Titration Calculations

Volumetric analysis — better known to students as titration — is the workhorse quantitative technique of A-Level wet chemistry. A solution of unknown concentration (the analyte) is reacted with a solution of known concentration (the standard, or titrant) until the stoichiometric end point is reached, signalled by a colour change in an indicator. From the recorded titre, the moles of titrant are calculated, the balanced equation supplies the mole ratio, and the unknown concentration follows from n = cV. This lesson covers the apparatus and procedure, indicator choice, simple monoprotic and diprotic acid-base titrations, finding the molar mass of an unknown acid, back titration (used when the analyte is insoluble or reacts slowly), multi-step problems combining empirical-formula and titration data, primary-standard standardisation, and propagation of equipment uncertainty. It anchors AQA Required Practical 1 (preparation of a standard solution and a back-titration), so the procedural detail and the apparatus tolerances are examinable in their own right.

Spec mapping (AQA 7405): This lesson maps to §3.1.2.4 (use of n = cV for solutions and the preparation of a standard solution from a primary standard) and §3.1.2.5 (use of balanced equations with reacting masses, volumes of gases, and volumes and concentrations of solutions, including back titrations and percentage purity). The procedural component anchors Required Practical 1 (preparation of a standard solution and its use in a simple acid-base or back titration), with associated CPAC competencies in apparatus selection, accurate measurement, and uncertainty assessment. The pH-curve theory that justifies indicator choice belongs to §3.1.12 (acids and bases, A2); redox titrations using KMnO₄ or Na₂S₂O₃ belong to §3.1.7 (oxidation, reduction and redox equations) and §3.1.11 (electrode potentials and electrochemical cells). Solutions and concentrations (the prerequisite material) are covered in lesson 5 of this course, and the underlying mole concept (n = m/M and the Avogadro constant) in lesson 4. Refer to the official AQA specification document for the exact wording of each section.

Assessment objectives: Description of titration apparatus, indicator behaviour, the distinction between end point and equivalence point, and the criterion for concordant titres are AO1 recall items. Application of n = cV combined with the balanced-equation mole ratio to find an unknown concentration, mass, or percentage purity is AO2 and features on every Paper 1 and Paper 2 in some form. AO3 reasoning is tested through multi-step problems that combine empirical-formula data with titration data, candidate-identification from molar-mass calculations, propagation of equipment uncertainty across multiple apparatus items to give a combined percentage uncertainty, and reasoned indicator choice based on the pH at the equivalence point.

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Apparatus

A standard titration uses four pieces of volumetric glassware plus an analytical balance. The manufacturer-stated tolerances (Class B borosilicate, the school standard) are the values examiners expect you to use in uncertainty calculations.

Apparatus	Typical value	Manufacturer tolerance
Burette (50 cm³, Class B)	each reading	± 0.05 cm³
Burette — total titre (two readings)	titre	± 0.10 cm³
Pipette (25 cm³, Class B)	25.00 cm³	± 0.06 cm³
Volumetric flask (250 cm³, Class B)	250.0 cm³	± 0.30 cm³
Conical flask (250 cm³)	rough vessel	no tolerance — only the apparatus measuring volumes contributes
Balance (2 d.p.)	mass to 0.01 g	± 0.005 g (last-digit rule)
Balance (4 d.p., analytical)	mass to 0.0001 g	± 0.0005 g

Weighing by difference is the recommended technique for primary standards. Weigh the weighing-bottle plus solid, tip the solid into the volumetric flask through a funnel, then re-weigh the empty bottle. The mass transferred is the difference. This is more accurate than weighing-by-addition because any solid retained in the weighing bottle is accounted for, and it eliminates the largest systematic error in school-level standard-solution preparation.

Why the conical flask has no tolerance. Only apparatus used to measure a volume contributes to the uncertainty. The conical flask is a reaction vessel — its graduations are nominal — so its tolerance is not entered into uncertainty propagation. The same logic applies to beakers used for transferring rinsings.

