Titrations & practical technique

11 practice questions

Carry out an acid–base titration

Learning goalExplain the apparatus, rinsing and endpoint technique used in a titration.

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A titration finds the volume of one solution needed to react with a measured portion of another. The burette delivers the titrant; the conical flask holds the solution being analysed.

Rinse the burette with deionised water, then its own solution. Fill it and run solution through the tip to remove air bubbles. Remove the filling funnel.
Rinse the pipette with deionised water, then the solution it will measure. Use a pipette filler to transfer a fixed aliquot (a measured portion of the solution) to the conical flask.
Let a standard transfer pipette drain. Do not blow out the drop designed to remain in it.
Add a few drops of a suitable indicator and place the flask on a white tile.
Read the burette meniscus at eye level and record the initial reading.
Do a rough titration to locate the endpoint. Repeat with fresh aliquots (measured portions), swirling and adding dropwise near the endpoint.
Wash splashes down the flask walls with deionised water. This changes volume but not the amount of solute being titrated.
Record the final reading and calculate final minus initial. Repeat until you obtain concordant titres (within 0.1 cm³ of each other).

Record burette readings to two decimal places, normally to the nearest 0.05 cm³ on a scale graduated every 0.10 cm³. Use eye protection, a pipette filler and the laboratory risk assessment.

End point: The observed indicator colour change.
Equivalence point: The point at which the reactants have been mixed in their exact stoichiometric ratio.

Worked example

Choose an indicator

Choose a suitable indicator for (a) HCl added to NaOH and (b) complete neutralisation of Na₂CO₃ by HCl. State the colour change at the endpoint for each, starting with the alkali in the conical flask.

Worked answer

HCl added to NaOH: phenolphthalein changes from pink to just colourless at the endpoint.
Complete neutralisation of Na₂CO₃ by HCl: methyl orange changes from yellow towards the first persistent orange endpoint.

Choose an indicator whose colour change is close to equivalence.

EXAM TRAPWater left in the pipette or burette dilutes the solution being measured. Water in the conical flask changes total volume but not the moles already delivered. A transfer pipette is calibrated to leave its final drop behind.

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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 35

Explain why a burette is rinsed with its titrant before filling. (1 mark)
Explain why a pipette is rinsed with its solution, but the conical flask may be rinsed or its walls washed with deionised water. (2 marks)
Explain why the filling funnel is removed before taking the initial burette reading. (1 mark)
State the purpose of the rough titration and describe how titrant should be added near the end point in accurate repeats. (2 marks)
Distinguish end point from equivalence point. (2 marks)
Explain why the final drop in a standard transfer pipette is not blown out. (1 mark)
Name a suitable indicator for HCl added to NaOH and state its end-point colour change. (2 marks)

Total: 11 marks

Show answer & marking points

(a) To avoid residual water diluting the titrant. [1]

(b) The pipette rinse prevents dilution of the aliquot. [1] Deionised water in the flask changes volume, but not the moles already delivered; washing the walls returns reacting droplets to the mixture. [1]

(d) The rough trial locates the approximate end point. [1] Add dropwise while swirling near that volume. [1]

(e) End point: observed indicator colour change. [1] Equivalence point: reactants mixed in their stoichiometric ratio. [1]

(f) The pipette is calibrated to deliver the stated volume while leaving that drop behind. [1]

(g) Phenolphthalein [1], pink to just colourless. [1] Accept methyl orange with yellow to the first persistent orange end point; other suitable indicators with correct observations are acceptable.

✓ DONE WHENExplain the apparatus, rinsing and endpoint technique used in a titration.

Read a burette and choose concordant titres

Learning goalCalculate titres from two readings and justify the mean used.

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A burette scale increases downwards. The titre is the volume delivered, so subtract the initial reading from the final reading.

Read the bottom of a clear solution's meniscus at eye level. For a burette graduated every 0.10 cm³, estimate to the nearest 0.05 cm³ and record two decimal places.

Concordant titres (within 0.1 cm³ of each other) agree closely enough to use in the mean. Use a different tolerance if the question supplies one. Compare the titres themselves, not just the final readings.

Leave out the rough titre and justified outliers when finding the mean. If no set is concordant, more titrations are needed; do not choose an unsupported average.

Titre = final burette reading − initial burette reading

Worked example

Select the reliable pair

The following burette readings were recorded during a titration. All readings are in cm³.

Reading / cm³	Rough	1	2	3
Initial	0.20	0.30	0.20	0.45
Final	25.00	24.45	24.70	24.65

Calculate all four titres. Select the concordant accurate titres and calculate their mean, reporting it to two decimal places. Treat accurate titres that differ by no more than 0.10 cm³ as concordant.

