Chemistry · Year 12 · Module 6 · Lesson 15
HSC Exam Practice
Indicators: Mechanism & Selecting the Right One
Short answer
1.Short answer
Define the term indicator as used in acid-base titrations, including reference to the equilibrium involved and why it produces a colour change.
Distinguish between the endpoint and the equivalence point of a titration. Explain how an incorrectly selected indicator causes these two points to differ significantly.
Identify the most suitable indicator for each of the following titrations and justify each choice with reference to the equivalence point pH.
(a) 0.100 mol/L HCl (aq) titrated with 0.100 mol/L NaOH (aq).
(b) 0.100 mol/L CH₃COOH (aq) titrated with 0.100 mol/L NaOH (aq).
(c) 0.100 mol/L HCl (aq) titrated with 0.100 mol/L NH₃ (aq).
Explain why the equivalence point pH of a weak acid / strong base titration is greater than 7. Your answer must include a relevant ionic equation.
Outline why no standard indicator is suitable for a weak acid / weak base titration.
Account for why the transition pH range of an indicator spans approximately one pH unit above and below its pKa(In).
Data response
2.Data response — titration curve analysis
The graph below shows the titration curve for 25.00 mL of an unknown weak acid solution titrated with 0.100 mol/L NaOH. Three horizontal bands mark the transition pH ranges of three standard indicators.
(a) Identify which indicator has its transition range overlapping the steep pH jump at the equivalence point. Justify your choice with reference to the graph.
(b) Calculate the concentration of the unknown weak acid solution. Show full working.
(c) Read the pKa of the unknown acid from the graph and determine Ka. Show the reasoning linking the half-equivalence volume to the pKa.
(d) Describe the outcome if bromothymol blue were used as the indicator for this titration, with reference to the graph.
A student titrates 25.00 mL of formic acid (HCOOH, Ka = 1.77 × 10–4) with 0.100 mol/L NaOH.
(a) Write the equation for the hydrolysis of the ion formed at the equivalence point and use it to explain why the equivalence point pH > 7.
(b) Calculate the equivalence point pH, given that the NaOH titre at equivalence is 25.00 mL of 0.100 mol/L NaOH. Show all steps. (Kw = 1.0 × 10–14)
(c) Select the appropriate indicator from methyl orange (3.1–4.4), bromothymol blue (6.0–7.6), and phenolphthalein (8.3–10.0). Justify in two sentences.
Extended response
3.Extended response
Evaluate the claim that “the choice of indicator in an acid-base titration is a minor procedural detail that has little impact on the accuracy of the result.” In your response, refer to the mechanism of indicator colour change, the concept of the equivalence point, at least two named indicators with their transition ranges, and a specific example of the error introduced by an incorrect indicator choice.
Chemistry · Year 12 · Module 6 · Lesson 15
Answer Key & Marking Guidelines
Section 1 · Short answer · 3 marks · Band 3
Sample response. An indicator is a weak acid (HIn) whose acid form (HIn) and conjugate base form (In–) have different colours. The equilibrium HIn ⇌ H¹ + In– shifts depending on the [H¹] of the solution: in acidic solution the equilibrium lies left (HIn dominates, one colour); in basic solution it shifts right (In– dominates, different colour). This colour change signals the endpoint.
Marking notes. 1 mark: indicator is a weak acid (HIn ⇌ H¹ + In–). 1 mark: acid and conjugate base forms have different colours. 1 mark: colour change occurs when the equilibrium shifts in response to pH change (or explicit mention of H¹ concentration shifting the equilibrium).
Section 1 · Short answer · 3 marks · Band 3
Sample response. The equivalence point is where stoichiometrically equivalent amounts of acid and base have reacted; its pH depends on the salt formed. The endpoint is the observed colour change of the indicator. If the indicator's transition range does not include the equivalence point pH, the colour change occurs at a different pH (e.g. in the buffer region or after equivalence), so the student stops the titration at a volume far from the true equivalence volume, causing a systematic error in the calculated concentration.
