Volumetric Analysis & Titration
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.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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?
Key facts
- Definitions: analyte, titrant, equivalence point, end point
- Equipment: burette, pipette, volumetric flask, conical flask
- Common indicators and their colour changes
- The 4-step titration calculation method
Concepts
- Why a standard solution is needed as the titrant
- Difference between equivalence point and end point
- Why concordant titres are averaged
Skills
- Describe the titration procedure with justifications
- Apply the 4-step method to find concentration, mass, or purity
- Calculate the average of concordant titres correctly
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
| Equipment | Purpose | Read to |
|---|---|---|
| Burette (50 mL) | Delivers variable volumes of titrant with precision | ±0.05 mL |
| Pipette (25 mL) | Delivers a fixed, precise volume of analyte | ±0.02 mL |
| Volumetric flask | Prepares standard solution to exact volume | ±0.1 mL |
| Conical flask | Holds analyte during titration (narrow neck, easy to swirl) | — |
| White tile | Placed under conical flask for contrast | — |
Indicators (the colour signal)
- Phenolphthalein: colourless → pink at pH 8.3 (use for strong acid–strong base or weak acid–strong base)
- Methyl orange: red → yellow at pH 3.1–4.4 (use for strong base–strong acid or weak base–strong acid)
- Universal indicator: rainbow gradient — too imprecise for titration; only good for estimating pH
The titration procedure (5 steps)
- Rinse and fill the burette with titrant. Rinse with titrant (not water) — water dilutes the titrant.
- Pipette the analyte into a conical flask. Rinse the pipette with analyte first for the same reason.
- Add indicator and perform a rough titration. Quick run to find approximate end point. Rough titre is discarded.
- Perform accurate titrations until concordant. Drop-by-drop near the end point. Two+ within 0.10 mL = concordant.
- Average the concordant titres. Never include the rough titre.
Titration: analyte (unknown, in flask) is titrated against titrant (standard, in burette) until the equivalence point — signalled by an indicator colour change (end point). Rinse burette with titrant, pipette with analyte. Procedure: rough titre (discard) → ≥2 concordant titres (within 0.10 mL) → average. Phenolphthalein (pH 8.3, colourless→pink); methyl orange (pH 3.1–4.4, red→yellow).
Pause — copy the highlighted procedure into your book before moving on.
Did you get this? True or false: the rough titre is always included in the average.
We just saw the titration procedure — how to accurately measure a titre using concordant results. That raises a question: once you have an average titre, how do you calculate the concentration of the unknown? This card answers it → with a four-step method that works for every acid-base titration.
Every titration calculation, regardless of complexity, follows the same four steps. Memorise this sequence — it will never fail you.
Step 2: Use the mole ratio from the balanced equation to find moles of the unknown
Step 3: Calculate the concentration of the unknown: c = n ÷ V (V in litres)
Step 4: If required, convert to mass or percentage
4-step titration calculation: (1) n(standard) = c × V (V in L); (2) n(unknown) = n(standard) × mole ratio from balanced equation; (3) c(unknown) = n ÷ V; (4) convert if needed. Always write the balanced equation first — the ratio is not always 1:1 (e.g. H₂SO₄:NaOH is 1:2).
Add the highlighted method to your notes before the check below.
Quick check: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. If n(NaOH) = 0.020 mol, what is n(H₂SO₄)?
Worked examples · reveal as you go
25.00 mL of NaOH solution is titrated against 0.1000 mol L⁻¹ HCl. Three concordant titres of 18.45, 18.50 and 18.48 mL are recorded. Calculate the concentration of the NaOH solution.
20.00 mL of 0.2500 mol L⁻¹ NaOH is titrated against H₂SO₄ solution. The average titre is 12.50 mL. Calculate c(H₂SO₄).
An antacid tablet is dissolved and made up to 250.0 mL. A 25.00 mL aliquot is titrated against 0.1000 mol L⁻¹ HCl. Average titre = 22.40 mL. Active ingredient: Mg(OH)₂. Calculate (a) moles of Mg(OH)₂ in the aliquot, (b) mass of Mg(OH)₂ in the whole tablet, (c) % purity if tablet mass = 1.500 g. (Mg = 24.305, O = 15.999, H = 1.008)
Two truths, one lie — about titration technique. Pick the lie.
