Titrations & practical skills
Titrations & practical skills
Titration apparatus and technique
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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.
| Titration | Suitable indicator |
|---|---|
| Strong acid + strong alkali | Methyl orange or phenolphthalein |
| Weak acid + strong alkali | Phenolphthalein |
| Strong acid + weak alkali | Methyl orange |
A pipette was left wet with water
A pipette is left wet with distilled water before it is used to measure the analyte solution into a conical flask. Explain how this affects the number of moles transferred and the titre, compared with rinsing the pipette with the analyte solution first.
Worked answer
Residual water dilutes the solution inside the pipette. The measured aliquot contains fewer moles of analyte.
For the burette diagram, the delivered volume is the final scale reading minus the initial scale reading. A single scale reading is not necessarily the delivered titre.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Describe how to carry out an accurate acid–base titration using a burette and a volumetric pipette. Include preparation of the apparatus, detection of the end point and how reliable results are obtained. (6 marks)
Name a suitable indicator for each titration: strong acid with a strong alkali; ethanoic acid with NaOH; strong acid with aqueous ammonia. (3 marks)
Total: 9 marks
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Rinse the burette and pipette with deionised water, then each with its own solution. [1] Fill the burette tip without an air bubble and remove the filling funnel. [1] Use a pipette filler to deliver the aliquot into a conical flask, allowing the pipette to drain without blowing out the retained drop. [1] Add a few drops of suitable indicator, using a white tile to see the change. [1] Perform a rough titration, then use fresh aliquots and add dropwise near the end point while swirling. [1] Read the meniscus at eye level; calculate titres from final minus initial readings and repeat for concordant titres (within 0.1 cm³ of each other). [1]
Strong acid/strong alkali: methyl orange or phenolphthalein. [1] Ethanoic acid/NaOH: phenolphthalein. [1] Strong acid/ammonia: methyl orange. [1] Equivalent correct operational detail is acceptable; do not require an exact prose script.
A student calculates the concentration of NaOH placed in the conical flask from a titration with 0.1000 mol dm−3 HCl in the burette. For each error, state whether the calculated NaOH concentration is too high, too low or unchanged, and explain why.
(a) The burette is rinsed only with water before it is filled with HCl. [2 marks]
(b) The pipette is rinsed only with water before the NaOH aliquot is measured. [2 marks]
(c) The conical flask is rinsed with deionised water immediately before use. [2 marks]
(d) An air bubble in the burette tip is expelled during the titration. [2 marks]
Total: 8 marks
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(a)
Calculated concentration is too high [1]
Residual water dilutes the HCl, so a larger recorded titre is required but the nominal 0.1000 mol dm−3 is still used [1]
(b)
Calculated concentration is too low [1]
Residual water dilutes the NaOH aliquot, so fewer moles enter the flask and a smaller titre is used [1]
(c)
Calculated concentration is unchanged [1]
Water changes the concentration in the flask but not the amount of NaOH delivered by the pipette [1]
(d)
Calculated concentration is too high [1]
Some recorded burette volume fills the tip instead of entering the flask, so the recorded titre is too large [1]
Concordant titres and percentage uncertainty
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The titre is final burette reading minus initial reading. Select concordant titres (within 0.1 cm³ of each other) before calculating a mean, and show which results you used.
Leave out the rough titre and justified non-concordant results. Use a different concordance tolerance if the question supplies one.
A titre depends on two burette readings, so add their absolute uncertainties in this treatment. Percentage uncertainty is absolute titre uncertainty divided by the titre, multiplied by 100.
For the same burette uncertainty, a larger titre gives a smaller percentage uncertainty. Use a different apparatus uncertainty if the question supplies one.
A directional error makes the result too high or too low. For example, consistently overshooting the end point gives too large a titre. When known acid is in the burette and unknown alkali is in the flask, this makes the calculated alkali concentration too high. Explain what changes, how the titre changes and how this changes the answer. Uncertainty alone does not tell you the direction of an error.