Worked answer

Reading / cm³	Rough	1	2	3
Final	25.00	24.45	24.70	24.65
Initial	0.20	0.30	0.20	0.45
Titre	24.80	24.15	24.50	24.20

Titres 1 and 3 differ by 0.05 cm³. The rough titre and run 2 are excluded.

Mean = (24.15 + 24.20)/2 = 24.175 cm³

Report 24.18 cm³. Retain the unrounded mean for later calculations.

EXAM TRAPA trailing zero in 24.20 records the measurement precision. Do not average the rough result simply because it looks close.

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Quick check 36

Record the initial and final readings. (2 marks)
Calculate the titre. (1 mark)
Explain why the meniscus must be read at eye level. (1 mark)

Total: 4 marks

Show answer & marking points

(a) Initial 18.15 cm³ [1]; final 25.45 cm³. [1]

(b) Titre = 25.45 − 18.15 = 7.30 cm³. [1]

Exam-style question 37

The table shows burette readings from one rough titration and four accurate titrations.

Reading / cm³	Rough	1	2	3	4
Final	25.10	24.80	49.30	24.75	49.60
Initial	0.20	0.15	24.60	0.25	24.90

(a) Calculate the five titres. (2 marks)

(b) Use a concordance limit of 0.10 cm³. Select the concordant accurate titres and calculate their mean. (3 marks)

Total: 8 marks

Show answer & marking points

(a) Titres are rough 24.90; run 1 24.65; run 2 24.70; run 3 24.50; run 4 24.70 cm³. Award [1] for at least three correct and [1] for all five correct.

(b) Use runs 1, 2 and 4 [1]. Mean = (24.65 + 24.70 + 24.70)/3 = 24.683… cm³ [1] = 24.68 cm³ [1].

(c) Exclude the rough titre because it was used only to locate the end point [1]. Exclude run 3 because 24.50 cm³ is not within 0.10 cm³ of the selected concordant set [1]. Two decimal places retain the resolution of a burette read to the nearest 0.05 cm³, including a trailing zero where needed [1].

EXAM TRAPCompare calculated titres, not final burette readings.

✓ DONE WHENCalculate titres from two readings and justify the mean used.

Calculate a concentration from a titre

Learning goalUse the balanced-equation ratio between known and unknown solution amounts.

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Use a grid with the reacting substances across the top and c (concentration), V (volume) and n (amount in moles) down the side. Add the coefficients from the balanced equation below the mole row.

Write the balanced equation and fill in the known concentrations and volumes. Convert each volume to dm³ by dividing its cm³ value by 1000.
Start in the column with both c and V known. Use n = cV to fill its mole row.
Work across the mole row using the equation ratio: divide by the known substance's coefficient, then multiply by the other substance's coefficient.
Work up the unknown substance's column: use c = n/V, dividing by that substance's own solution volume in dm³.
Give the concentration unit and round only the final answer.

Keep substance names beside the numbers. The burette volume and pipette volume belong to different solutions and cannot be swapped.

Worked example

Sodium carbonate and hydrochloric acid

25.0 cm³ of Na₂CO₃ solution requires 20.0 cm³ of 0.100 mol dm⁻³ HCl solution for complete neutralisation. Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O.

Calculate the concentration of the sodium carbonate solution in mol dm⁻³.

Worked answer

Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O. Start with HCl, whose concentration and volume are known. The 1 : 2 ratio means the carbonate has half as many moles.

Quantity	Na₂CO₃	HCl
Concentration, c / mol dm⁻³	c = n/V 0.00100/0.0250 = 0.0400	0.100
Volume, V / dm³	25.0/1000 = 0.0250	20.0/1000 = 0.0200
Amount, n / mol	0.00200/2 = 0.00100	n = cV 0.100 × 0.0200 = 0.00200
Coefficient (mole ratio)	1	2

The Na₂CO₃ concentration is 0.0400 mol dm⁻³.

Worked example

A one-to-one neutralisation

25.0 cm³ of HNO₃ solution requires 22.4 cm³ of 0.125 mol dm⁻³ KOH solution for complete neutralisation. HNO₃ + KOH → KNO₃ + H₂O.

Calculate the concentration of the nitric acid solution in mol dm⁻³.

Worked answer

HNO₃ + KOH → KNO₃ + H₂O. The 1 : 1 ratio gives equal moles at the end point.

