Chemistry • Year 12 • Module 8 • Lesson 1
Acid-Base Titrations & Indicators
Build HSC Band 5–6 extended-response technique on titration as an analytical method: accuracy, reliability, indicator selection, and evaluation of real-world data.
1. Stimulus-based extended response — Orica industrial acid analysis (Band 5–6)
8 marks Band 5–6
Stimulus. Orica, an Australian industrial chemicals manufacturer, produces large batches of concentrated sulfuric acid (H2SO4) for use in mining and fertiliser production. Their quality control laboratory performs routine titration analysis on each batch to verify the H2SO4 concentration against the product specification of 18.0 mol L−1. Technicians use a primary standard of anhydrous Na2CO3 to prepare the NaOH standard solution, then titrate small diluted aliquots of the acid. The balanced equation for the titration is:
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
A technician performs four titrations of a 1.00 mL aliquot of the diluted acid (1 in 1000 dilution) with 0.2000 mol L−1 NaOH:
| Trial | Titre / mL | Include in average? |
|---|---|---|
| Rough | 18.20 | No |
| 1 | 17.90 | Yes |
| 2 | 17.85 | Yes |
| 3 | 17.92 | Yes |
Q1. Analyse and evaluate the quality control titration above. In your response you must:
- Define primary standard and explain why Na2CO3 is suitable for this role.
- Calculate the concentration of H2SO4 in the original undiluted batch using the concordant titres; show all steps.
- Evaluate whether the batch meets the 18.0 mol L−1 specification and calculate the percentage difference from the target.
- Identify one source of systematic error in this method and explain how it would affect the calculated concentration (whether it would produce a result that is too high or too low).
- Explain why using a primary standard to prepare the NaOH solution — rather than simply using the labelled concentration on a reagent bottle — is critical to the reliability of the result in a NATA-accredited laboratory.
2. Evaluate a titration experimental design — TGA pharmaceutical analysis (Band 5–6)
7 marks Band 5–6
Scenario. A TGA (Therapeutic Goods Administration) laboratory is evaluating a new back-titration method proposed for measuring the active calcium carbonate (CaCO3) content in a chewable antacid tablet. The proposed method is:
- Add 50.00 mL of 0.5000 mol L−1 HCl to the crushed tablet and stir for 5 minutes.
- Filter the mixture to remove insoluble excipients.
- Titrate the filtrate with 0.2000 mol L−1 NaOH using phenolphthalein indicator.
Relevant equations:
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
In a trial run, the NaOH titre for the excess acid was 28.40 mL.
Q2. Evaluate the suitability of this back-titration method for determining the CaCO3 content in the antacid tablet. In your response you must:
- Calculate the number of moles of CaCO3 in the tablet using the trial titre, and hence the mass of CaCO3. (M(CaCO3) = 100.09 g mol−1)
- Explain why back titration is more appropriate than direct titration for this sample, using at least two chemical reasons.
- Evaluate the choice of phenolphthalein as the indicator for step 3, with specific reference to what species are present in the filtrate at the equivalence point.
- Identify one modification to the method that would improve the reliability of the result and justify your suggestion.
Q1 — Sample Band 6 response (8 marks), annotated
Primary standard definition and Na2CO3 suitability: A primary standard is a highly pure, chemically stable solid with a known molar mass that can be accurately weighed to prepare or verify a standard solution of precisely known concentration. Anhydrous Na2CO3 is suitable because it is non-hygroscopic under normal storage, has a reasonably high molar mass (105.99 g mol−1) that reduces the relative error in weighing, and reacts completely and stoichiometrically with HCl (and indirectly validates the NaOH used in subsequent titrations). [2 marks: 1 definition; 1 Na2CO3 justification with at least two properties]
Calculation: Average concordant titre = (17.90 + 17.85 + 17.92) / 3 = 17.89 mL = 0.01789 L. n(NaOH) = 0.2000 × 0.01789 = 0.003578 mol. Mole ratio H2SO4 : NaOH = 1:2, so n(H2SO4) = 0.003578 / 2 = 0.001789 mol. This is in the 1.00 mL diluted aliquot, so c(diluted) = 0.001789 / 0.001000 = 1.789 mol L−1. Multiply by dilution factor 1000: c(undiluted H2SO4) = 1.789 × 1000 = 17.9 mol L−1. [2 marks: 1 correct mole calculation with ratio applied; 1 correct application of dilution factor]
Evaluation against specification: The calculated concentration of 17.9 mol L−1 is below the 18.0 mol L−1 specification. Percentage difference = |(17.9 − 18.0)| / 18.0 × 100 = 0.56%. While this is a small discrepancy, in an industrial context it means the batch is slightly below specification and would need to be flagged for review. [1 mark]
Systematic error: One systematic error would be using a burette with a calibration error (e.g. scale displaced by 0.1 mL throughout), which would consistently produce titres that are slightly too large or too small in every trial, shifting the calculated concentration in one direction. Another example: if the NaOH solution absorbed CO2 from the air, forming Na2CO3, the effective NaOH concentration would be lower than assumed, causing the calculated H2SO4 concentration to be underestimated (since less NaOH reacts per mole than expected). [1 mark: identified error + direction of effect]
Primary standard vs reagent bottle label: The labelled concentration on a reagent bottle is often nominal or reflects the manufactured concentration at the time of production, not the actual concentration at time of use. NaOH solutions in particular absorb CO2 and water vapour from the air over time, reducing their effective concentration. Standardising against a primary standard ensures that the concentration used in calculations is verified experimentally rather than assumed, which is a fundamental requirement in accredited laboratories where results must be traceable to a known measurement standard. [2 marks]
Marking criteria summary:
- 1 mark — Correct definition of primary standard (pure, stable, known molar mass, accurately weighed).
