You are viewing a free preview of this lesson.
Subscribe to unlock all 10 lessons in this course and every other course on LearningBro.
An acid-base indicator is a substance that changes colour depending on the pH of the solution it is in. Indicators are themselves weak acids (or weak bases), where the undissociated form (HIn) and the dissociated form (In⁻) have different colours.
The equilibrium is:
HIn(aq) ⇌ H⁺(aq) + In⁻(aq)
| Species | Colour |
|---|---|
| HIn (acid form) | Colour A |
| In⁻ (base form) | Colour B |
For example, for methyl orange: HIn is red and In⁻ is yellow. For phenolphthalein: HIn is colourless and In⁻ is pink.
In acidic solution (high [H⁺]), Le Chatelier's principle pushes the equilibrium to the left, favouring HIn. You see colour A.
In alkaline solution (low [H⁺]), the equilibrium shifts to the right, favouring In⁻. You see colour B.
The colour change occurs over a range of about 2 pH units, centred on the pKin of the indicator (where pKin = -log₁₀KIn, and KIn is the dissociation constant of the indicator).
At pH = pKin, the concentrations of HIn and In⁻ are equal, and you see a mixture of both colours.
The 1:10 and 10:1 rule: The human eye perceives a colour change when the ratio [In⁻]/[HIn] shifts from about 1:10 (predominantly HIn colour) to 10:1 (predominantly In⁻ colour). Using Henderson-Hasselbalch:
So the visible colour change spans approximately pKin ± 1, which is about 2 pH units.
| Indicator | Acid colour | Base colour | pH range | pKin |
|---|---|---|---|---|
| Methyl orange | Red | Yellow | 3.1 – 4.4 | ~3.7 |
| Bromophenol blue | Yellow | Blue | 3.0 – 4.6 | ~4.0 |
| Methyl red | Red | Yellow | 4.4 – 6.2 | ~5.1 |
| Litmus | Red | Blue | 5.0 – 8.0 | ~6.5 |
| Bromothymol blue | Yellow | Blue | 6.0 – 7.6 | ~7.0 |
| Phenolphthalein | Colourless | Pink | 8.3 – 10.0 | ~9.3 |
| Alizarin yellow R | Yellow | Red | 10.1 – 12.0 | ~11.0 |
The indicator must change colour within the steep section of the titration curve. If the indicator's pH range falls outside the steep section, the colour change will occur either before or after the equivalence point, giving an inaccurate result.
flowchart TD
A["What type of titration?"] --> B{"Strong acid + Strong base?"}
B -->|Yes| C["Steep section pH 3-11: Use methyl orange OR phenolphthalein"]
B -->|No| D{"Strong acid + Weak base?"}
D -->|Yes| E["Steep section pH 3-7: Use methyl orange"]
D -->|No| F{"Weak acid + Strong base?"}
F -->|Yes| G["Steep section pH 7-11: Use phenolphthalein"]
F -->|No| H["Weak acid + Weak base: NO suitable indicator, use pH meter"]
Strong acid + Strong base (steep section: pH 3–11):
Strong acid + Weak base (steep section: pH 3–7):
Weak acid + Strong base (steep section: pH 7–11):
Weak acid + Weak base (no steep section):
Using methyl orange for a WA/SB titration: Methyl orange changes colour at pH 3.1–4.4. For a WA/SB titration, the steep section starts at about pH 7. The indicator changes colour in the buffer region, well before the equivalence point. The student would stop adding base too early, recording a titre that is too small.
Using phenolphthalein for a SA/WB titration: Phenolphthalein changes colour at pH 8.3–10.0. For a SA/WB titration, the steep section ends at about pH 7. The indicator would not change colour until well past the equivalence point. The student would add too much base, recording a titre that is too large.
The fact that indicators are weak acids is not just a curiosity — it is essential to their function. If an indicator were a strong acid, it would be fully dissociated at all pH values above about 1, and you would only ever see the In⁻ colour. There would be no colour change over a useful pH range.
Because indicators are weak acids, they exist in equilibrium between HIn and In⁻. The position of this equilibrium, and hence the colour, depends on the pH. This is exactly what makes them useful.
You must add only a few drops of indicator, not large volumes. There are two reasons:
The equivalence point is the theoretical point where acid and base have reacted in exact stoichiometric ratio. It is determined by the chemistry.
The end point is the experimental observation — the point at which the indicator changes colour.
A well-chosen indicator gives an end point that is very close to the equivalence point. A poorly chosen indicator gives an inaccurate end point. The difference between the end point and equivalence point is called the titration error.
Universal indicator is a mixture of several indicators that produces a continuous range of colours across the pH scale:
| pH range | Colour |
|---|---|
| 1–3 | Red |
| 3–5 | Orange |
| 5–7 | Yellow/green |
| 7 | Green |
| 8–10 | Blue |
| 11–14 | Purple |
Universal indicator is useful for estimating pH but is NOT suitable for titrations because it does not give a sharp, single colour change at a specific pH. It changes gradually over the whole range, making it impossible to identify a precise end point.
A student titrates 25.0 cm³ of 0.10 mol dm⁻³ CH₃COOH with 0.10 mol dm⁻³ NaOH. Which indicator should they use?
This is a weak acid + strong base titration. The equivalence point pH is approximately 8.7, and the steep section spans roughly pH 7 to pH 11.
The correct choice is phenolphthalein.
HCl is titrated with NH₃. Which indicator should be used?