Quantity	HNO₃	KOH
Concentration, c / mol dm⁻³	c = n/V 0.00280/0.0250 = 0.112	0.125
Volume, V / dm³	25.0/1000 = 0.0250	22.4/1000 = 0.0224
Amount, n / mol	0.00280	n = cV 0.125 × 0.0224 = 0.00280
Coefficient (mole ratio)	1	1

The HNO₃ concentration is 0.112 mol dm⁻³.

EXAM TRAPDo not set acid and alkali moles equal unless the equation ratio is 1 : 1. Use the volume of the solution whose concentration you are finding in the final c = n/V step.

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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 38

25.00 cm³ sulfuric acid reacts exactly with 21.35 cm³ of 0.1200 mol dm⁻³ NaOH. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Use Mᵣ(H₂SO₄) = 98.1.

Calculate the sulfuric acid concentration in mol dm⁻³. (4 marks)
Calculate its mass concentration in g dm⁻³. (2 marks)

Total: 6 marks

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(a) H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O: acid : alkali = 1 : 2. [1]

Quantity	H₂SO₄	NaOH
Concentration, c / mol dm⁻³	c = n/V 0.001281/0.02500 = 0.05124 [1]	0.1200
Volume, V / dm³	25.00/1000 = 0.02500	21.35/1000 = 0.02135
Amount, n / mol	0.002562/2 = 0.001281 [1]	n = cV 0.1200 × 0.02135 = 0.002562 [1]
Coefficient (mole ratio)	1	2

(b) Mass concentration = c × Mᵣ. [1] 0.05124 × 98.1 = 5.03 g dm⁻³. [1]

✓ DONE WHENUse the balanced-equation ratio between known and unknown solution amounts.

Titrate a diluted sample

Learning goalDistinguish the original sample, final flask volume and titrated aliquot.

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A titration of a diluted sample first tells you about the diluted solution. You still need to work back to the original sample.

An aliquot is a measured portion taken from a well-mixed solution. To scale its mole amount to the full flask, multiply by whole-flask volume divided by aliquot volume.

For example, a 100.0 cm³ flask contains four 25.0 cm³ aliquots. If one contains 0.00200 mol acid, the whole flask contains 4 × 0.00200 = 0.00800 mol.

Those moles came from the original 0.0100 dm³ sample, so its concentration is 0.00800/0.0100 = 0.800 mol dm⁻³. Alternatively, multiply the diluted concentration by the dilution factor.

c₁V₁ = c₂V₂

V₁ volume taken from the original solution to make the dilution, in cm³ or dm³; V₂ final total volume of the diluted solution, in the same units as V₁; c₁ concentration of the original solution before dilution, in mol dm⁻³; c₂ concentration of the diluted solution, in mol dm⁻³.

1 means before dilution; 2 means after dilution. Adding water changes the concentration and volume, but not the number of moles of solute. Use the same units for both volumes: either both cm³ or both dm³. The final volume is the total volume of the diluted solution, not the volume of water added.

First use the titration and balanced-equation mole ratio to find the diluted concentration. Then use the dilution equation to work back to the original concentration. The volume of the aliquot titrated is not the final dilution volume.

Worked example

Vinegar after dilution

10.0 cm³ vinegar is diluted to 100.0 cm³. A 25.0 cm³ aliquot of the diluted vinegar requires 20.0 cm³ of 0.100 mol dm⁻³ NaOH solution for complete neutralisation. CH₃COOH + NaOH → CH₃COONa + H₂O. M_r(CH₃COOH) = 60.0.

Calculate the concentration of ethanoic acid in the original vinegar in mol dm⁻³ and g dm⁻³.

Worked answer

An aliquot is a measured portion: here it is 25.0 cm³ of the diluted vinegar. CH₃COOH + NaOH → CH₃COONa + H₂O.

Quantity	CH₃COOH (diluted)	NaOH
Concentration, c / mol dm⁻³	c = n/V 0.00200/0.0250 = 0.0800	0.100
Volume, V / dm³	25.0/1000 = 0.0250	20.0/1000 = 0.0200
Amount, n / mol	0.00200 (1 : 1 ratio)	n = cV 0.100 × 0.0200 = 0.00200
Coefficient (mole ratio)	1	1

Dilution factor = final dilution volume/original sample volume = 100.0/10.0 = 10.0.

Original concentration = diluted concentration × dilution factor = 0.0800 × 10.0 = 0.800 mol dm⁻³.

Mass concentration = c × Mᵣ = 0.800 × 60.0 = 48.0 g dm⁻³.

EXAM TRAPDo not apply both the dilution factor and the whole-flask factor to the same quantity without tracking what that quantity represents.