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Indicators

An indicator is a weak organic acid or base whose conjugate forms have different colours. The colour change occurs over a narrow pH range (typically about 2 pH units centred on the indicator's pKa). For a titration to give a sharp end point, the vertical region of the pH curve at the equivalence point must overlap the indicator's pH-range.

Indicator	pH range (colour change)	Colour in acid	Colour in alkali	Best suited to
Methyl orange	3.1 – 4.4	red	yellow	strong acid + weak base (e.g. HCl titrated against NH₃, Na₂CO₃)
Bromothymol blue	6.0 – 7.6	yellow	blue	strong acid + strong base (alternative to phenolphthalein or methyl orange when a sharp endpoint near pH 7 is wanted)
Phenolphthalein	8.2 – 10.0	colourless	pink	weak acid + strong base (e.g. ethanoic acid titrated against NaOH)

Either methyl orange or phenolphthalein is acceptable for a strong-acid/strong-base titration because the pH change at equivalence is large (typically pH 3 to pH 11 over a single drop of titrant near the end point), so both indicators change colour within that vertical region.

Neither methyl orange nor phenolphthalein is suitable for a weak acid + weak base titration: the pH curve has no near-vertical region at the equivalence point and no indicator gives a sharp end point. Such titrations are performed by pH meter, conductometric monitoring, or thermometric monitoring instead. The theoretical justification — that the vertical-region pH range is set by the pKa of the weak species and the salt-hydrolysis equilibria — belongs to §3.1.12 (pH curves of strong and weak acids and bases) and is taught in the A2 acids-and-bases course.

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End Point versus Equivalence Point

These two terms are often used interchangeably in casual lab speech but are conceptually distinct, and AQA examiners do test the distinction.

• Equivalence point: the volume of titrant at which the moles of titrant added exactly equal the stoichiometric amount required by the balanced equation. This is a theoretical point defined by the reaction stoichiometry — for HCl + NaOH (1:1) it is the volume at which n(HCl)_added = n(NaOH)_in_flask.
• End point: the volume of titrant at which the indicator changes colour. This is an empirical, observed point.

A well-chosen indicator has its colour change centred close to the equivalence pH, so the end point is a very close approximation to the equivalence point and the systematic error introduced by reading the end point is negligible. A badly-chosen indicator (e.g. phenolphthalein for a strong-acid + weak-base titration, which equilibrates around pH 5) gives an end-point volume that differs systematically from the equivalence-point volume — the titre is wrong by a fixed amount irrespective of how carefully the burette is read. This is the largest avoidable systematic error in school titration.

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The Titration Procedure (Required Practical 1 anchor)

The following procedure is the AQA-aligned protocol for a standard acid-base titration. It applies whether you are determining the concentration of an unknown alkali (direct), the purity of a solid base such as Na₂CO₃ (direct, using a standardised acid), or the percentage of an insoluble base in a tablet (back titration).

1. Rinse the burette with distilled water, then with a small portion of the titrant. Discard each rinse. This prevents the titrant being diluted by residual water and prevents contamination by previous titrants.
2. Fill the burette above the zero mark, then run titrant out through the tap to fill the tip below the tap (no air bubble) and to bring the level just below zero. Read the initial volume to the nearest 0.05 cm³ from the bottom of the meniscus at eye level.
3. Rinse the pipette with distilled water, then with a small portion of the analyte. Discard each rinse. Pipette 25.0 cm³ of analyte into a clean conical flask. (The flask does not need rinsing with analyte — water in the flask only dilutes the analyte, which does not change the moles present.)
4. Add a few drops (typically 3–4) of indicator to the conical flask. More indicator is not better — the indicator itself consumes a small amount of titrant.
5. Run a rough titration quickly. Note the approximate titre to the nearest 0.5 cm³. Discard this result; it serves only to bracket the end point for the accurate runs.
6. Run accurate titrations, adding titrant drop-wise within ~1 cm³ of the expected end point. Swirl continuously. Read the final volume from the bottom of the meniscus at eye level. The titre is final – initial.
7. Repeat until you obtain at least two concordant titres (within ±0.10 cm³ of each other).
8. Calculate the mean titre from the concordant runs only. Discard the rough titration and any anomalous accurate run.

