH suits any indicator with pKa 4–10 (very tolerant); HCN + NaOH requires phenolphthalein or thymolphthalein only (narrow range matching basic equivalence).
• B1 (AO3.1) — Underlying chemistry: differences arise from HCN being a weak acid; the buffering region exists because its dissociation is partial, and the equivalence pH > 7 because CN⁻ hydrolyses.

Total: 6 marks all AO2/AO3. Edexcel uses the strong-vs-weak comparison to test whether candidates can decouple the four titration archetypes.

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Synoptic links

Connects to:

1. Topic 12.4 — Buffers (Lessons 5–6): the buffering region of a weak-acid/strong-base curve is the Henderson-Hasselbalch regime. Reading pKa from half-equivalence is the canonical experimental Ka determination.

2. Topic 12.5 — Indicators (Lesson 9): the steep portion of the curve sets the constraint for indicator pKa. Lesson 8 generates the curve; Lesson 9 picks the indicator.