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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 39

10.00 cm³ of household ammonia solution is diluted to 250.0 cm³. A 25.00 cm³ aliquot of the diluted solution requires 18.40 cm³ of 0.1000 mol dm⁻³ HCl solution for complete neutralisation. NH₃ + HCl → NH₄Cl

(a) Calculate the concentration of NH₃ in the original household solution, in mol dm⁻³. (5 marks)

(b) Calculate the mass concentration of NH₃ in the original solution, in g dm⁻³. M_r(NH₃) = 17.0. (2 marks)

Total: 7 marks

Show answer & marking points

(a) NH₃ + HCl → NH₄Cl; use the 25.00 cm³ aliquot of diluted ammonia in the grid.

Quantity	NH₃ (diluted)	HCl
Concentration, c / mol dm⁻³	c = n/V 0.001840/0.02500 = 0.07360 [1]	0.1000
Volume, V / dm³	25.00/1000 = 0.02500	18.40/1000 = 0.01840
Amount, n / mol	0.001840 (1 : 1 ratio) [1]	n = cV 0.1000 × 0.01840 = 0.001840 [1]
Coefficient (mole ratio)	1	1

Dilution factor = final dilution volume/original sample volume = 250.0/10.00 = 25.0. [1] Original concentration = 0.07360 × 25.0 = 1.840 mol dm⁻³. [1]

(b) Mass concentration = c × Mᵣ. [1] 1.840 × 17.0 = 31.3 g dm⁻³. [1]

The 25.00 cm³ aliquot is a portion of the diluted solution; 10.00 cm³ is the original sample volume.

✓ DONE WHENDistinguish the original sample, final flask volume and titrated aliquot.

Titration uncertainty and error direction

Learning goalCalculate uncertainty and connect practical errors to the reported result.

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A titre uses two measurements: an initial and a final burette reading. In this treatment, add their absolute uncertainties to obtain the titre uncertainty.

To find percentage uncertainty, divide the absolute uncertainty by the measured titre and multiply by 100. With the same apparatus, a larger titre gives a smaller percentage uncertainty.

A directional error is a mistake that makes a result too high or too low. For example, consistently adding acid past the end point makes the recorded titre too large. If that titre is used to calculate an unknown alkali concentration, the concentration comes out too high.

Explain the chain: what went wrong → whether the titre is too large or too small → how that changes the calculated result. This is different from uncertainty, which describes the range of values consistent with a measurement; it does not by itself tell you which way the result is wrong.

Use the uncertainty stated in the question. Two reading uncertainties alone give ±0.10 cm³ when each is ±0.05 cm³. A total including a further ±0.05 cm³ end-point judgement is ±0.15 cm³. Do not add the end-point term twice.

Percentage uncertainty = absolute uncertainty / measured value × 100

Worked example

Two readings contribute

A titre is 20.00 cm³ and each reading has uncertainty ±0.05 cm³.

Calculate the absolute and percentage uncertainty in the titre from the two burette readings. Then calculate the percentage uncertainty when a further endpoint uncertainty of ±0.05 cm³ is included.

Worked answer

Absolute uncertainty = 0.05 + 0.05 = 0.10 cm³

Percentage uncertainty = (absolute uncertainty ÷ measured value) × 100
Percentage uncertainty = 0.10/20.00 × 100 = 0.50%

If endpoint judgement contributes a further ±0.05 cm³, use total ±0.15 cm³.

0.15/20.00 × 100 = 0.75%

0.50% for the readings alone; 0.75% including the stated endpoint uncertainty.

Precision: How closely repeated measurements agree.
Accuracy: How close a result is to the true or accepted value.
Random error: A varying effect that causes scatter.
Systematic error: A consistent effect that shifts results in one direction.

Known acid in burette; unknown alkali in flask	Effect	Calculated alkali concentration
Burette wet with water	Acid diluted; larger titre needed; stated acid concentration overestimates moles	Too high
Pipette wet with water	Aliquot diluted; fewer alkali moles; smaller titre	Too low
Conical flask wet with deionised water	Aliquot moles unchanged	Unchanged
Initial air bubble disappears from burette jet	Some measured volume fills jet rather than reaching flask	Too high
Endpoint consistently overshot	Recorded titre larger than needed	Too high

EXAM TRAPConcordant titres (within 0.1 cm³ of each other) show precision. They do not remove a systematic error or prove accuracy. Explain the chain: what changes → effect on titre → effect on calculated quantity.