- 1 mark — Explains why Na2CO3 is suitable (at least two properties: non-hygroscopic, high molar mass, stoichiometric reaction).
- 1 mark — Correct average titre and mole calculation with 1:2 ratio applied.
- 1 mark — Correct application of the 1000× dilution factor to give 17.9 mol L−1.
- 1 mark — Evaluates result against 18.0 mol L−1 specification with percentage difference or clear comparison.
- 1 mark — Identifies a named systematic error and correctly states the direction of its effect on the result.
- 2 marks — Explains why standardisation against a primary standard is essential for reliable results (distinguishes from nominal bottle label; links to traceability or absorptive degradation of NaOH).
Q2 — Sample Band 6 response (7 marks), annotated
Calculation: n(NaOH) = 0.2000 × 0.02840 = 0.005680 mol. HCl + NaOH react 1:1, so n(excess HCl) = 0.005680 mol. n(initial HCl) = 0.5000 × 0.05000 = 0.02500 mol. n(HCl reacted with CaCO3) = 0.02500 − 0.005680 = 0.01932 mol. CaCO3 : HCl = 1:2, so n(CaCO3) = 0.01932 / 2 = 0.009660 mol. m(CaCO3) = 0.009660 × 100.09 = 0.967 g. [2 marks: 1 for correctly subtracting excess; 1 for correct 1:2 ratio and mass]
Why back titration is more appropriate: CaCO3 is an insoluble solid that cannot be dissolved and directly pipetted as a precise aliquot, making direct titration impractical. Additionally, the reaction of CaCO3 with HCl produces CO2 gas which can cause bubbling that makes it difficult to detect a precise endpoint directly. Back titration avoids both problems by adding a known excess of HCl to dissolve the solid and allow complete reaction, then measuring only the remaining excess acid in a clear, CO2-free solution. [2 marks: 1 per distinct chemical reason]
Indicator choice evaluation: At the equivalence point of the second titration (HCl + NaOH), the solution contains NaCl(aq), a neutral salt that produces a solution at approximately pH 7. Phenolphthalein transitions between pH 8.2 and 10.0, which is above pH 7. For a strong acid (excess HCl) being titrated by a strong base (NaOH), the equivalence point is at pH 7, so the steep pH jump spans from about pH 4 to pH 10. Phenolphthalein’s transition range (pH 8.2–10.0) falls within this steep region, so it is an acceptable indicator, though bromothymol blue (pH 6.0–7.6) would also work and would give an endpoint closer to pH 7. [2 marks: 1 for identifying equivalence pH; 1 for evaluating whether phenolphthalein is within the steep jump]
Modification to improve reliability: Stir the mixture for longer (e.g. 15–20 minutes) or gently heat the mixture after adding HCl to ensure the CaCO3 reacts completely before filtering. If any CaCO3 remains unreacted, the calculated amount of acid consumed will be too low, leading to an underestimate of the CaCO3 content in the tablet. Complete reaction is the most critical validity requirement for back titration. [1 mark]
Marking criteria summary:
- 1 mark — Correctly calculates n(excess HCl) using 1:1 NaOH:HCl ratio.
- 1 mark — Correctly subtracts excess HCl, applies 1:2 CaCO3:HCl ratio, and calculates mass of CaCO3 as 0.967 g.
- 1 mark — Explains one chemical reason why back titration is preferred (insoluble solid or CO2 gas production).
- 1 mark — Explains a second distinct chemical reason.
- 1 mark — Identifies that the equivalence point in step 3 is at pH ≈ 7 (strong acid + strong base).
- 1 mark — Evaluates phenolphthalein against the steep jump and states whether it is acceptable (yes, falls within the steep region; alternative indicators also acceptable).
- 1 mark — Identifies a specific modification to improve reliability with a valid justification (e.g. longer stirring/heating to ensure complete reaction of CaCO3; conducting multiple trials and averaging; filtering carefully to prevent any CaCO3 passing into the filtrate).