Sample results table

Run	Initial / cm³	Final / cm³	Titre / cm³	Use in mean?
Rough	0.00	24.10	24.10	No (rough)
1	0.00	23.55	23.55	No (anomalous, > 0.10 cm³ from runs 2/3)
2	0.00	23.35	23.35	Yes
3	0.00	23.40	23.40	Yes
4	0.00	23.30	23.30	Yes (within 0.10 cm³ of runs 2 and 3)

Mean titre = (23.35 + 23.40 + 23.30) / 3 = 23.35 cm³

Key Point: "Concordant" means within 0.10 cm³ of each other — this matches the per-titre burette uncertainty (two readings × ±0.05 cm³). Runs that lie outside this tolerance are outside the equipment's resolution; mixing them into the mean inflates uncertainty without improving accuracy. Always state explicitly which runs you used and which you discarded — this is an examinable practical-skills point.

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Standard Titration Calculations

General method

1. Write the balanced equation for the reaction.
2. Calculate the moles of the substance whose concentration and volume are both known (usually one of the two solutions) using n = cV (V in dm³).
3. Use the mole ratio from the equation to find moles of the unknown.
4. Calculate the concentration, mass, or percentage purity of the unknown.

Worked Example 1 — simple acid-base titration

25.0 cm³ of sodium hydroxide solution was titrated against 0.100 mol dm⁻³ hydrochloric acid. The mean titre was 22.5 cm³. Calculate the concentration of the sodium hydroxide.

NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

Step 1 (moles of HCl, known c and V): n(HCl) = c × V = 0.100 × (22.5 / 1000) = 2.25 × 10⁻³ mol.

Step 2 (mole ratio NaOH : HCl = 1 : 1): n(NaOH) = 2.25 × 10⁻³ mol.

Step 3 (concentration of NaOH): c(NaOH) = n / V = 2.25 × 10⁻³ / 0.0250 = 0.0900 mol dm⁻³.

Worked Example 2 — diprotic carbonate

25.0 cm³ of 0.120 mol dm⁻³ sodium carbonate solution was titrated against hydrochloric acid. The mean titre was 30.0 cm³. Calculate the concentration of the HCl.

Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)

Step 1: n(Na₂CO₃) = 0.120 × 0.0250 = 3.00 × 10⁻³ mol.

Step 2 (mole ratio Na₂CO₃ : HCl = 1 : 2): n(HCl) = 2 × 3.00 × 10⁻³ = 6.00 × 10⁻³ mol.

Step 3: c(HCl) = 6.00 × 10⁻³ / 0.0300 = 0.200 mol dm⁻³.

Worked Example 3 — sulfuric acid against NaOH

25.0 cm³ of sulfuric acid was neutralised by 20.0 cm³ of 0.150 mol dm⁻³ NaOH. Calculate the concentration of the sulfuric acid.

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

Step 1: n(NaOH) = 0.150 × 0.0200 = 3.00 × 10⁻³ mol.

Step 2 (mole ratio H₂SO₄ : NaOH = 1 : 2): n(H₂SO₄) = 3.00 × 10⁻³ / 2 = 1.50 × 10⁻³ mol.

Step 3: c(H₂SO₄) = 1.50 × 10⁻³ / 0.0250 = 0.0600 mol dm⁻³.

Exam Tip: Mark schemes routinely award independent marks for (i) the correct mole ratio and (ii) the correct conversion of cm³ to dm³. Write each step out — partial credit accumulates even if the final number is wrong. Skipping working to a one-line calculation is the easiest way to lose 2 of 4 marks.