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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 40

A mean titre is 23.40 cm³. Each burette reading has uncertainty ±0.05 cm³.

(a) Calculate the percentage uncertainty in the titre due to the two burette readings only. (2 marks)

(b) The end-point judgement adds a further ±0.05 cm³. Calculate the total percentage uncertainty when this is included. (2 marks)

(c) Known acid is in the burette and unknown alkali is in the conical flask. Explain the effect on the calculated alkali concentration if the burette is left wet with water before filling. (3 marks)

(d) Explain why concordant titres do not prove that the final concentration is accurate. (1 mark)

Total: 8 marks

Show answer & marking points

Percentage uncertainty = (absolute uncertainty ÷ measured value) × 100
(a) Absolute uncertainty = 0.05 + 0.05 = 0.10 cm³ [1]. Percentage uncertainty = 0.10/23.40 × 100 = 0.43% [1].

Percentage uncertainty = (absolute uncertainty ÷ measured value) × 100
(b) Total absolute uncertainty = 0.10 + 0.05 = 0.15 cm³ [1]. Percentage uncertainty = 0.15/23.40 × 100 = 0.64% [1].

(c) Residual water dilutes the known acid [1], so a larger titre is needed to neutralise the alkali [1]. Using the stated undiluted acid concentration then overestimates acid moles and therefore overestimates the calculated alkali concentration [1].

(d) Concordance indicates precision; the same systematic error could shift every titre in one direction [1].

Exam-style question 41

Known hydrochloric acid is delivered from a burette to an aliquot of unknown sodium hydroxide in a conical flask.

(a) The pipette is left wet with water before the sodium hydroxide aliquot is measured. Explain the effect on the titre and on the calculated sodium hydroxide concentration. (3 marks)

(b) The conical flask is left wet with deionised water before the aliquot is added. Explain why this does not change the calculated concentration. (2 marks)

(c) An air bubble in the burette jet is present at the initial reading and disappears during the titration. Explain the effect on the titre and on the calculated sodium hydroxide concentration. (3 marks)

Total: 8 marks

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(a) Water dilutes the NaOH aliquot, so it contains fewer moles of NaOH [1]. A smaller acid titre is required [1], giving a calculated NaOH concentration that is too low [1].

(b) The pipette has already fixed the moles of NaOH delivered [1]. Water in the flask changes concentration during the titration but not those moles or the equivalence-point amount of acid [1].

(c) Some of the measured volume fills the jet instead of reaching the flask, so the apparent titre is too large [1]. The calculation treats all that volume as reacted acid [1], so the calculated NaOH concentration is too high [1].

✓ DONE WHENCalculate uncertainty and connect practical errors to the reported result.

Use a titration to calculate percentage purity

Learning goalScale an aliquot result to the whole sample before calculating purity.

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A titration can find how much pure reacting substance was present in an impure sample. Keep track of whether each amount refers to an aliquot or the whole prepared solution.

Use the titre and known concentration to find moles of titrant.
Apply the equation ratio to find moles of the sample substance in the aliquot.
Scale up to the whole flask using the flask-volume/aliquot-volume ratio.
Convert to mass of pure substance, then divide by the original sample mass and multiply by 100.

This method assumes that the impurities do not react with the titrant or interfere with the endpoint.

Worked example

Potassium carbonate purity

2.00 g of an impure K₂CO₃ sample is made up to 250.0 cm³. A 25.0 cm³ aliquot of this solution requires 20.0 cm³ of 0.100 mol dm⁻³ HCl solution for complete neutralisation. K₂CO₃ + 2HCl → 2KCl + CO₂ + H₂O. M_r(K₂CO₃) = 138.2.

Calculate the percentage purity by mass of the potassium carbonate sample. Assume the impurities do not react with hydrochloric acid.

Worked answer

K₂CO₃ + 2HCl → 2KCl + CO₂ + H₂O. The grid describes the titrated portion.

Quantity	K₂CO₃ aliquot	HCl
Concentration, c / mol dm⁻³	Not needed for purity	0.100
Volume, V / dm³	25.0/1000 = 0.0250	20.0/1000 = 0.0200
Amount, n / mol	0.00200/2 = 0.00100	n = cV 0.100 × 0.0200 = 0.00200
Coefficient (mole ratio)	1	2

Moles in the full flask = moles in aliquot × flask volume/aliquot volume = 0.00100 × 250.0/25.0 = 0.0100 mol.

Mass of K₂CO₃ = n × Mᵣ = 0.0100 × 138.2 = 1.382 g.

Percentage purity = pure mass/sample mass × 100 = 1.382/2.00 × 100 = 69.1%.