This is a SA + WB titration. The equivalence point pH is approximately 5.3, and the steep section covers pH 3 to 7.
| Titration type | Equivalence pH | Suitable indicator(s) | Unsuitable |
|---|---|---|---|
| SA + SB | 7.0 | Methyl orange or phenolphthalein | - |
| SA + WB | < 7 | Methyl orange | Phenolphthalein |
| WA + SB | > 7 | Phenolphthalein | Methyl orange |
| WA + WB | ≈ 7 | None (use pH meter) | All |
| Mistake | Correction |
|---|---|
| Choosing phenolphthalein for a SA + WB titration | Phenolphthalein is above the steep section; use methyl orange |
| Choosing methyl orange for a WA + SB titration | Methyl orange is below the steep section; use phenolphthalein |
| Using universal indicator in a titration | Universal indicator does not give a sharp colour change |
| Adding too much indicator | Indicator is a weak acid and will affect the pH if too much is added |
| Confusing end point with equivalence point | End point = observed colour change; equivalence point = theoretical stoichiometric point |
| Thinking all indicators change at pH 7 | Each indicator has its own pKin and changes at pKin ± 1 |
Edexcel 9CH0 specification, Topic 12: Acid-base equilibria, sub-strand 12.5 requires you to describe acid-base indicators as weak acids (HIn) or weak bases whose protonated and deprotonated forms have distinct colours; recognise that the indicator's colour change occurs over a pH range of approximately ±1 unit around its pKa (the transition range); select an appropriate indicator for a given titration based on matching its transition range to the pH at the titration's equivalence point and the steep portion of the curve (refer to the official specification document for exact wording). Examined in Paper 1 (selection, justification) and Paper 3 (CP13). Linked synoptically to titration curves (Lesson 8) and weak-acid equilibria (Lesson 3).
Question (8 marks):
(a) Methyl orange is a weak acid indicator with pKa ≈ 3.7. Its acid form (HIn) is red and its base form (In⁻) is yellow. Use the equilibrium HIn ⇌ H⁺ + In⁻ and Le Chatelier's principle to explain how methyl orange changes colour with pH. (3)
(b) State the approximate pH range over which methyl orange changes colour and justify the rule of thumb that an indicator changes colour over ≈ pKa ± 1. (3)
(c) Suggest, with reasoning, the most suitable indicator from methyl orange (pKa 3.7), bromothymol blue (pKa 7.1), and phenolphthalein (pKa 9.3) for each of the following titrations: (i) 0.10 mol dm⁻³ HCl + 0.10 mol dm⁻³ NaOH; (ii) 0.10 mol dm⁻³ CH₃COOH + 0.10 mol dm⁻³ NaOH; (iii) 0.10 mol dm⁻³ HCl + 0.10 mol dm⁻³ NH₃. (2)
Solution with mark scheme:
(a) The indicator equilibrium is HIn(aq) ⇌ H⁺(aq) + In⁻(aq), where HIn is red and In⁻ is yellow.
In acidic solution (high [H⁺]), Le Chatelier shifts the equilibrium to the left, so HIn predominates and the solution is red.
In basic solution (low [H⁺]), the equilibrium shifts to the right, so In⁻ predominates and the solution is yellow.
The colour transition occurs at the pH where [HIn] ≈ [In⁻], which by the indicator's own Henderson-Hasselbalch is when pH = pKa (= 3.7).
M1 — Le Chatelier mechanism. M1 — distinct colours of HIn and In⁻. A1 — colour transition at pH ≈ pKa.
(b) Approximate range: 3.1–4.4 (the human eye perceives a colour change roughly when one form exceeds the other by ≥ 10:1).
Justification: at pH = pKa − 1, the ratio [In⁻]/[HIn] = 10⁻¹ = 0.1, so HIn predominates 10:1 — visible as the acid colour. At pH = pKa + 1, the ratio = 10, so In⁻ predominates 10:1 — visible as the base colour. The transition is therefore confined to roughly pKa ± 1 (i.e. ≈ 2 pH units wide).
M1 — range stated. M1 — 10:1 ratio reasoning. A1 — links to Henderson-Hasselbalch.
(c) (i) HCl + NaOH (strong-strong): equivalence pH = 7. Any of the three indicators works because the steep portion spans pH 4–10. Bromothymol blue (pKa 7.1) is the closest match, but mark scheme accepts methyl orange or phenolphthalein with reasoning.
(ii) CH₃COOH + NaOH (weak-strong): equivalence pH ≈ 8.7. Phenolphthalein (pKa 9.3, range 8.2–10.0) is the only match.
(iii) HCl + NH₃ (strong-weak): equivalence pH ≈ 5.3. Methyl orange (pKa 3.7, range 3.1–4.4) is the closest, although the equivalence is slightly above the upper end. Mark scheme accepts methyl orange with reasoning.
M1 + A1 — three indicator choices with brief reasoning.
Total: 8 marks (M5 A3, split as shown).
Question (6 marks): Bromothymol blue is a weak acid indicator with pKa ≈ 7.1. The acid form is yellow; the base form is blue. The indicator is used at concentration ≈ 1 × 10⁻⁵ mol dm⁻³ in titration (≈ 1–2 drops in 25 cm³).
(a) Write the equilibrium and Ka expression for bromothymol blue. (2)
(b) Calculate the [HIn]/[In⁻] ratio at pH 6.0 and at pH 8.0. State which colour predominates at each pH. (3)
(c) Explain why a small quantity of indicator is used rather than a large quantity. (1)
Mark scheme decomposition by AO:
(a)
(b)
(c)
Total: 6 marks split AO1 = 3, AO2 = 3.
Connects to:
Subscribe to continue reading
Get full access to this lesson and all 10 lessons in this course.