Sort the steps — to find the concentration of NaOH from a titration with standard HCl. Click two steps to swap them, then check the order.
- n(HCl) = c(HCl) × V(HCl, average titre in L)
- Write balanced equation: HCl + NaOH → NaCl + H₂O | identify mole ratio (1:1)
- c(NaOH) = n(NaOH) ÷ V(NaOH aliquot, in L)
- Average the concordant titres (ignore the rough)
- n(NaOH) = n(HCl) × (1 ÷ 1) | apply mole ratio
Common errors · the 3 traps that cost marks
Assuming 1:1 mole ratio without checking
H₂SO₄ reacts with 2 NaOH. H₃PO₄ reacts with 3 NaOH. Na₂CO₃ reacts with 2 HCl. If you blindly apply 1:1, you'll get answers exactly 2× or 3× off.
Fix: Write the balanced equation before every calculation. Circle the coefficients. Write "ratio: X:Y" explicitly.
Including the rough titre in the average
The rough titre is always larger than accurate titres (you overshoot intentionally). Including it inflates the average → too low a calculated concentration.
Fix: Identify concordant titres first (within 0.10 mL of each other). Average only those. Never include the rough.
Forgetting to scale up from aliquot to whole sample
When a tablet is dissolved in 250 mL and a 25 mL aliquot is titrated, the titration gives n for just that aliquot. To find the total in the tablet, multiply by (250÷25) = 10.
Fix: After finding n in the aliquot, always ask: was the whole sample titrated, or just a portion? If a portion, scale up.
Quick-fire practice · 5 reps +2 XP per reveal
25.00 mL of NaOH titrated against 0.0500 mol L⁻¹ HCl. Concordant titres: 20.10 and 20.15 mL. Find c(NaOH). (HCl + NaOH → NaCl + H₂O)
20.00 mL of Na₂CO₃ (0.1500 mol L⁻¹) titrated against HCl. Average titre = 30.00 mL. Find c(HCl). (Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂)
25.00 mL of H₂SO₄ titrated against 0.1200 mol L⁻¹ NaOH. Average titre = 24.00 mL. Find c(H₂SO₄). (H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O)
Vinegar diluted to 100.0 mL. 20.00 mL aliquot titrated against 0.1000 mol L⁻¹ NaOH. Average titre = 16.80 mL. Find mass of CH₃COOH in original 100 mL. (MM = 60.05) (CH₃COOH + NaOH → CH₃COONa + H₂O)
Four titres recorded: 24.50 (rough), 23.80, 23.75, 23.82 mL. Identify concordant titres and calculate the average.
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). Measure the titre. Knowing n(NaOH) = c × V and using the 1:1 mole ratio, find n(HCl) = n(NaOH). Then c(HCl) = n(HCl) ÷ V(HCl). Key info needed: exact volumes + balanced equation for the mole ratio.
Pick your answer, then rate your confidence — that tells the system what to drill next.
Q1. A student titrates 25.00 mL of a Na₂CO₃ solution against 0.1000 mol L⁻¹ HCl, using methyl orange. Results: rough = 26.50 mL, accurate = 25.30, 25.25, 25.80. (a) Select the concordant titres and calculate the average. (b) Calculate c(Na₂CO₃). Equation: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂.