EXAM TRAPDividing the pure mass in one aliquot by the whole sample mass misses the flask/aliquot factor. Overshooting the endpoint gives too much apparent carbonate and a purity that is too high.

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Exam-style question 42

A 1.80 g sample contains Na₂CO₃ and impurities. The sample is dissolved and made up to 250.0 cm³. A 25.00 cm³ aliquot of this solution requires 20.40 cm³ of 0.1000 mol dm⁻³ HCl solution for complete neutralisation. Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O

(a) Calculate the percentage by mass of Na₂CO₃ in the sample. M_r(Na₂CO₃) = 106.0. (6 marks)

(b) State one assumption about the impurities required by this calculation. (1 mark)

(c) A student overshoots the end point in every accurate titration. Explain the effect on the calculated percentage of Na₂CO₃. (2 marks)

Total: 9 marks

Show answer & marking points

(a) Na₂CO₃ : HCl = 1 : 2.

Quantity	Na₂CO₃ aliquot	HCl
Concentration, c / mol dm⁻³	Not needed for purity	0.1000
Volume, V / dm³	25.00/1000 = 0.02500	20.40/1000 = 0.02040
Amount, n / mol	0.002040/2 = 0.001020 [1]	n = cV 0.1000 × 0.02040 = 0.002040 [1]
Coefficient (mole ratio)	1	2

Moles in full flask = moles in aliquot × flask volume/aliquot volume = 0.001020 × 250.0/25.00 = 0.01020 mol. [1]

Pure mass = n × Mᵣ = 0.01020 × 106.0 = 1.0812 g. [1] Percentage purity = pure mass/sample mass × 100. [1] Purity = 1.0812/1.80 × 100 = 60.1%. [1]

(b) The impurities do not react with HCl / do not affect the titration [1].

✓ DONE WHENScale an aliquot result to the whole sample before calculating purity.

Link titration, purity, gas and process calculations

Learning goalChoose a valid sequence, preserve units and audit the final result.

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A long calculation becomes more manageable when every amount has a label: which substance, and which part of the sample? Distinguish an aliquot, the whole flask, the original sample and a production batch.

Start with the requested quantity and write an equation containing it. Work backwards through any missing quantities, then carry full calculator values through the forward calculation.

In a back titration, find the reagent left over and subtract it from the known amount originally added. Only the difference reacted with the sample. Apply the sample-reaction ratio after that subtraction.

In a standardisation chain, the concentration found in the first titration becomes data for the second. Do not round it prematurely.

Quantity	Solution amount or concentration	Another substance’s amount	Gas amount or volume	Purity	Yield	Atom economy
Start with	n = cV or c = n/V	Balanced-equation coefficient ratio	pV = nRT	Pure mass/sample mass × 100	Actual/theoretical × 100	Balanced formula masses and coefficients
Check	Solution volume in dm³	Limiting reactant used	Pa, m³, K and supplied R	Whole sample, not one aliquot	Same product and units	All reactants; desired product identified

Worked example

Diagnose a plausible-looking answer

A student writes n = 0.100 × 25.0 = 2.50 mol for 25.0 cm³ of 0.100 mol dm⁻³ solution.

Identify the error, calculate the correct number of moles and state the factor by which the student’s answer is too large.

Worked answer

The equation is suitable, but the volume must be in dm³.

Correct n = cV
n = 0.100 × 0.0250 = 0.00250 mol

The student’s amount is 1000 times too large.

EXAM TRAPKeep theoretical and measured gas amounts separate. For a production batch, first remove any inert fraction before calculating a theoretical product mass. Do not include that inert fraction in an equation-based atom-economy denominator.

Worked example

Carry a concentration into a second titration

25.00 cm³ of 0.04000 mol dm⁻³ Na₂CO₃ solution requires 20.00 cm³ of HCl solution for complete neutralisation. Then 18.00 cm³ of this HCl solution completely neutralises 25.00 cm³ of NaOH solution.

Calculate the concentrations of the HCl and NaOH solutions in mol dm⁻³. The reactions are Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O and HCl + NaOH → NaCl + H₂O.

Worked answer

First titration: find the HCl concentration. Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O.

Quantity	Na₂CO₃	HCl
Concentration, c / mol dm⁻³	0.04000	c = n/V 0.002000/0.02000 = 0.1000
Volume, V / dm³	25.00/1000 = 0.02500	20.00/1000 = 0.02000
Amount, n / mol	n = cV 0.04000 × 0.02500 = 0.001000	0.001000 × 2 = 0.002000
Coefficient (mole ratio)	1	2

Second titration: use that HCl to find the NaOH concentration. HCl + NaOH → NaCl + H₂O.