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Finding Mᵣ from Titration Data

Worked Example 4 — molar mass of a monobasic acid

1.325 g of a monobasic acid HA was dissolved in water to make 250 cm³ of solution. 25.0 cm³ of this solution required 18.5 cm³ of 0.108 mol dm⁻³ NaOH for complete neutralisation. Calculate the molar mass of HA and suggest its identity.

HA(aq) + NaOH(aq) → NaA(aq) + H₂O(l)

Step 1: n(NaOH) = 0.108 × (18.5 / 1000) = 2.00 × 10⁻³ mol.

Step 2 (mole ratio HA : NaOH = 1 : 1): n(HA) in 25.0 cm³ aliquot = 2.00 × 10⁻³ mol.

Step 3 (scale up to volumetric flask): n(HA) in 250 cm³ = 2.00 × 10⁻³ × (250 / 25.0) = 0.0200 mol.

Step 4: M(HA) = m / n = 1.325 / 0.0200 = 66.3 g mol⁻¹.

Candidate identification: 66.3 g mol⁻¹ for a monobasic acid. Methanoic acid HCOOH has M = 46.0 (too low); ethanoic acid CH₃COOH has M = 60.0 (close but ~10% low); propanoic acid C₂H₅COOH has M = 74.0 (~12% high). Within ±3% experimental tolerance the closest credible monobasic candidate is a substituted ethanoic acid such as chloroethanoic acid (ClCH₂COOH, M = 94.5 — too high) or, more plausibly, a small unsaturated acid such as 2-propenoic acid (acrylic acid, CH₂=CHCOOH, M = 72.1). Without an independent piece of data (combustion analysis, IR carbonyl band, ¹H NMR), the calculation alone cannot uniquely identify the acid — this is the typical AO3 limitation that mark schemes call out.

Worked Example 4b — diprotic acid (second Mᵣ example)

A 0.250 g sample of a diprotic acid H₂X was dissolved in water and titrated against 0.100 mol dm⁻³ NaOH; the mean titre was 25.0 cm³. Calculate Mᵣ and identify the acid.

H₂X(aq) + 2NaOH(aq) → Na₂X(aq) + 2H₂O(l)

Step 1: n(NaOH) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol.

Step 2 (mole ratio H₂X : NaOH = 1 : 2): n(H₂X) = 2.50 × 10⁻³ / 2 = 1.25 × 10⁻³ mol.

Step 3: M(H₂X) = 0.250 / 1.25 × 10⁻³ = 200.0 g mol⁻¹.

Candidate identification: A diprotic acid with M = 200 corresponds most closely to dihydrated oxalic acid (COOH)₂·2H₂O with M = 126.1 (too low) — so the candidate is more plausibly a higher dicarboxylic acid such as sebacic acid (HOOC(CH₂)₈COOH, M = 202.3) within ~1% tolerance. Note the importance of stating the acid as a hydrate or anhydrate — primary standards such as oxalic acid are almost always supplied as the dihydrate and must be entered into the calculation with its full hydrated formula mass.

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Back Titration

A back titration is used when one of the following applies:

• The analyte is insoluble in water (e.g. metal carbonates such as CaCO₃ in limestone; insoluble metal oxides; MgO in milk-of-magnesia tablets).
• The analyte reacts slowly with the titrant (kinetic control prevents a sharp end point being detected by indicator colour change).
• The end point of the direct titration is unclear (mineral suspension obscures the indicator colour change).

Procedure

1. Add a known excess of standard reagent A (typically a strong acid such as HCl) to the analyte. Let the reaction go to completion.
2. Titrate the unreacted excess of reagent A against a standard solution of reagent B (typically a strong base such as NaOH).
3. n(A reacted with analyte) = n(A originally added) − n(A unreacted).
4. Convert to moles, then mass, of analyte using the analyte-A balanced equation.