Choose the readings to average
The following burette readings were recorded during a titration. All readings are in cm³.
| Reading / cm³ | Rough | 1 | 2 | 3 |
|---|---|---|---|---|
| Initial | 0.00 | 0.10 | 0.20 | 0.00 |
| Final | 24.85 | 24.20 | 24.40 | 24.60 |
Calculate all four titres, select the concordant accurate titres and calculate their mean. Treat accurate titres that differ by no more than 0.10 cm³ as concordant. Each burette reading has uncertainty ±0.05 cm³. Calculate the absolute and percentage uncertainty in a titre equal to that mean.
Worked answer
| Titration | Initial | Final | Titre |
|---|---|---|---|
| Rough | 0.00 | 24.85 | 24.85 |
| 1 | 0.10 | 24.20 | 24.10 |
| 2 | 0.20 | 24.40 | 24.20 |
| 3 | 0.00 | 24.60 | 24.60 |
Percentage uncertainty = (0.10/24.15) × 100 = 0.414…%
This example includes the two reading uncertainties only. Include any additional contribution specified in a question; do not assume that one uncertainty value applies to every burette or experiment.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
A rough titre is 24.30 cm3. Four accurate titres are 24.05, 24.00, 24.45 and 24.05 cm3. Use a concordance tolerance of 0.10 cm3. A 25.00 cm3 Na2CO3 sample is titrated with 0.1000 mol dm−3 HCl. Na2CO3 + 2HCl → 2NaCl + CO2 + H2O. Each burette reading has uncertainty ±0.05 cm3.
(a) Select the concordant accurate titres. (1 mark)
(b) Calculate the mean titre. (1 mark)
(c) Calculate the carbonate concentration in mol dm−3. (3 marks)
(d) Calculate the percentage uncertainty of a titre at this mean volume. (2 marks)
Total: 7 marks
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(a) Use 24.05, 24.00 and 24.05 cm3; exclude the rough and 24.45 cm3. [1]
(b) Mean = (24.05 + 24.00 + 24.05)/3 = 24.03333… cm3, recorded as 24.03 cm3. [1]
(c) Use the unrounded mean titre in the calculation. Na₂CO₃ : HCl = 1 : 2.
| Quantity | Na₂CO₃ | HCl |
|---|---|---|
| Concentration, c / mol dm⁻³ | c = n/V 0.001201667…/0.02500 = 0.04807 [1] | 0.1000 |
| Volume, V / dm³ | 25.00/1000 = 0.02500 | 24.03333…/1000 = 0.02403333… |
| Amount, n / mol | 0.002403333…/2 = 0.001201667… [1] | n = cV 0.1000 × 0.02403333… = 0.002403333… [1] |
| Coefficient (mole ratio) | 1 | 2 |
Using the recorded 24.03 cm³ mean gives 0.04806 mol dm⁻³ and is accepted at this precision.
(d) Absolute titre uncertainty = 0.05 + 0.05 = 0.10 cm3. [1] Percentage = (0.10/24.03333…) × 100 = 0.4161…%, or 0.42%. [1] This uses the specified two-reading convention and does not divide the apparatus uncertainty by the number of repeats.
Titration calculations
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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.
Hydrochloric acid against carbonate
25.00 cm3 of 0.0500 mol dm−3 Na2CO3 solution requires 23.60 cm3 of HCl solution for complete neutralisation.
Write the balanced neutralisation equation and calculate the HCl concentration in mol dm⁻³.
Worked answer
Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O. Twice as many moles of HCl react as carbonate.
| Quantity | Na₂CO₃ | HCl |
|---|---|---|
| Concentration, c / mol dm⁻³ | 0.0500 | c = n/V 0.002500/0.02360 = 0.106 |
| Volume, V / dm³ | 25.00/1000 = 0.02500 | 23.60/1000 = 0.02360 |
| Amount, n / mol | n = cV 0.0500 × 0.02500 = 0.001250 | 0.001250 × 2 = 0.002500 |
| Coefficient (mole ratio) | 1 | 2 |
The HCl concentration is 0.106 mol dm⁻³ (3 significant figures).