Quantity	HCl	NaOH
Concentration, c / mol dm⁻³	0.1000	c = n/V 0.001800/0.02500 = 0.07200
Volume, V / dm³	18.00/1000 = 0.01800	25.00/1000 = 0.02500
Amount, n / mol	n = cV 0.1000 × 0.01800 = 0.001800	0.001800 (1 : 1 ratio)
Coefficient (mole ratio)	1	1

The NaOH concentration is 0.07200 mol dm⁻³.

Worked example

Find what reacted by measuring what is left

0.500 g impure MgO reacts with 40.00 cm³ of 0.5000 mol dm⁻³ HCl. All the hydrochloric acid remaining after the reaction is completely neutralised by 12.00 cm³ of 0.2000 mol dm⁻³ NaOH solution. Use Mᵣ(MgO) = 40.3.

Calculate the percentage purity by mass of MgO in the sample. Assume the impurities do not react. The reactions are MgO + 2HCl → MgCl₂ + H₂O and HCl + NaOH → NaCl + H₂O.

Worked answer

HCl + NaOH → NaCl + H₂O. The back titration measures the acid left after MgO reacts.

Quantity	HCl added	NaOH titre	HCl left
Concentration, c / mol dm⁻³	0.5000	0.2000	Not needed
Volume, V / dm³	40.00/1000 = 0.04000	12.00/1000 = 0.01200	Not needed
Amount, n / mol	n = cV 0.5000 × 0.04000 = 0.02000	n = cV 0.2000 × 0.01200 = 0.002400	Same moles as NaOH 0.002400
Coefficient (mole ratio)	Not compared	1	1

Moles of HCl consumed = initial − remaining = 0.02000 − 0.002400 = 0.01760 mol.

MgO + 2HCl → MgCl₂ + H₂O: MgO : HCl = 1 : 2. Moles of MgO = 0.01760/2 = 0.008800 mol.

Pure mass = n × Mᵣ = 0.008800 × 40.3 = 0.35464 g. Percentage purity = pure mass/sample mass × 100 = 0.35464/0.500 × 100 = 70.9%.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 43

1.590 g anhydrous Na₂CO₃ is made up to 250.0 cm³. A 25.00 cm³ aliquot of this solution requires 30.20 cm³ of HCl solution for complete neutralisation. Then 22.65 cm³ of this HCl solution completely neutralises 25.00 cm³ of NaOH solution. Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O; HCl + NaOH → NaCl + H₂O. Mᵣ(Na₂CO₃) = 106.0.

Calculate the HCl concentration in mol dm⁻³. (5 marks)
Calculate the NaOH concentration in mol dm⁻³, retaining the full first answer in your working. (3 marks)

Total: 8 marks

Show answer & marking points

(a) Moles of Na₂CO₃ in full flask = m/Mᵣ = 1.590/106.0 = 0.01500 mol. [1]

Quantity	Na₂CO₃ aliquot	HCl
Concentration, c / mol dm⁻³	0.01500/0.2500 = 0.06000	c = n/V 0.003000/0.03020 = 0.09934 [1]
Volume, V / dm³	25.00/1000 = 0.02500	30.20/1000 = 0.03020 [1]
Amount, n / mol	0.01500 × 25.00/250.0 = 0.001500 [1]	0.001500 × 2 = 0.003000 [1]
Coefficient (mole ratio)	1	2

(b) Retain the full HCl concentration, 0.003000/0.03020, for the second titration. HCl + NaOH → NaCl + H₂O.

Quantity	HCl	NaOH
Concentration, c / mol dm⁻³	0.003000/0.03020 = 0.0993377…	c = n/V 0.002250/0.02500 = 0.09000 [1]
Volume, V / dm³	22.65/1000 = 0.02265	25.00/1000 = 0.02500
Amount, n / mol	n = cV (0.003000/0.03020) × 0.02265 = 0.002250 [1]	0.002250 (1 : 1 ratio) [1]
Coefficient (mole ratio)	1	1

Exam-style question 44

A 1.25 g antacid tablet containing CaCO₃ and inert binder reacts with excess HCl: 50.00 cm³ of 0.5000 mol dm⁻³. All the HCl remaining after complete reaction is completely neutralised by 15.00 cm³ of 0.2000 mol dm⁻³ NaOH solution. CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O; HCl + NaOH → NaCl + H₂O. Mᵣ(CaCO₃) = 100.1.

