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Titration Curves and Indicators

Learn and apply Titration Curves and Indicators in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Acid–Base Titration Curves and Indicators: Orientation

Titration curve questions are mainly “read the shape, then justify”: identify the titration type (strong/weak), locate the equivalence point, and choose an indicator whose transition range sits inside the steep pH jump. This lesson also explains the half-equivalence shortcut (pH = pKₐ) for weak acid–strong base titrations.

If the acid-base foundations feel rusty, revisit Acids and Bases (Theories) and keep the Aqueous Equilibria hub open for linked methods.

Definitions (Must Know)

A. Equivalence point

The equivalence point is the point where acid and base have reacted in the exact stoichiometric ratio.

B. End-point

The end-point is the point where the indicator changes colour.

C. Half-equivalence point (weak acid/weak base titrations)

The half-equivalence point is when half the initial acid/base has been neutralised. For a weak acid–strong base titration, pH = pKₐ at the half-equivalence point.

Detailed Explanations

A. What curve features mean (and why equivalence pH shifts)

Common features:

  • Initial pH depends on the analyte in the flask (strong acids start lower pH than weak acids at the same concentration).
  • Buffer region appears when a weak acid/base and its conjugate are both present (gentle slope).
  • Equivalence point is the stoichiometric neutralisation point (steep region).

Because a weak acid–strong base equivalence mixture contains the conjugate base A⁻, therefore A⁻ hydrolyses water to form OH⁻ and the equivalence pH is > 7.

B. Workflow: interpreting a titration curve question

  1. Identify the titration type using the initial pH and the equivalence pH (strong/weak clues).
  2. Locate the equivalence point at the midpoint of the steep vertical section.
  3. For a weak acid/weak base titration, find the half-equivalence point (half the equivalence volume).
  4. Choose an indicator whose transition range sits inside the steep vertical section near equivalence.

Mini example: If a curve starts around pH 3 and the equivalence point is above pH 7, it is consistent with a weak acid titrated by a strong base.

C. Why pH = pKₐ at the half-equivalence point (weak acid + strong base)

At half-equivalence, half the weak acid has been converted to its conjugate base, so [HA] = [A⁻].

Substitute into Henderson–Hasselbalch:

pH = pKₐ + log ₁₀([A⁻]/[HA]) = pKₐ + log ₁₀(1) = pKₐ

Worked Examples

Modelled example 1

Explain an alkaline equivalence point

Core

Problem

For a weak acid–strong base titration, is the equivalence point above or below pH 7? Explain.
Study the worked solution

  1. Identify the equivalence mixture

    Method

    Recognise that the weak acid has been converted to its conjugate base, A⁻.

    Reason

    Stoichiometric neutralisation removes the original weak acid at equivalence.

    Working

    The solution contains the salt supplying A⁻.

  2. Apply salt hydrolysis

    Method

    Show the conjugate base reacting with water to form hydroxide.

    Reason

    The conjugate base of a weak acid accepts a proton from water.

    Working

    A⁻ + H₂O ⇌ HA + OH⁻, so equivalence pH is above 7.

Guided practice 2

Choose phenolphthalein from curve position

About 6 min

Problem

Why is phenolphthalein, with a transition in the alkaline range, often suitable for a weak acid–strong base titration?

Try this before viewing the solution

Hints

Hint 1: locate equivalence
The conjugate-base salt makes the equivalence region alkaline.
Hint 2: indicator rule
A useful indicator completes its colour transition within the steep pH jump.
View solution step by step

  1. Locate the rapid change

    Method

    Place the equivalence point above pH 7.

    Reason

    The weak acid’s conjugate base hydrolyses at equivalence.

    Working

    The steep region extends into alkaline pH.

  2. Match the transition

    Method

    Place phenolphthalein’s range within that steep region.

    Reason

    Only a small titrant-volume change then completes the colour change.

    Working

    Its end-point closely estimates the equivalence volume.

Common misconception 3

Separate equivalence from neutrality

Find and correct the mistake

Learner claim

A learner marks pH 7 as the equivalence point on every acid–base titration curve because equal reacting amounts must make a neutral solution. Diagnose the claim.

Choose what fixes equivalence pH

Equivalence pH depends on

View solution step by step

  1. Define equivalence

    Method

    Use stoichiometric reacting amounts.

    Reason

    Equivalence is an amount condition, not a pH definition.

    Working

    Acid and base have reacted in the equation ratio.

  2. Inspect the resulting salt

    Method

    Consider hydrolysis of a weak partner’s conjugate.

    Reason

    It can form H₃O⁺ or OH⁻ after neutralisation.

    Working

    Strong acid–weak base is acidic; weak acid–strong base is alkaline; strong–strong is near 7 at 25°C.

Examiner practice 4

(curve interpretation)

4 marks

Problem

An acid is titrated with NaOH(aq). The curve starts at pH 3.0 and has equivalence at pH 8.8. Identify the likely titration type and assess an indicator whose full transition range lies around pH 9 within the steep region. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Classify the acid

    1 mark

    Method

    Use the moderately acidic initial pH.

    Reason

    A weak acid begins at a higher pH than an equal-concentration strong acid.

    Working

    The analyte is most consistent with a weak acid.

  2. Classify the titrant

    1 mark

    Method

    Use the stated sodium hydroxide titrant.

    Reason

    NaOH is a strong base under the lesson model.

    Working

    Weak acid–strong base titration.

  3. Use equivalence evidence

    1 mark

    Method

    Link pH 8.8 to conjugate-base hydrolysis.

    Reason

    An alkaline equivalence point supports the classification.

    Working

    Equivalence above 7 is consistent.

  4. Assess the indicator

    1 mark

    Method

    Accept the pH-9 indicator because its full range lies in the steep region.

    Reason

    A small volume change carries it through the colour transition.

    Working

    The indicator is suitable.

Challenge 5

(half-equivalence link)

Minimal support

Problem

In a weak acid–strong base titration, the pH at half-equivalence is 4.76. Find Kₐ for the acid and explain the half-equivalence link.

Try this before viewing the solution

Hints

Hint 1: equal components
At half-equivalence, half the acid has become conjugate base, so [HA] = [A⁻].
Hint 2: invert pKa
This makes pH = pKₐ; then use Kₐ = 10^(-pKₐ).
View solution step by step

  1. Use the buffer ratio

    Method

    Set the acid and conjugate-base concentrations equal.

    Reason

    Half of the initial weak acid has been neutralised.

    Working

    log ₁₀([A⁻]/[HA]) = log ₁₀(1) = 0.

  2. Read pKa

    Method

    Set pKₐ equal to the measured pH.

    Reason

    The concentration-ratio term vanishes in Henderson–Hasselbalch.

    Working

    pKₐ = 4.76.

  3. Convert to Ka

    Method

    Apply the inverse logarithm.

    Reason

    pKₐ = - log ₁₀Kₐ.

    Working

    Kₐ = 10^(-4.76) = 1.74 × 10⁻⁵.

Mind Stretchers

Mind stretcher 1Extension

Explain why a weak acid–weak base titration is hard to do accurately with an indicator.

Show Answer

Mark scheme:

  • The pH change near equivalence is small (no steep vertical region).
  • An indicator transition range cannot “fit” cleanly inside a sharp pH jump, so the end-point has a large uncertainty.
  • Therefore a pH meter is preferred for this titration type.