Worked Example 5 — CaCO₃ back titration (corrected)

0.750 g of impure calcium carbonate was added to 50.0 cm³ of 0.300 mol dm⁻³ HCl (excess). The excess acid was titrated with 0.200 mol dm⁻³ NaOH, requiring 25.0 cm³ for neutralisation. Calculate the percentage purity of the calcium carbonate sample.

Step 1 (total moles of HCl added): n(HCl)_total = 0.300 × 0.0500 = 0.0150 mol.

Step 2 (moles of HCl unreacted — back titrate with NaOH): NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l) n(NaOH) = 0.200 × 0.0250 = 5.00 × 10⁻³ mol. 1:1 ratio ⇒ n(HCl)_excess = 5.00 × 10⁻³ mol.

Step 3 (moles of HCl that reacted with CaCO₃): n(HCl)_reacted = 0.0150 − 5.00 × 10⁻³ = 0.0100 mol.

Step 4 (moles of CaCO₃): CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g) 1:2 ratio ⇒ n(CaCO₃) = 0.0100 / 2 = 5.00 × 10⁻³ mol.

Step 5 (mass and percentage purity): m(CaCO₃) = 5.00 × 10⁻³ × 100.1 = 0.5005 g. Percentage purity = (0.5005 / 0.750) × 100 = 66.8%.

Exam Tip: Label moles explicitly — n(HCl)_total, n(HCl)_excess, n(HCl)_reacted. The single commonest examiner-noted error in back titration is conflating "moles of HCl reacted with NaOH" with "moles of HCl reacted with the analyte" — they are complementary, not the same.

Worked Example 5b — NH₃ in a fertiliser (second back titration)

A 1.50 g sample of fertiliser containing ammonium sulfate (NH₄)₂SO₄ was warmed with 50.0 cm³ of 0.500 mol dm⁻³ NaOH (excess). All the ammonia was driven off and the residual NaOH was back-titrated with 0.250 mol dm⁻³ HCl, requiring 22.0 cm³. Calculate the percentage by mass of (NH₄)₂SO₄ in the fertiliser.

Step 1: n(NaOH)_total = 0.500 × 0.0500 = 0.0250 mol.

Step 2 (back-titrate residual NaOH): n(HCl) = 0.250 × 0.0220 = 5.50 × 10⁻³ mol; 1:1 ratio ⇒ n(NaOH)_excess = 5.50 × 10⁻³ mol.

Step 3: n(NaOH)_reacted = 0.0250 − 5.50 × 10⁻³ = 0.0195 mol.

Step 4: (NH₄)₂SO₄ + 2NaOH → Na₂SO₄ + 2NH₃ + 2H₂O. Mole ratio NaOH : (NH₄)₂SO₄ = 2 : 1, so n((NH₄)₂SO₄) = 0.0195 / 2 = 9.75 × 10⁻³ mol.

Step 5: M((NH₄)₂SO₄) = 132.1 g mol⁻¹; m = 9.75 × 10⁻³ × 132.1 = 1.288 g. Percentage by mass = (1.288 / 1.50) × 100 = 85.8%.

The 14% non-(NH₄)₂SO₄ residue would typically be inert filler (sand, ground limestone) — this is realistic for agricultural-grade fertiliser.

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Multi-Step Problems

Worked Example 6 — identifying a Group 2 metal from its hydroxide (corrected)

A metal hydroxide M(OH)₂ reacts with HCl. 0.370 g of M(OH)₂ required 20.0 cm³ of 0.500 mol dm⁻³ HCl for complete neutralisation. Find the molar mass of M(OH)₂, the Aᵣ of M, and identify M.

M(OH)₂(aq) + 2HCl(aq) → MCl₂(aq) + 2H₂O(l)

Step 1: n(HCl) = 0.500 × 0.0200 = 0.0100 mol.

