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.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A chemist has a solution of hydrochloric acid (HCl) but doesn't know its exact concentration. They have a standard solution of NaOH (exactly 0.1000 mol L⁻¹). How could they use the NaOH to find the concentration of the HCl? What would they need to measure, and what would they need to know about the reaction?
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📚 Core Content
Wrong: Concentration and amount of solute are the same thing.
Right: Concentration is amount per unit volume; the same amount of solute can produce different concentrations in different volumes.
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
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At the start of this lesson, you thought about how to use a standard NaOH solution to find the concentration of HCl.
The answer: place a known volume of HCl in the conical flask and titrate with the standard NaOH from the burette until the indicator changes colour (endpoint). You measure the titre (volume of NaOH used). Knowing n(NaOH) = c × V and using the 1:1 mole ratio (HCl:NaOH = 1:1), you find n(HCl) = n(NaOH). Then c(HCl) = n(HCl) ÷ V(HCl). The key information needed: the exact volume of each solution and the balanced equation to apply the mole ratio.
Reflect: how close was your initial strategy to the actual titration procedure?
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5 random questions from a replayable lesson bank — feedback shown immediately
✍️ 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
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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
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8. A student titrates 25.00 mL of vinegar (dilute acetic acid, CH₃COOH) against 0.1000 mol L⁻¹ NaOH using phenolphthalein. Equation: CH₃COOH + NaOH → CH₃COONa + H₂O. The student records titres of 21.50 (rough), 22.10, 22.05, 22.12 mL. They calculate c(CH₃COOH) = 0.0882 mol L⁻¹. A classmate says: "You made an error — you included the non-concordant titre." Evaluate the student's calculation and identify whether the classmate's claim is correct. (Assume concordant means within 0.10 mL) 5 MARKS
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9. Design a complete back-titration procedure to determine the percentage purity of a sample of CaCO₃ (limestone). You have: the limestone sample, 1.000 mol L⁻¹ HCl (excess), 0.1000 mol L⁻¹ NaOH (standard), phenolphthalein indicator, and a 250 mL volumetric flask. The reaction is: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Include all steps and show how you would calculate % purity from the titration data. (Ca = 40.078, C = 12.011, O = 15.999) 7 MARKS
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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.
6. B — All four titres (15.20, 15.15, 15.22, 15.17) are concordant: range = 15.22 − 15.15 = 0.07 mL ≤ 0.10 mL. Both averaging methods are valid. Mean of 4 = 15.185 mL; mean of 3 most concordant (excluding 15.22) = (15.20+15.15+15.17)÷3 = 15.173 mL. c(HCl) difference = 0.2000 × 0.02000 ÷ (0.01519 − 0.01517) ≈ negligible. Both approaches are acceptable.
7. C — n(NaOH) ≈ 0.10 × 0.020 = 0.0020 mol. Since KHP:NaOH = 1:1, n(KHP) = 0.0020 mol. m(KHP) = 0.0020 × 204.22 = 0.408 g. This gives a titre in the 20 mL range with 0.10 mol L⁻¹ NaOH.
Q8 (5 marks):
Concordance check: 22.12 − 22.05 = 0.07; 22.12 − 22.10 = 0.02; all three within 0.10 mL ✓ [1]. Average = (22.10 + 22.05 + 22.12) ÷ 3 = 22.09 mL [1]. n(NaOH) = 0.1000 × 0.02209 = 2.209 × 10⁻³ mol [1]. 1:1 ratio → n(CH₃COOH) = 2.209 × 10⁻³ mol. c(CH₃COOH) = 2.209 × 10⁻³ ÷ 0.02500 = 0.0884 mol L⁻¹ [1]. The student's answer of 0.0882 mol L⁻¹ is slightly off — they likely used a slightly different average (perhaps included the rough titre or rounded differently). The classmate is wrong about a non-concordant titre — all three accurate titres are within 0.10 mL [1].
Q9 (7 marks):
Step 1: Weigh ~1 g of limestone sample accurately. Record mass m₁ [1].
Step 2: Add exactly 50.00 mL of 1.000 mol L⁻¹ HCl (n(HCl)initial = 0.05000 mol). React until fizzing stops [1].
Step 3: Transfer quantitatively to 250 mL volumetric flask and make up to the mark [1].
Step 4: Pipette 25.00 mL aliquot into conical flask. Add phenolphthalein. Titrate with 0.1000 mol L⁻¹ NaOH. Record titre V(NaOH) [1].
Calculation:
n(HCl) in aliquot = n(NaOH) × titre volume × 0.1000 n(HCl) total excess = n(NaOH) × (250÷25) [×10 scale factor] [1] n(HCl) reacted with CaCO₃ = n(HCl)initial − n(HCl)excess n(CaCO₃) = n(HCl reacted) ÷ 2 [from 1:2 ratio] [1] m(CaCO₃) = n × MM(CaCO₃) = n × 100.09 g mol⁻¹; % purity = m(CaCO₃) ÷ m₁ × 100 [1]Put your knowledge of titration technique, equivalence point and volumetric analysis calculations to the test. Answer correctly to deal damage — get it wrong and the boss hits back. Pool: lessons 1–10.
Tick when you've finished all activities and checked your answers.