Choose indicators from titration data

Key idea: An indicator's acid and conjugate base have different colours. Choose one whose entire transition range lies within the steep pH change near equivalence, so its end-point is close to equivalence.

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  • GCE A-Level H2 Chemistry 2027
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An indicator shows the end of a titration by changing colour. It only works if its colour change happens in the steep part of the titration curve, where a single drop of titrant moves the pH a long way. This lesson explains how an indicator works and how to choose one from a titration curve or from pH data.

How an indicator works

An indicator is a weak acid, HIn, whose undissociated form and conjugate base have different colours:

HIn(aq) ⇌ H⁺(aq) + In⁻(aq)

In acid, the high [H⁺] pushes the equilibrium to the left, so the colour of HIn shows. In alkali, hydroxide ions remove H⁺, the equilibrium moves to the right, and the colour of In⁻ shows. The colour changes gradually over a range of about two pH units, centred near the indicator’s pKₐ.

IndicatorColour in acidColour in alkalipH range of colour change
methyl orangeredyellow3.1–4.4
phenolphthaleincolourlesspink8.3–10.0

The end-point is where the indicator changes colour. A good indicator gives an end-point close to the equivalence point.

Choosing an indicator

Near equivalence, the pH changes by several units for a tiny volume of titrant. Choose an indicator whose whole colour-change range lies inside that steep section. Then the colour change happens over a small volume interval close to equivalence. If the range lies outside the steep section, the colour changes gradually over several cm³, too early or too late.

Worked example 1

Compare both ends of each range

Problem

A titration curve rises steeply from pH 6.0 to 10.5 near equivalence. Indicator X changes colour over pH 8.0–9.0; indicator Y over pH 4.0–6.5. Which is suitable?
Show full solution
  1. Compare each range with the steep section

    Method

    Check both the lower and upper limits of each range.

    Reason

    The whole colour change must happen in the steep section.

    Working

    Answer

    X: 6.0 < 8.0 and 9.0 < 10.5, so all of X’s range is inside. Y: 4.0 is below 6.0, so most of Y’s range is outside.
  2. Decide

    Method

    Choose X.

    Reason

    X changes colour completely in the steep section near equivalence. Y starts changing well before equivalence, in the gently sloping part of the curve.

    Working

    Answer

    X is suitable; Y is not.

Check your understanding 1

The steep interval is pH 7.14–10.30. Choose phenolphthalein (8.3–10.0) or methyl orange (3.1–4.4), then explain the data-to-volume link in the worked problem.

Suitable indicator

Show answer

Phenolphthalein lies entirely within 7.14–10.30. Methyl orange lies outside, giving an earlier end-point. Use the next response to explain the small volume over which the suitable colour change happens.

Guided practice 2

Ethanoic acid and sodium hydroxide

About 6 min

Problem

When 0.100 mol dm⁻³ sodium hydroxide is added to 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid, the pH rises from 7.14 at 24.90 cm³ to 10.30 at 25.10 cm³. Is phenolphthalein (8.3–10.0) or methyl orange (3.1–4.4) suitable? Explain using both ranges.

Try this before viewing the solution

Hints

Hint 1

steep section

Between 24.90 and 25.10 cm³ the pH jumps from 7.14 to 10.30.

Hint 2

compare

Check whether each indicator’s whole range lies between 7.14 and 10.30.

Show solution step by step
  1. Find the steep section

    Method

    The pH rises from 7.14 to 10.30 over only 0.20 cm³.

    Reason

    The equivalence point is alkaline, because ethanoate ions are a weak base.

    Working

    Answer

    Steep section: pH 7.14 to 10.30.
  2. Choose the indicator

    Method

    Phenolphthalein is suitable; methyl orange is not.

    Reason

    Phenolphthalein’s whole range, 8.3–10.0, lies inside the steep section. Methyl orange’s range, 3.1–4.4, is reached long before equivalence, in the buffer region.

    Working

    Answer

    7.14 < 8.3 and 10.0 < 10.30

Check your understanding 2

For 25.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid titrated with 0.100 mol dm⁻³ sodium hydroxide, the pH rises from 3.09 at 24.60 cm³ to 10.90 at 25.40 cm³. Which of methyl orange and phenolphthalein can be used? Why is the choice wider here than for ethanoic acid?

Suitable indicator(s)

Show answer

Both ranges, 3.1–4.4 and 8.3–10.0, lie inside the steep section from pH 3.09 to 10.90, so either indicator changes colour completely between 24.6 and 25.4 cm³, around equivalence. A strong acid–strong base curve has a much longer steep section, from about pH 3 to 11, so more indicators fit. With ethanoic acid the steep section starts at about pH 7, which rules out methyl orange.

Check your understanding 3

Explain why the strong acid–strong base data above allow both indicators, while the ethanoic acid data exclude methyl orange. Use the full ranges and the pH change over a small added volume.