Step 2 (mole ratio M(OH)₂ : HCl = 1 : 2): n(M(OH)₂) = 0.0100 / 2 = 5.00 × 10⁻³ mol.

Step 3: M(M(OH)₂) = 0.370 / 5.00 × 10⁻³ = 74.0 g mol⁻¹.

Step 4 (extract Aᵣ of M): M(M(OH)₂) = Aᵣ(M) + 2 × (16.0 + 1.0) = Aᵣ(M) + 34.0. Aᵣ(M) = 74.0 − 34.0 = 40.0.

Step 5 (identify): Aᵣ = 40.0 corresponds to calcium. The hydroxide is Ca(OH)₂ — consistent with a Group 2 dihydroxide.

Worked Example 7 — volumetric flask aliquot scaling

A student dissolved 2.65 g of anhydrous sodium carbonate in water and made the solution up to 250 cm³ in a volumetric flask. 25.0 cm³ portions were titrated against hydrochloric acid. The mean titre was 24.8 cm³. Calculate the concentration of the HCl.

Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)

Step 1: n(Na₂CO₃) in 250 cm³ flask = 2.65 / 106.0 = 0.0250 mol.

Step 2 (scale down to 25 cm³ aliquot): n(Na₂CO₃) in 25.0 cm³ = 0.0250 × (25.0 / 250) = 2.50 × 10⁻³ mol.

Step 3 (mole ratio Na₂CO₃ : HCl = 1 : 2): n(HCl) = 2 × 2.50 × 10⁻³ = 5.00 × 10⁻³ mol.

Step 4: c(HCl) = 5.00 × 10⁻³ / 0.0248 = 0.202 mol dm⁻³.

Worked Example 7b — primary-standard standardisation of NaOH using oxalic acid dihydrate

NaOH cannot be used as a primary standard because solid NaOH absorbs water from the atmosphere (deliquescence) and reacts with atmospheric CO₂ to form Na₂CO₃; its concentration must therefore be determined by titration against a true primary standard. Oxalic acid dihydrate (COOH)₂·2H₂O (M = 126.07 g mol⁻¹) is the standard choice — it is non-deliquescent, stable in air, available at high purity, and dissolves cleanly in water.

A student weighed 1.575 g of (COOH)₂·2H₂O on a 4-d.p. balance and dissolved it in distilled water in a 250 cm³ volumetric flask, making up to the mark. 25.0 cm³ aliquots were titrated against NaOH solution of unknown concentration; the mean titre was 24.95 cm³. Calculate the concentration of the NaOH.

(COOH)₂(aq) + 2NaOH(aq) → (COONa)₂(aq) + 2H₂O(l)

Step 1: n((COOH)₂·2H₂O) in 250 cm³ flask = 1.575 / 126.07 = 0.01250 mol. (Note: the dihydrate formula mass is used because that is the form that was weighed.)

Step 2: n((COOH)₂) in 25.0 cm³ aliquot = 0.01250 × (25.0 / 250) = 1.250 × 10⁻³ mol.

Step 3 (mole ratio (COOH)₂ : NaOH = 1 : 2): n(NaOH) = 2 × 1.250 × 10⁻³ = 2.500 × 10⁻³ mol.

Step 4: c(NaOH) = 2.500 × 10⁻³ / 0.02495 = 0.1002 mol dm⁻³.

The standardised NaOH can now be used for subsequent titrations to determine the concentrations of unknown weak acids (Example 4-type problems).

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Uncertainty

Typical equipment uncertainties (re-stated)

Apparatus	Reading uncertainty
25.0 cm³ pipette	±0.06 cm³
50.0 cm³ burette	±0.05 cm³ per reading; ±0.10 cm³ per titre
250 cm³ volumetric flask	±0.30 cm³
Balance, 2 d.p.	±0.005 g
Balance, 4 d.p.	±0.0005 g

Percentage uncertainty