Marking notes. 1 mark: correct definition of equivalence point. 1 mark: correct definition of endpoint. 1 mark: explaining how an incorrect indicator makes the endpoint occur at a pH different from the equivalence point pH, causing error in the titre.
Section 1 · Short answer · 3 marks · Band 4
Sample response. (a) Strong acid / strong base: EP pH = 7; any of the three indicators is valid (large pH jump ~4–10 encompasses all); bromothymol blue or phenolphthalein are common choices. (b) Weak acid / strong base: EP pH > 7 (~8.7 for CH₃COOH/NaOH); phenolphthalein (8.3–10.0) — its range encompasses the EP pH. (c) Strong acid / weak base: EP pH < 7 (~5.3 for HCl/NH₃); methyl orange (3.1–4.4) — the sharp pH jump in the acidic region is within MO's range.
Marking notes. 1 mark per correctly named indicator with EP pH direction stated or justified (max 3). No mark for indicator only without justification.
Section 1 · Short answer · 3 marks · Band 4
Sample response. At the equivalence point, the salt of a weak acid and strong base (e.g. CH₃COONa) is present. The conjugate base ion (CH₃COO–) partially hydrolyses: CH₃COO– + H₂O ⇌ CH₃COOH + OH–. The production of OH– makes the solution basic, so pH > 7.
Marking notes. 1 mark: identifies the conjugate base ion as the species present at EP. 1 mark: correct hydrolysis equation with OH– product. 1 mark: concludes that OH– production gives pH > 7.
Section 1 · Short answer · 2 marks · Band 3–4
Sample response. For a weak acid / weak base titration, there is no sharp pH jump at the equivalence point — the pH changes gradually over many mL of added base. Any indicator would change colour gradually over a wide volume range, making it impossible to identify a precise endpoint. The error in the titre would be too large to be useful.
Marking notes. 1 mark: no sharp pH jump at equivalence. 1 mark: colour change is gradual / no precise endpoint can be located.
Section 1 · Short answer · 3 marks · Band 4–5
Sample response. Using the Henderson–Hasselbalch equation for the indicator: pH = pKa(In) + log([In–]/[HIn]). The eye perceives predominantly the HIn colour when [HIn]/[In–] ≥ 10 (i.e. pH ≤ pKa – 1) and predominantly the In– colour when [In–]/[HIn] ≥ 10 (i.e. pH ≥ pKa + 1). The transition range — where both species are present in comparable amounts and a mixed colour is visible — therefore spans approximately pKa(In) – 1 to pKa(In) + 1, a range of about 2 pH units.
Marking notes. 1 mark: states that the transition occurs where both HIn and In– are present in significant concentrations. 1 mark: explains that the eye favours one colour when the ratio is ≥ 10:1 (or equivalent). 1 mark: concludes that the range spans approximately ± 1 pH unit around pKa(In).
Section 2 · Data response · 10 marks · Band 4–5
(a) Indicator choice — 2 marks. Phenolphthalein (8.3–10.0). From the graph, the EP pH ≈ 8.9 and the sharp pH jump runs approximately pH 7.5–12. Phenolphthalein's band (purple/violet shading) overlaps this jump. BTB (6.0–7.6) falls below the jump; MO (3.1–4.4) falls in the buffer region. (1 mark indicator; 1 mark correct graph-based justification.)
(b) Concentration — 3 marks. n(NaOH) at EP = 0.100 × 0.02000 = 2.00 × 10–³ mol = n(acid). c(acid) = 2.00 × 10–³ / 0.02500 = 0.0800 mol/L. (1 mark: n(NaOH); 1 mark: equimolar ratio; 1 mark: correct concentration.)
(c) pKa — 2 marks. Half-equivalence volume = 20.00 / 2 = 10.00 mL. At this volume, [acid] = [conjugate base], so log([A–]/[HA]) = 0, therefore pH = pKa. From graph at 10.00 mL: pH ≈ 4.8. Ka = 10–⁴‧⁸ ≈ 1.6 × 10–⁵. (1 mark reasoning; 1 mark Ka value or correct pKa from graph.)