Show answer

Both 3.1–4.4 and 8.3–10.0 fit within 3.09–10.90, traversed over 0.80 cm³ around equivalence. The ethanoic-acid steep interval starts at 7.14, above the whole methyl-orange range; methyl orange therefore changes before equivalence. The range comparison connects colour change with volume error, rather than just an acidic or alkaline label.

Check your understanding 4

At 25 °C, a 0.100 mol dm⁻³ monobasic acid titrated with NaOH starts at pH 3.0 and has equivalence pH 8.8. Its steep interval is 7.0–10.5. Classify the acid and decide whether thymol blue (8.0–9.6) is suitable.

Acid classification
Thymol blue

Show answer

This is a weak acid–strong base titration. Thymol blue is suitable. In the next problem, connect the starting concentration/pH, salt hydrolysis and complete range to your conclusion.

Exam-style question 1

Judge an indicator from a curve

4 marks

Problem

An unknown 0.100 mol dm⁻³ monobasic acid is titrated with sodium hydroxide at 25 °C. The curve starts at pH 3.0, rises steeply from pH 7.0 to 10.5 near equivalence, and the equivalence point is at pH 8.8. Identify the type of titration, and decide whether thymol blue, which changes colour over pH 8.0–9.6, is a suitable indicator. [4 marks]

Try this before viewing the solution

Show solution step by step
  1. Classify the acid

    1 mark

    Method

    The acid is weak.

    Reason

    A strong acid at 0.100 mol dm⁻³ would start at pH 1.0, not 3.0.

    Working

    Answer

    Starting pH 3.0 means only about 1% is dissociated.
  2. Classify the base

    1 mark

    Reason

    Sodium hydroxide is fully dissociated.

    Working

    Answer

    Weak acid–strong base titration.
  3. Use the equivalence pH

    1 mark

    Method

    An equivalence point at pH 8.8 supports this.

    Reason

    The conjugate base of a weak acid makes the solution alkaline.

    Working

    Answer

    Equivalence above 7 fits a weak acid and a strong base.
  4. Judge the indicator

    1 mark

    Method

    Thymol blue is suitable.

    Reason

    Its whole range, 8.0–9.6, lies inside the steep section from 7.0 to 10.5.

    Working

    Answer

    Its colour-change range lies completely inside the steep section near equivalence.

When the pH falls

When acid is added to a base, the curve falls instead of rising, but the rule is the same: the whole colour-change range must lie inside the steep section.

Try it yourself 3

Ammonia and hydrochloric acid

Minimal support

Problem

25.0 cm³ of 0.100 mol dm⁻³ ammonia is titrated with 0.100 mol dm⁻³ hydrochloric acid at 25 °C:

Acid added / cm³pH
24.906.85
25.005.28
25.103.70

Indicator R changes colour over pH 4.0–6.0; indicator S over pH 8.0–10.0. Choose an indicator. In the next response, explain why the alkaline-range indicator that suits ethanoic acid with sodium hydroxide is unsuitable here.

Try this before viewing the solution

Suitable indicator

Hints

Hint 1

direction

The pH falls as acid is added. Find the pH values on either side of 25.00 cm³.
Show solution step by step
  1. Find the steep section

    Method

    The pH falls from 6.85 to 3.70 between 24.90 and 25.10 cm³.

    Reason

    That is a fall of over 3 pH units for 0.20 cm³ of acid.

    Working

    Answer

    Steep section: pH 6.85 to 3.70.
  2. Choose

    Method

    Choose R.

    Reason

    Its whole range, 4.0–6.0, lies inside the steep section. S changes between pH 10 and 8, well before equivalence.

    Working

    Answer

    3.70 < 4.0 and 6.0 < 6.85
  3. Explain the difference

    Method

    This equivalence point is acidic; the ethanoic acid one was alkaline.

    Reason

    Ammonium ions are a weak acid, while ethanoate ions are a weak base.

    Working

    Answer

    The steep section moves to lower pH, so a different indicator fits.

Check your understanding 5

Explain the R/S decision from the descending data above, then compare the salt formed here with the salt in the ethanoic acid–NaOH titration.

Show answer

The steep interval spans pH 3.70–6.85 in a 0.20 cm³ added-volume interval. R’s whole 4.0–6.0 range fits; S’s 8.0–10.0 range is reached before equivalence. Here NH₄⁺ is a weak acid and gives acidic equivalence. Ethanoate is a weak base and gives alkaline equivalence, placing its steep interval higher. Curve direction changes the order of readings, rather than the whole-range criterion.

Common mistakes

  • Choosing an indicator because the equivalence point is “alkaline” or “acidic” without comparing ranges with the steep section.
  • Accepting an indicator when only part of its range overlaps the steep section.
  • Treating the colour change as happening at one exact pH. It happens over a range.
  • Confusing the end-point, where the colour changes, with the equivalence point, where the amounts match.
  • Expecting any indicator to work for a weak acid–weak base titration, which has no steep section.

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