Every antacid tablet you've ever taken was tested by titration before it left the factory. Every blood pH reading, every wine acidity measurement, every batch of pharmaceutical drugs — titration is the technique that underpins quantitative chemistry in the real world. It's also the most commonly examined calculation type in NSW HSC Chemistry.
📚 Core Content
Volumetric analysis (titration) determines the concentration of an unknown solution (the analyte) by reacting it with a solution of known concentration (the titrant). The titrant is added slowly from a burette until the reaction is exactly complete — this is the equivalence point.
Because you can't always see the equivalence point directly, an indicator is added — a chemical that changes colour near the equivalence point. The moment the colour change occurs is the end point. Ideally, the end point and equivalence point coincide.
| Equipment | Purpose | Read to | Why this piece? |
|---|---|---|---|
| Burette (50 mL) | Delivers variable volumes of titrant with precision | ±0.05 mL | Only piece that can dispense continuously and be read mid-titration |
| Pipette (25 mL) | Delivers a fixed, precise volume of analyte | ±0.02 mL | More accurate than measuring cylinder; delivers exact aliquot each time |
| Volumetric flask | Prepares standard solution to exact volume | ±0.1 mL | Narrow neck allows precise reading of the calibration mark |
| Conical flask | Holds analyte during titration | — | Narrow neck prevents splashing; easy to swirl without spilling |
| White tile | Placed under conical flask | — | Provides contrast to see indicator colour change clearly |
Every titration calculation, regardless of complexity, follows the same four steps. Memorise this sequence — it will never fail you.
🧮 Worked Examples
🧪 Activities
1 25.00 mL of NaOH is titrated against 0.0500 mol L⁻¹ HCl. The concordant titres are 20.10 mL and 20.15 mL. Calculate c(NaOH).
HCl + NaOH → NaCl + H₂O
2 20.00 mL of Na₂CO₃ (0.1500 mol L⁻¹) is titrated against HCl. The average titre is 30.00 mL. Calculate c(HCl).
Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂
3 25.00 mL of H₂SO₄ is titrated against 0.1200 mol L⁻¹ NaOH, using an average titre of 24.00 mL. Calculate c(H₂SO₄).
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
4 A vinegar sample is diluted to 100.0 mL. A 20.00 mL aliquot is titrated against 0.1000 mol L⁻¹ NaOH. The average titre is 16.80 mL. Calculate the mass of acetic acid (CH₃COOH) in the original 100 mL sample. (MM = 60.05 g mol⁻¹)
CH₃COOH + NaOH → CH₃COONa + H₂O
Show full 4-step working for each problem:
Complete in your workbook.
❓ Multiple Choice
1. Why is the burette rinsed with titrant solution before filling?
2. In a titration, 25.00 mL of 0.200 mol L⁻¹ NaOH reacts with HCl. The average titre is 20.00 mL. Using HCl + NaOH → NaCl + H₂O, what is c(HCl)?
3. What is the difference between the equivalence point and the end point?
4. Titres of 24.50 mL (rough), 23.80 mL, 23.75 mL, and 23.82 mL are recorded. What volume should be used for calculations?
5. In the reaction H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O, if 0.0100 mol of H₂SO₄ reacts, how many moles of KOH are needed?
✍️ Short Answer
6. A student titrates 25.00 mL of a Na₂CO₃ solution against 0.1000 mol L⁻¹ HCl, using methyl orange as an indicator. The titration results are: rough = 26.50 mL, accurate = 25.30 mL, 25.25 mL, 25.80 mL. (a) Select the concordant titres and calculate the average. (b) Calculate c(Na₂CO₃). Equation: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. 5 MARKS
Type your answer:
Answer in workbook.
7. A student dissolves a sample of impure oxalic acid (H₂C₂O₄) in water and makes up to 250.0 mL. A 25.00 mL aliquot is titrated against 0.1000 mol L⁻¹ NaOH. The average titre is 18.60 mL. Equation: H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O. (a) Calculate the moles of H₂C₂O₄ in the aliquot. (b) Calculate the mass of H₂C₂O₄ in the original 250 mL solution. (c) If the original sample had a mass of 0.400 g, calculate the percentage purity. (MM of H₂C₂O₄ = 90.03 g mol⁻¹) 6 MARKS
Type your answer:
Answer in workbook.
1. B — Residual water in the burette dilutes the titrant, reducing its concentration and causing more to be used than expected — making the unknown appear more concentrated than it is.
2. C — n(NaOH) = 0.200 × 0.02500 = 5.00 × 10⁻³ mol. 1:1 ratio → n(HCl) = 5.00 × 10⁻³. c(HCl) = 5.00 × 10⁻³ ÷ 0.02000 = 0.250 mol L⁻¹.
3. A — The equivalence point is a stoichiometric concept (equal moles reacted); the end point is an experimental observation (indicator colour change). A good indicator choice makes them nearly coincide.
4. D — Rough titre (24.50) excluded. Accurate titres: 23.80, 23.75, 23.82. Check concordance: 23.82 − 23.75 = 0.07 ✓. Average = (23.80 + 23.75 + 23.82) ÷ 3 = 23.79 mL.
5. B — The ratio is 1:2 (H₂SO₄:KOH), so 0.0100 × 2 = 0.0200 mol KOH.
Q6 (5 marks):
(a) Concordant: 25.30 and 25.25 mL (diff = 0.05 mL ✓). 25.80 excluded (diff from 25.30 = 0.50 mL ✗). Average = (25.30 + 25.25) ÷ 2 = 25.28 mL = 0.02528 L (b) n(HCl) = 0.1000 × 0.02528 = 2.528 × 10⁻³ mol n(Na₂CO₃) = 2.528 × 10⁻³ ÷ 2 = 1.264 × 10⁻³ mol c(Na₂CO₃) = 1.264 × 10⁻³ ÷ 0.02500 = 0.05056 mol L⁻¹ ≈ 0.0506 mol L⁻¹Q7 (6 marks):
(a) n(NaOH) = 0.1000 × 0.01860 = 1.860 × 10⁻³ mol n(H₂C₂O₄) in aliquot = 1.860 × 10⁻³ ÷ 2 = 9.300 × 10⁻⁴ mol (b) n(total) = 9.300 × 10⁻⁴ × (250.0 ÷ 25.00) = 9.300 × 10⁻³ mol m = 9.300 × 10⁻³ × 90.03 = 0.8373 g (c) % purity = (0.8373 ÷ 0.400) × 100 = 209%Note: A purity >100% is physically impossible and indicates an error in the given data (sample mass too small, or titre too large). In an exam, present the calculation correctly and note the anomaly.
Tick when you've finished all activities and checked your answers.