Chemistry • Year 12 • Module 8 • Lesson 1
Acid-Base Titrations & Indicators
Apply titration analysis to real data sets, titration curve interpretation, and Australian laboratory contexts including AWRI wine acidity analysis.
1. Interpret a laboratory titration data set
A Year 12 student is standardising a sodium hydroxide solution against a primary standard of oxalic acid (H2C2O4). A 25.00 mL aliquot of 0.0500 mol L−1 oxalic acid solution is pipetted into each conical flask. The student records the following burette readings. 8 marks
| Trial | Initial reading / mL | Final reading / mL | Titre / mL |
|---|---|---|---|
| Rough | 0.10 | 21.85 | 21.75 |
| 1 | 0.05 | 20.70 | 20.65 |
| 2 | 0.15 | 20.80 | 20.65 |
| 3 | 0.20 | 20.95 | 20.75 |
| 4 | 0.10 | 20.78 | 20.68 |
Balanced equation: H2C2O4(aq) + 2NaOH(aq) → Na2C2O4(aq) + 2H2O(l)
1.1 Identify which trials should be used in the average and explain why trials 3 and the rough are excluded. 3 marks
1.2 Calculate the average concordant titre. 1 mark
1.3 Using your average titre, calculate the concentration of the NaOH solution. Show every step. 3 marks
1.4 Suggest one reason why oxalic acid dihydrate (H2C2O4·2H2O) rather than a different acid is commonly used as a primary standard to standardise NaOH. 1 mark
2. Titration curve analysis
The graph below shows a pH–volume titration curve for 25.00 mL of a weak acid titrated with 0.100 mol L−1 NaOH. Study the curve carefully, then answer the sub-questions. 9 marks
Figure 2.1. Titration curve for 25.00 mL of a weak monoprotic acid with 0.100 mol L−1 NaOH. Stylised model for HSC analysis practice.
2.1 Read the graph and state the volume of NaOH at the equivalence point and the approximate pH at that point. 2 marks
2.2 Explain why phenolphthalein (pH transition 8.2–10.0) is the correct indicator choice for this titration, while methyl orange (pH 3.1–4.4) would give an error. 3 marks
2.3 The graph shows the equivalence point at approximately pH 8.8. Explain why phenolphthalein (pH 8.2–10.0) is a better indicator choice than methyl orange (pH 3.1–4.4) for this weak acid–strong base titration. 2 marks
2.4 The equivalence point of a weak acid–strong base titration is above pH 7. Explain why this means the endpoint detected by methyl orange would occur before the true equivalence point, and state the effect this would have on the calculated analyte concentration. 2 marks
3. Case study — AWRI wine acidity analysis
The Australian Wine Research Institute (AWRI) in Adelaide routinely measures the titratable acidity (TA) of wine to ensure product quality and regulatory compliance. In a standard AWRI TA analysis, a 5.00 mL aliquot of wine is diluted and titrated against standardised 0.1000 mol L−1 NaOH solution to pH 8.2 (the phenolphthalein endpoint). The dominant acid in wine is tartaric acid (H2C4H4O6, M = 150.09 g mol−1), a diprotic weak acid. 6 marks
Titration results for a Barossa Valley shiraz sample:
| Trial | Volume NaOH / mL | Include in average? |
|---|---|---|
| Rough | 14.60 | No |
| 1 | 13.25 | Yes |
| 2 | 13.30 | Yes |
| 3 | 13.20 | Yes |
Balanced equation: H2C4H4O6(aq) + 2NaOH(aq) → Na2C4H4O6(aq) + 2H2O(l)
3.1 Using the concordant titres, calculate the concentration of tartaric acid in the original wine in g L−1. Show all steps. 4 marks
3.2 A NATA-accredited laboratory requires results to be reported to three significant figures and concordant within ±0.10 mL. Explain why these two requirements both contribute to the reliability of the result. 2 marks
4. Compare and contrast — direct vs back titration
Complete the comparison table for the two titration approaches used to analyse the acid-neutralising capacity of antacid tablets. 6 marks
| Feature | Direct titration | Back titration |
|---|---|---|
| How the analyte is introduced | ||
| Number of titration steps | ||
| When it is preferred | ||
| Key calculation step (after titration) | ||
| Main source of error | ||
| Suitable for slow-dissolving solids? |
Q1 — Titration data set
1.1 Trials 1, 2 and 4 (titres 20.65, 20.65, 20.68 mL) are concordant — maximum spread is 0.03 mL, well within 0.10 mL. Trial 3 (20.75 mL) is excluded because it differs from trials 1 and 2 by 0.10 mL and from trial 4 by 0.07 mL — it sits just outside the concordant group. The rough trial is excluded because it was used only to find the endpoint region.