Q2. 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. Average titre = 18.60 mL. Equation: H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O. (a) Calculate moles of H₂C₂O₄ in the aliquot. (b) Calculate mass of H₂C₂O₄ in the original 250 mL solution. (c) If the original sample mass = 0.400 g, calculate % purity. (MM H₂C₂O₄ = 90.03)
Q3. A student titrates 25.00 mL of vinegar (CH₃COOH) against 0.1000 mol L⁻¹ NaOH using phenolphthalein. Equation: CH₃COOH + NaOH → CH₃COONa + H₂O. Titres: 21.50 (rough), 22.10, 22.05, 22.12 mL. Student calculates c = 0.0882 mol L⁻¹. A classmate says: "You included a non-concordant titre." Evaluate the calculation and the classmate's claim. (Concordant = within 0.10 mL)
Q4. Design a complete back-titration procedure to determine the % purity of a CaCO₃ (limestone) sample. You have: limestone, 1.000 mol L⁻¹ HCl (excess), 0.1000 mol L⁻¹ NaOH (standard), phenolphthalein, 250 mL volumetric flask. Reaction: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Include all steps + calculation method. (Ca = 40.078, C = 12.011, O = 15.999)
📖 Comprehensive answers (click to reveal)
Multiple choice
1. B — Residual water in the burette dilutes the titrant, reducing its concentration → more titrant needed → unknown appears more concentrated than it is.
2. C — n(NaOH) = 0.200 × 0.02500 = 5.00 × 10⁻³ mol. 1:1 → n(HCl) = 5.00 × 10⁻³. c(HCl) = 5.00 × 10⁻³ ÷ 0.02000 = 0.250 mol L⁻¹.
3. A — Equivalence point is stoichiometric; end point is experimental (indicator). A good indicator makes them nearly coincide.
4. D — Rough (24.50) excluded. 23.82 − 23.75 = 0.07 ✓. Average = (23.80 + 23.75 + 23.82) ÷ 3 = 23.79 mL.
5. B — Ratio 1:2 (H₂SO₄:KOH) → 0.0100 × 2 = 0.0200 mol KOH.
Short answer model answers
Q1 (5 marks): (a) Concordant: 25.30 and 25.25 mL (diff = 0.05 ✓). 25.80 excluded. Average = 25.28 mL = 0.02528 L [1]. (b) n(HCl) = 0.1000 × 0.02528 = 2.528 × 10⁻³ mol [1]. n(Na₂CO₃) = ÷2 = 1.264 × 10⁻³ mol [1]. c(Na₂CO₃) = 1.264 × 10⁻³ ÷ 0.02500 = 0.0506 mol L⁻¹ [2].
Q2 (6 marks): (a) n(NaOH) = 0.1000 × 0.01860 = 1.860 × 10⁻³ mol [1]. n(H₂C₂O₄) in aliquot = ÷2 = 9.300 × 10⁻⁴ mol [1]. (b) n(total) = 9.300 × 10⁻⁴ × (250/25) = 9.300 × 10⁻³ mol [1]. m = 9.300 × 10⁻³ × 90.03 = 0.8373 g [1]. (c) % purity = (0.8373 ÷ 0.400) × 100 = 209% [1]. Note: purity > 100% is physically impossible → indicates an error in given data [1].
Q3 (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 → c(CH₃COOH) = 2.209 × 10⁻³ ÷ 0.02500 = 0.0884 mol L⁻¹ [1]. Student's 0.0882 is slightly off — they likely used a different average or rounded differently. Classmate is wrong — all three accurate titres are concordant [1].
Q4 (7 marks): Step 1: Weigh ~1 g of limestone accurately; record m₁ [1]. Step 2: Add exactly 50.00 mL of 1.000 mol L⁻¹ HCl (n initial = 0.05000 mol). React until fizzing stops [1]. Step 3: Transfer to 250 mL volumetric flask; make up to mark [1]. Step 4: Pipette 25.00 mL aliquot to conical flask. Add phenolphthalein. Titrate with 0.1000 mol L⁻¹ NaOH; record titre [1]. Calculation: n(HCl excess in aliquot) = 0.1000 × V(NaOH). Scale × (250/25) → total n(HCl) excess [1]. n(HCl reacted with CaCO₃) = 0.05000 − n(excess). n(CaCO₃) = ÷2 (from 1:2 ratio) [1]. m(CaCO₃) = n × 100.09. % purity = m(CaCO₃) ÷ m₁ × 100 [1].
Five timed questions on titration. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
⚔ Enter the arenaClimb platforms, hit checkpoints, and answer titration questions. Quick recall from lessons 1–10.
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