Module 2 · L10 of 20~40 min⚡ +50 XP in Learn · +25 to complete
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.
Today's hook, You've got HCl of unknown concentration and a perfect NaOH standard. How do you turn that pair into a number?
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You're here
Warm up and think first
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Worksheets
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?
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The titration pathway, and the words for it
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The titration calculation pathway
core method
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What you'll master
Know
Key facts
Definitions: analyte, titrant, equivalence point, end point
Difference between equivalence point and end point
Why concordant titres are averaged
Can do
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
Beyond the syllabus. Module 2 requires you to determine concentration by practical investigation, to use c = n/V, and to prepare standard solutions and dilutions. All of that is Core, all of it is examinable, and a titration is one legitimate way to do the investigation. What is not separately prescribed is formal titration mastery: the apparatus vocabulary, indicator choice, concordant titres and the full four-step titre calculation go past what Module 2 names, and none of it is a mastery gate here. Do the lesson, it is the best Year 12 preparation in this module, but if you are short of time then Lessons 6 and 7, Concentration and Standard Solutions and Dilutions, are the ones carrying the marks.
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Key terms
Titration
A volumetric technique to determine the concentration of a solution by reacting it with a standard solution of known concentration.
Standard solution
A solution of accurately known concentration prepared from a primary standard or calibrated by titration.
Equivalence point
The point where stoichiometrically equivalent amounts of acid and base have reacted; pH changes sharply.
Endpoint
The point where the indicator changes colour; should coincide with the equivalence point for accurate results.
Titre
The volume of standard solution (from burette) required to reach the endpoint in a titration.
Concordant titres
Titres that agree within ±0.10 mL; the mean of at least two concordant titres is used for calculation.
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Apparatus and procedure
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Apparatus & procedure
core concept
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.
Concordant titres:
Two or more titres are concordant if they agree within 0.10 mL. Only concordant titres are averaged. If your first two accurate titres are 24.35 and 24.28 mL, they're concordant (diff = 0.07 mL) → average = 24.32 mL.
Safety boundary for titrations. Use only dilute, school-approved acids and bases (typically ≤1 mol L⁻¹) prepared by the technician. Wear safety glasses; wipe up spills and rinse glassware; neutralise and dilute waste before disposal as directed. Filling the burette is done at bench height using a funnel, never above eye level. Back titrations (adding a measured excess of one reagent, then titrating what remains) are an extension method, carried out only with teacher supervision and approved concentrations.
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.
A correct titration setup links careful volume reading with dropwise endpoint technique.
Did you get this? True or false: the rough titre is always included in the average.
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The 4-step calculation method
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The 4-step calculation method
core concept
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 1: Calculate moles of the known/standard solution: n = c × V (V in litres) 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
The mole ratio is not always 1:1.
For example, H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O gives a 1:2 ratio. If you assume 1:1 when it's 1:2, your answer will be exactly double the correct value. Always write the balanced equation first.
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)?
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Three worked examples
Worked examples · reveal as you go
Worked example 1 · finding c(NaOH) · 1:1 ratio+5 XP on full reveal
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.
Worked example 3 · antacid purity (multi-step)+5 XP on full reveal
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)
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$\mathrm{Mg(OH)_2} + 2\mathrm{HCl} \rightarrow \mathrm{MgCl_2} + 2\mathrm{H_2O}$ | ratio 1:2
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.
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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.
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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.
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Quick-fire drill, then revisit your thinking
Quick-fire practice · 5 reps +2 XP per reveal
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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)
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)
n(NaOH) = 0.1000 × 0.01680 = 1.680 × 10⁻³ mol = n(CH₃COOH) in aliquot (1:1). n(total) = 1.680 × 10⁻³ × (100/20) = 8.400 × 10⁻³ mol. m = 8.400 × 10⁻³ × 60.05 = 0.5044 g
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Four titres recorded: 24.50 (rough), 23.80, 23.75, 23.82 mL. Identify concordant titres and calculate the average.
Rough (24.50) excluded. Check: 23.82 − 23.75 = 0.07 mL ✓ all within 0.10 mL → concordant. Average = (23.80 + 23.75 + 23.82) ÷ 3 = 23.79 mL
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Revisit your thinking
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.
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Practice questions
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Multiple choice
+2 XP per correct · +5 bonus if perfect
Pick your answer, then rate your confidence. That tells the system what to drill next.
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Short answer
ApplyMid-order5 marks
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.
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AnalyseMid-order6 marks
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 = 1.000 g, calculate % purity. (MM H₂C₂O₄ = 90.03)
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EvaluateHigher-order5 marks
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)
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CreateHigher-order7 marks
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)
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📖 Comprehensive answers (click to reveal)
Multiple choice
Wet burette. Residual water in the burette dilutes the titrant, reducing its concentration → more titrant needed → unknown appears more concentrated than it is.
Equivalence point vs end point. Equivalence point is stoichiometric; end point is experimental (indicator). A good indicator makes them nearly coincide.
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 ÷ 1.000) × 100 = 83.7% [1]. The remaining ~16% is inert impurity that did not react with the NaOH [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].
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Retrieve, then finish
Check what actually stuck
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Take the full module quiz
quiz
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
