Chemistry • Year 11 • Module 2 • Lesson 10
Volumetric Analysis & Titration
Apply the 4-step titration calculation method to real data, interpret titration results, and reason through experimental scenarios.
1. Interpret titration results — concordance and averaging
A student titrated 25.00 mL aliquots of NaOH solution with 0.1000 mol L−1 HCl. The results are shown below. 8 marks
| Run | Initial burette reading (mL) | Final burette reading (mL) | Titre (mL) | Include in average? |
|---|---|---|---|---|
| Rough | 0.00 | 20.90 | ||
| 1 | 0.10 | 20.45 | ||
| 2 | 0.20 | 20.58 | ||
| 3 | 0.15 | 20.48 |
(a) Calculate the titre for each accurate run and determine which are concordant. Show your concordance check. (3 marks)
(b) Calculate the average titre to use in the calculation. (1 mark)
(c) Using the balanced equation HCl + NaOH → NaCl + H2O, apply all 4 steps to find c(NaOH). Show all working. (3 marks)
(d) A student in the next group accidentally uses Run 2’s titre in isolation rather than the concordant average. Identify the error this introduces and whether it makes c(NaOH) higher or lower than correct. (1 mark)
2. Cause-and-effect chain — the 4-step calculation
The boxes on the left describe the inputs at each step of a titration calculation. Fill in the corresponding output / effect boxes on the right. The scenario: 20.00 mL of 0.2500 mol L−1 NaOH is titrated against H2SO4; average titre = 12.50 mL. Equation: H2SO4 + 2NaOH → Na2SO4 + 2H2O. 5 marks
Cause: c(NaOH) = 0.2500 mol L−1; V(NaOH) = 20.00 mL = 0.02000 L
Effect (Step 1): n(NaOH) = _______________________
Cause: Equation gives 1 mol H2SO4 : 2 mol NaOH
Effect (Step 2): n(H2SO4) = _______________________
Cause: V(H2SO4) = 12.50 mL = 0.01250 L; n(H2SO4) from Step 2
Effect (Step 3): c(H2SO4) = _______________________
Common error check: A student gets 0.4000 mol L−1. What ratio mistake did they make?
Overall outcome: Error =
3. Interpret a titration curve
The graph below shows a simulated pH–volume titration curve for a strong acid–strong base titration. Use it to answer the questions. 7 marks
(a) Identify the equivalence point volume from the graph and state the pH at that point. (2 marks)
(b) Describe the shape of the curve in the region 0–20 mL and explain why the pH changes slowly in this region compared to the region near 25 mL. (2 marks)
(c) Identify which indicator — phenolphthalein or methyl orange — would be more appropriate for this titration and justify your choice with reference to pH transition ranges. (2 marks)
(d) Estimate the initial concentration of the HCl solution, showing your reasoning. (Assume the pipette volume was 25.00 mL.) (1 mark)
4. Case study — antacid tablet analysis
7 marks
Scenario. A Quality Assurance chemist at a pharmaceutical plant needs to verify that each antacid tablet contains the correct amount of magnesium hydroxide (Mg(OH)2). The tablet is crushed and dissolved, then made up to 250.0 mL in a volumetric flask. A 25.00 mL aliquot is titrated against 0.1000 mol L−1 HCl using methyl orange. Three accurate titres are recorded: 22.38, 22.40, 22.35 mL. The equation is: Mg(OH)2 + 2HCl → MgCl2 + 2H2O. Molar mass of Mg(OH)2 = 58.32 g mol−1. The labelled amount per tablet is 400 mg.
(a) Determine the average titre and explain why the rough titration (26.10 mL) was excluded. (2 marks)
(b) Apply the 4-step method to calculate the mass of Mg(OH)2 in the whole tablet. Show all steps. (4 marks)
(c) State whether the tablet meets the labelled specification and calculate the percentage difference from 400 mg. (1 mark)
Q1 — Concordance and 4-step calculation
(a) Titres: Rough = 20.90 mL (discarded); Run 1 = 20.45 − 0.10 = 20.35 mL; Run 2 = 20.58 − 0.20 = 20.38 mL; Run 3 = 20.48 − 0.15 = 20.33 mL. Concordance check (all three): max–min = 20.38 − 20.33 = 0.05 mL ≤ 0.10 mL → all three are concordant.