Calculate the percentage by mass of CaCO₃ in the tablet. (6 marks)
Explain the effect on the calculated percentage if some CaCO₃ does not react. (2 marks)

Total: 8 marks

Show answer & marking points

(a) The remaining HCl reacts with NaOH in a 1 : 1 ratio.

Quantity	HCl added	NaOH titre	HCl left
Concentration, c / mol dm⁻³	0.5000	0.2000	Not needed
Volume, V / dm³	50.00/1000 = 0.05000	15.00/1000 = 0.01500	Not needed
Amount, n / mol	n = cV 0.5000 × 0.05000 = 0.02500 [1]	n = cV 0.2000 × 0.01500 = 0.003000	Same moles as NaOH 0.003000 [1]
Coefficient (mole ratio)	Not compared	1	1

Moles of HCl consumed = initial − remaining = 0.02500 − 0.003000 = 0.02200 mol. [1] CaCO₃ : HCl = 1 : 2, so moles of CaCO₃ = 0.02200/2 = 0.01100 mol. [1]

Pure mass = n × Mᵣ = 0.01100 × 100.1 = 1.1011 g. [1] Percentage purity = pure mass/sample mass × 100 = 1.1011/1.25 × 100 = 88.1%. [1]

(b) More HCl is measured as left over, so less is inferred to have reacted. [1] The calculated carbonate amount and purity are too low. [1]

Exam-style question 45

An industrial feed contains NaHCO₃ and inert material. A 1.200 g feed sample is dissolved and made up to 250.0 cm³. A 25.00 cm³ aliquot of this solution requires 12.80 cm³ of 0.1000 mol dm⁻³ HCl solution for complete neutralisation. A production batch uses 50.0 kg of the same feed and obtains 24.6 kg Na₂CO₃. NaHCO₃ + HCl → NaCl + CO₂ + H₂O 2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O M_r: NaHCO₃ = 84.0; Na₂CO₃ = 106.0.

(a) Use the titration to calculate the percentage by mass of NaHCO₃ in the feed. (5 marks)

(b) Calculate the theoretical mass of Na₂CO₃ obtainable from the 50.0 kg feed batch. (4 marks)

(d) Calculate the atom economy for making Na₂CO₃ by the decomposition reaction. (2 marks)

(e) Explain why the percentage yield and atom economy have different values, and state one further item of information needed to judge the process. (3 marks)

Total: 16 marks

Show answer & marking points

(a) NaHCO₃ + HCl → NaCl + CO₂ + H₂O.

Quantity	NaHCO₃ aliquot	HCl
Concentration, c / mol dm⁻³	Not needed for purity	0.1000
Volume, V / dm³	25.00/1000 = 0.02500	12.80/1000 = 0.01280
Amount, n / mol	0.001280 (1 : 1 ratio) [1]	n = cV 0.1000 × 0.01280 = 0.001280 [1]
Coefficient (mole ratio)	1	1

Moles in full flask = moles in aliquot × flask volume/aliquot volume = 0.001280 × 250.0/25.00 = 0.01280 mol. [1]

Pure mass = n × Mᵣ = 0.01280 × 84.0 = 1.0752 g. [1] Percentage purity = pure mass/sample mass × 100 = 1.0752/1.200 × 100 = 89.6%. [1]

(b) Pure NaHCO₃ in the batch = 50.0 × 0.896 = 44.8 kg [1]. Amount = 44 800/84.0 = 533.333… mol [1]. The 2:1 ratio gives 266.666… mol Na₂CO₃ [1]. Theoretical mass = 266.666… × 106.0 = 28 266.666… g = 28.3 kg [1].

Percentage yield = (actual yield ÷ theoretical yield) × 100
(c) Percentage yield = 24.6/28.2667 × 100 [1] = 87.0% [1].

Atom economy = (total relative mass of desired product ÷ total relative mass of reactants) × 100
(d) Atom economy = 106.0/(2 × 84.0) × 100 [1] = 63.1% [1].

(e) Yield measures the actual Na₂CO₃ obtained relative to its theoretical maximum [1], whereas atom economy measures the fraction of reactant mass that can enter Na₂CO₃ according to the equation, with CO₂ and H₂O as by-products [1]. One further factor such as energy demand, feedstock cost, hazards, emissions or use of the by-products is needed [1].

EXAM TRAPThe inert fraction affects how much pure NaHCO₃ enters the batch, but it is not included in the equation-based atom-economy denominator.

✓ DONE WHENChoose a valid sequence, preserve units and audit the final result.

Titrations & practical technique: AQA A-level Chemistry