Chemistry • Year 12 • Module 6 • Lesson 17
Titration & Indicator Mastery
Apply titration curve analysis, indicator selection, and multi-step concentration calculations to real-data scenarios including an Australian food-industry context.
1. Interpret the titration curve — acetic acid quality control
The SVG graph below shows a titration curve obtained by a NATA-accredited food laboratory in Sydney testing a batch of white wine vinegar for the Australian food industry (FSANZ Standard 2.10.1 requires minimum 4.0% w/v acetic acid). The titration adds 0.1000 mol/L NaOH to 20.00 mL of diluted vinegar sample. 12 marks
1.1 Identify the acid type (weak or strong) and justify using two features of the curve. 2 marks
1.2 State the equivalence point volume and equivalence point pH. Identify the pH at the equivalence point and explain whether it is above, at, or below 7. 3 marks
1.3 Determine the pKa of acetic acid from the curve and calculate Ka. Show working. 2 marks
1.4 Calculate the concentration of acetic acid in the diluted sample (mol/L). Show full working. 2 marks
1.5 Identify the appropriate indicator for this titration. Write a complete three-step justification. 3 marks
2. Concordant results — mining wastewater neutralisation titration
A mining company in Western Australia must comply with Department of Water regulations by monthly titration of acid mine drainage (AMD) against standard NaOH solution. A technician performs six titration trials. The results are shown below. 8 marks
| Trial | Initial burette reading (mL) | Final burette reading (mL) | Titre volume (mL) | Include in mean? |
|---|---|---|---|---|
| 1 (rough) | 0.00 | 24.85 | 24.85 | No — rough trial |
| 2 | 0.00 | 25.60 | 25.60 | |
| 3 | 1.20 | 26.75 | 25.55 | |
| 4 | 0.45 | 26.02 | 25.57 | |
| 5 | 2.10 | 27.68 | 25.58 | |
| 6 | 0.00 | 26.98 | 26.98 |
2.1 Identify which trials give concordant results (within 0.10 mL of each other). Mark them in the table and calculate the mean concordant titre. Show full working. 3 marks
2.2 The NaOH titrant is 0.09850 mol/L. The AMD sample volume is 20.00 mL. Calculate the concentration of H⁺ (as HCl equivalent) in the AMD sample. Show full working. 2 marks
2.3 Trial 2 and Trial 6 were excluded. Give a distinct reason for excluding each. 2 marks
2.4 NATA laboratory accreditation requires that burette readings are recorded to 0.05 mL precision and that the mean titre is calculated only from concordant results. Explain why both requirements improve the reliability of the reported acid concentration. 1 mark
3. Cause-and-effect chain — methyl orange error propagation
A student incorrectly selects methyl orange for a weak acid + strong base titration. Complete the cause-and-effect chain by filling in each effect box. 5 marks
| Cause | Effect (fill in) |
|---|---|
| Methyl orange (range 3.1–4.4) is used for a weak acid + strong base titration (EP pH > 7). | |
| The colour change occurs in the buffer region of the titration curve, far before the equivalence point. | |
| The student records a titre volume far smaller than the true equivalence volume. | |
| The calculated moles of acid is severely underestimated. | |
| Overall outcome: The reported acid concentration is a large underestimate of the true value. | Estimated % error if true EP = 25.00 mL and MO endpoint = 5.50 mL: _______% |
4. Predict and justify
A student performs a titration of 0.100 mol/L HCl with 0.100 mol/L NaOH using phenolphthalein. The student then repeats the exact experiment but switches to methyl orange as the indicator. 4 marks
4.1 Predict whether the two titres will be significantly different or essentially the same, and justify your prediction with reference to the titration curve type. 2 marks
4.2 Explain how this result demonstrates that the strong acid + strong base scenario differs fundamentally from the weak acid + strong base scenario in terms of indicator selection constraints. 2 marks
Q1 — Titration curve interpretation
1.1 Weak acid. Feature 1: starting pH is approximately 2.9, well above what 0.1000 mol/L strong acid would give (pH 1.0) — indicates partial ionisation characteristic of a weak acid. Feature 2: there is a visible buffer plateau (gradual pH rise) before the equivalence point, caused by coexistence of CH₃COOH and CH₃COO⁻. (Award 1 mark per feature with justification.)