(d) BTB outcome — 3 marks. From the graph, BTB's range (6.0–7.6) intersects the gradually rising section of the curve before the sharp jump, at approximately 17–18 mL NaOH. Using BTB, the student would observe a colour change at ≈17–18 mL, recording an endpoint 2–3 mL before the true equivalence point (20.00 mL). The calculated concentration of the weak acid would be underestimated. (1 mark: BTB falls below EP; 1 mark: endpoint recorded early from graph; 1 mark: consequence — titre too small / concentration underestimated.)
Section 2 · Data response · 7 marks · Band 4–5
(a) Hydrolysis equation — 2 marks. At equivalence, HCOONa is present. Hydrolysis: HCOO– + H₂O ⇌ HCOOH + OH–. OH– is produced → basic solution → EP pH > 7. (1 mark equation; 1 mark reasoning.)
(b) EP pH calculation — 4 marks. At EP: equal volumes mixed → [HCOO–] = 0.0500 mol/L.
Kb(HCOO–) = Kw/Ka = 1.0 × 10–¹´ / 1.77 × 10–´ = 5.65 × 10–¹¹.
[OH–] = √(5.65 × 10–¹¹ × 0.0500) = √(2.82 × 10–¹²) = 1.68 × 10–⁶ mol/L.
pOH = –log(1.68 × 10–⁶) = 5.77. EP pH = 14.00 – 5.77 = 8.23.
(1 mark: [HCOO–] after dilution; 1 mark: Kb; 1 mark: [OH–]; 1 mark: pH = 8.23.)
(c) Indicator — 1 mark. Phenolphthalein (8.3–10.0) — EP pH 8.23 is at the lower edge of phenolphthalein's range; the sharp pH jump encompasses this range and the student will observe a clear colour change close to equivalence. (Accept phenolphthalein with correct reasoning. Note: BTB would also give a colour change close to EP in this case but is below the EP pH; phenolphthalein is the more reliable choice.)
Section 3 · Extended response · 7 marks · Band 5–6
Sample response. The claim is incorrect: indicator selection critically determines the accuracy of an acid-base titration result. Indicators are weak acids (HIn) in equilibrium: HIn ⇌ H¹ + In–. The acid form carries one colour; the conjugate base a different colour. The transition pH range is approximately pKa(In) ± 1. For a result to be accurate, the indicator's range must overlap the sharp pH jump that occurs at the equivalence point — not simply any region of the curve. The equivalence point pH is not universally pH 7: it is determined by the salt formed. For a weak acid / strong base titration (e.g. ethanoic acid + NaOH), the conjugate base CH₃COO– hydrolyses to produce OH–, giving an EP pH of approximately 8.7. Phenolphthalein (range 8.3–10.0) is valid here; methyl orange (range 3.1–4.4) is not. If methyl orange is used for this titration, the colour change occurs in the buffer region — at approximately pH 3.7–4.4 — when only ~10–20% of the acid has been neutralised. The recorded titre would be far smaller than the true equivalence volume, causing a severe underestimate of the calculated acid concentration. In an analytical chemistry laboratory context (e.g. NATA-accredited testing for acidity of wine as per AWRI protocols), such an error would render the result worthless and potentially dangerous. Therefore, indicator choice is not a minor detail — it is the variable that determines whether the endpoint coincides with the equivalence point, and an incorrect choice introduces a large, systematic, non-random error that cannot be detected by replication alone.
Marking notes. 1 mark: correct mechanism (HIn ⇌ H¹ + In–; different colours; equilibrium shifts with pH). 1 mark: defines equivalence point with correct EP pH for at least one specific titration type. 1 mark: explains that the indicator range must overlap the sharp pH jump at equivalence (not just any region). 1 mark: names at least two indicators with correct transition ranges. 1 mark: specific quantified example of error from wrong indicator (direction and mechanism of error). 1 mark: applies real-world context (analytical lab, AWRI, QC, NATA, or equivalent). 1 mark: clear evaluative conclusion explicitly rejecting the claim with reasoned justification.