(b) Average = (20.35 + 20.38 + 20.33) ÷ 3 = 20.35 mL
(c) Step 1: n(HCl) = 0.1000 × 0.02035 = 2.035 × 10−3 mol. Step 2: 1:1 ratio → n(NaOH) = 2.035 × 10−3 mol. Step 3: c(NaOH) = 2.035 × 10−3 ÷ 0.02500 = 0.0814 mol L−1.
(d) Run 2 titre (20.38 mL) is the largest concordant titre — using it alone gives a slightly higher n(HCl) and therefore a slightly higher c(NaOH) compared to the correct average. The error is small (<0.2%) but using a single non-averaged titre is poor quantitative technique.
Q2 — Cause-and-effect chain
Step 1: n(NaOH) = 0.2500 × 0.02000 = 5.000 × 10−3 mol
Step 2: n(H2SO4) = 5.000 × 10−3 ÷ 2 = 2.500 × 10−3 mol (ratio 1:2)
Step 3: c(H2SO4) = 2.500 × 10−3 ÷ 0.01250 = 0.2000 mol L−1
Common error: The student used a 1:1 ratio instead of 1:2, doubling n(H2SO4) and therefore doubling the final concentration.
Q3 — Titration curve
(a) Equivalence point volume = 25 mL (where the curve is steepest / centre of the vertical jump); pH at equivalence point = 7 (strong acid + strong base → neutral salt).
(b) From 0–20 mL the pH rises slowly and gradually (gentle slope) because excess HCl buffers the solution — the large reservoir of H+ ions means small additions of OH− cause negligible pH change. Near 25 mL, almost all H+ has been neutralised, so each drop of NaOH now dramatically shifts the pH.
(c) Either indicator is suitable for a strong acid–strong base titration because the vertical pH jump spans both transition zones (phenolphthalein: pH 8.3–10; methyl orange: pH 3.1–4.4). The sharp jump >8 pH units means both would change colour within a single drop of titrant at the equivalence point. Either is acceptable; phenolphthalein is often preferred because its colourless → pink change is easier to judge against a white tile.
(d) At the equivalence point, n(HCl) = n(NaOH) = 0.1000 × 0.02500 = 2.500 × 10−3 mol. Volume of HCl = 25.00 mL = 0.02500 L. c(HCl) = 2.500 × 10−3 ÷ 0.02500 = 0.1000 mol L−1.
Q4 — Antacid tablet case study
(a) Average titre = (22.38 + 22.40 + 22.35) ÷ 3 = 22.38 mL. The rough titre (26.10 mL) is excluded because it intentionally overshoots the endpoint and is used only to locate the approximate end point for subsequent accurate runs; it is never concordant with accurate titres.
(b)
Step 1: n(HCl) = 0.1000 × 0.02238 = 2.238 × 10−3 mol
Step 2: n(Mg(OH)2) in aliquot = 2.238 × 10−3 ÷ 2 = 1.119 × 10−3 mol (ratio 1:2)
Step 3 (scale-up): n(Mg(OH)2) in whole tablet = 1.119 × 10−3 × (250.0 ÷ 25.00) = 1.119 × 10−2 mol
Step 4: m(Mg(OH)2) = 1.119 × 10−2 × 58.32 = 0.6526 g = 652.6 mg
(c) 652.6 mg is greater than the labelled 400 mg. % difference = |(652.6 − 400) ÷ 400| × 100 = 63.2% above specification. The tablet exceeds the labelled amount significantly and would fail QA. Note: this discrepancy may indicate an error in the illustrative data; present the calculation correctly and flag the anomaly.