1.2 Average = (20.65 + 20.65 + 20.68) / 3 = 20.66 mL.
1.3 n(H2C2O4) = cV = 0.0500 × 0.02500 = 0.001250 mol. Mole ratio NaOH : H2C2O4 = 2:1, so n(NaOH) = 2 × 0.001250 = 0.002500 mol. c(NaOH) = n/V = 0.002500 / 0.02066 = 0.1210 mol L−1.
1.4 Oxalic acid dihydrate is a solid primary standard with a known, stable molar mass that can be accurately weighed; it is not hygroscopic under normal storage conditions and reacts completely with NaOH, making it suitable for preparing a solution of accurately known concentration.
Q2 — Titration curve analysis
2.1 Equivalence point: 25.0 mL of NaOH; pH ≈ 8.8.
2.2 The equivalence point in this weak acid–strong base titration occurs at approximately pH 8.8, which is above pH 7. Phenolphthalein transitions between pH 8.2 and 10.0, which overlaps the steep pH jump region around the equivalence point; therefore the endpoint occurs close to pH 8.8 and gives a reliable result. Methyl orange changes colour in the range pH 3.1–4.4, which is far below the equivalence point. If methyl orange were used, the endpoint would signal too early — long before stoichiometric neutralisation is reached — giving a titre that is too small and an underestimated analyte concentration.
2.3 The equivalence point at pH ≈ 8.8 is above pH 7, which means it falls inside the phenolphthalein transition range (pH 8.2–10.0). The endpoint detected by phenolphthalein therefore occurs very close to the equivalence point, making it a reliable indicator. Methyl orange transitions at pH 3.1–4.4, which is far below the equivalence point — it would change colour early, long before stoichiometric neutralisation is complete, giving an incorrect titre.
2.4 If methyl orange is used, the endpoint is signalled at around pH 3.1–4.4, well before the weak acid–strong base equivalence point at ≈ pH 8.8. The titre recorded would be too small because the student stops adding NaOH too early. This means the calculated moles of NaOH used would be smaller than the true amount needed for equivalence, so the calculated concentration of the weak acid analyte would be underestimated (too low).
Q3 — AWRI wine analysis
3.1 Average concordant titre = (13.25 + 13.30 + 13.20) / 3 = 13.25 mL = 0.01325 L. n(NaOH) = 0.1000 × 0.01325 = 0.001325 mol. n(H2C4H4O6) = 0.001325 / 2 = 6.625 × 10−4 mol. This is in 5.00 mL of wine, so c(tartaric acid) = 6.625 × 10−4 / 0.005000 = 0.1325 mol L−1. Mass concentration = 0.1325 × 150.09 = 19.9 g L−1.
3.2 Reporting to three significant figures limits the effect of rounding errors, ensuring the result accurately represents the measurement precision achievable with burettes (which read to 0.05 mL). Requiring concordant titres within ±0.10 mL ensures reproducibility — it confirms the measurement can be reliably repeated, which is a key criterion for NATA accreditation where traceability and consistency must be demonstrated.
Q4 — Direct vs back titration comparison
| Feature | Direct titration | Back titration |
|---|---|---|
| How the analyte is introduced | Dissolved analyte placed directly in the flask as an aliquot | Excess standard reagent is added to the sample first; unreacted excess is then titrated |
| Number of titration steps | One | Two (add excess, then titrate excess) |
| When it is preferred | When the analyte dissolves readily and reacts quickly to a clear endpoint | When the analyte is slow to dissolve, a solid, or lacks a clear endpoint for direct titration |
| Key calculation step | Use n = cV of standard directly | Subtract excess moles from initial moles to find reacted moles |
| Main source of error | Overshooting the endpoint; poor indicator choice | Incomplete reaction of sample with the added excess reagent |
| Suitable for slow-dissolving solids? | Generally no | Yes — the tablet can be given time to react before the second titration |