1.2 Vₛₖ = 25.00 mL (midpoint of the steepest section). EP pH ≈ 9.2 (above pH 7). The EP pH is above 7 because the salt formed at equivalence is sodium acetate; the acetate ion (CH₃COO⁻) is the conjugate base of a weak acid and undergoes base hydrolysis: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, producing OH⁻ ions and raising pH above 7. (1 mark for Vₛₖ and pH; 1 mark for EP pH above 7; 1 mark for hydrolysis explanation.)
1.3 Half-EP at V = 12.50 mL; pH at that point = 4.74. Therefore pKa = 4.74. Ka = 10⁻⁴⋅⁷⁴ = 1.82 × 10⁻⁵. (1 mark pKa; 1 mark Ka.)
1.4 n(NaOH) at EP = 0.1000 × 0.02500 = 2.500 × 10⁻³ mol. n(CH₃COOH) = n(NaOH) = 2.500 × 10⁻³ mol. c(CH₃COOH) = 2.500 × 10⁻³ / 0.02000 = 0.1250 mol/L. (1 mark for moles; 1 mark for concentration with correct units.)
1.5 Step 1: EP pH ≈ 9.2 — above 7 because acetate ion undergoes base hydrolysis, producing OH⁻. Step 2: Phenolphthalein — transition range pH 8.3–10.0. Step 3: The EP pH (≈ 9.2) falls within phenolphthalein’s range (8.3–10.0), so the indicator will change colour when the sharp pH jump passes through this range at equivalence, giving an accurate endpoint. (1 mark per step.)
Q2 — Concordant results
2.1 Trials 3, 4, 5 are concordant: 25.55, 25.57, 25.58 mL — all within 0.03 mL of each other (well within 0.10 mL). Mean = (25.55 + 25.57 + 25.58) / 3 = 25.57 mL. (1 mark identification; 1 mark mean with working.)
2.2 n(NaOH) = 0.09850 × 0.02557 = 2.519 × 10⁻³ mol. n(H⁺) = n(NaOH) = 2.519 × 10⁻³ mol. c(H⁺) = 2.519 × 10⁻³ / 0.02000 = 0.1260 mol/L. (1 mark moles; 1 mark concentration.)
2.3 Trial 2 (25.60 mL): outside the concordant group by more than 0.10 mL from Trials 3–5; likely an early endpoint or initial burette not rinsed correctly. Trial 6 (26.98 mL): large discrepancy of ~1.4 mL from the concordant group; likely a technique error (overshot endpoint or burette leak).
2.4 Recording to 0.05 mL reduces reading uncertainty; using only concordant results removes outliers caused by technique errors, so the mean titre better approximates the true equivalence volume and yields a more reliable concentration.
Q3 — Cause-and-effect chain
Row 1: The MO transition range (pH 3.1–4.4) does not encompass the equivalence point pH (> 7), so the indicator will change colour at the wrong point in the titration. Row 2: The student observes the colour change well before stoichiometric neutralisation is achieved. Row 3: The calculated moles of NaOH used is far less than the moles needed to reach the true equivalence point. Row 4: Since n(acid) = n(NaOH used), the reported moles of acid is much less than the true moles of acid. Overall % error: endpoint at 5.50 mL vs true 25.00 mL; reported concentration = (5.50/25.00) × 100% = 22% of true value; error = 78%.
Q4 — Predict and justify
4.1 The two titres will be essentially the same. A strong acid + strong base titration has a very large pH jump (approximately pH 4–10), so both phenolphthalein (8.3–10.0) and methyl orange (3.1–4.4) transition within the jump. Both indicators detect the equivalence point accurately, giving effectively identical titres. [1 mark prediction; 1 mark justification referencing jump range.]
4.2 For a weak acid + strong base, the jump is smaller and centred above pH 7; the buffer region occupies pH 3.1–5.0 before equivalence, so methyl orange transitions in the buffer region (far from EP) and gives a drastically underestimated titre. The fundamental difference is that the large jump of strong/strong encompasses all indicator ranges, while the smaller, higher jump of weak/strong only encompasses indicators whose range is above ~7. [1 mark each concept: why SA/SB is unconstrained; why WA/SB is constrained.]