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
25.00 cm3 KOH is neutralised by 19.60 cm3 of 0.1250 mol dm−3 HNO3.
(a) Write the balanced equation. (1 mark)
(b) Calculate the KOH concentration. (3 marks)
Total: 4 marks
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(a) HNO₃ + KOH → KNO₃ + H₂O. [1]
| Quantity | HNO₃ | KOH |
|---|---|---|
| Concentration, c / mol dm⁻³ | 0.1250 | c = n/V 0.002450/0.02500 = 0.09800 [1] |
| Volume, V / dm³ | 19.60/1000 = 0.01960 | 25.00/1000 = 0.02500 |
| Amount, n / mol | (b) n = cV 0.1250 × 0.01960 = 0.002450 [1] | 0.002450 (1 : 1 ratio) [1] |
| Coefficient (mole ratio) | 1 | 1 |
Citric acid reacts as shown: C6H8O7 + 3NaOH → Na3C6H5O7 + 3H2O. A 25.00 cm3 sample of citric acid solution requires 24.80 cm3 of 0.1500 mol dm−3 NaOH solution for complete neutralisation. M(C6H8O7) = 192.1 g mol−1. Calculate the acid concentration in mol dm−3 and g dm−3. (4 marks)
Total: 4 marks
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One mole of citric acid reacts with three moles of NaOH.
| Quantity | Citric acid | NaOH |
|---|---|---|
| Concentration, c / mol dm⁻³ | c = n/V 0.001240/0.02500 = 0.04960 [1] | 0.1500 |
| Volume, V / dm³ | 25.00/1000 = 0.02500 | 24.80/1000 = 0.02480 |
| Amount, n / mol | 0.003720/3 = 0.001240 [1] | n = cV 0.1500 × 0.02480 = 0.003720 [1] |
| Coefficient (mole ratio) | 1 | 3 |
Mass concentration = c × M = 0.04960 × 192.1 = 9.528 g dm⁻³. [1]
Aliquots and dilution factors
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An aliquot is a measured portion of a well-mixed solution. A titration of that portion gives the amount in the aliquot, not automatically the amount in the whole flask.
When only the desired component reacts, calculate its amount and mass from the reaction. Percentage purity = (mass of pure component / mass of original sample) × 100. Use the whole-sample amount, not the aliquot amount.
Multiply the aliquot mole amount by whole-flask volume divided by aliquot volume to find the total. Use matching volume units in that ratio.
If the flask was made by diluting an original sample, those total solute moles came from that original sample. Divide by the original sample volume to recover its concentration. Label each stage so the dilution factor is applied only once.
Find the original acid concentration
10.00 cm3 ethanoic acid is diluted to 250.0 cm3. A 25.00 cm3 aliquot of the diluted ethanoic acid solution requires 20.00 cm3 of 0.100 mol dm−3 NaOH solution for complete neutralisation.
Calculate the ethanoic acid concentration in the original 10.00 cm³ sample, in mol dm⁻³. The reaction is CH₃COOH + NaOH → CH₃COONa + H₂O.
Worked answer
The aliquot is the measured 25.00 cm³ portion of diluted acid. CH₃COOH + NaOH → CH₃COONa + H₂O.
| Quantity | CH₃COOH (diluted) | NaOH |
|---|---|---|
| Concentration, c / mol dm⁻³ | c = n/V 0.00200/0.02500 = 0.0800 | 0.100 |
| Volume, V / dm³ | 25.00/1000 = 0.02500 | 20.00/1000 = 0.02000 |
| Amount, n / mol | 0.00200 (1 : 1 ratio) | n = cV 0.100 × 0.02000 = 0.00200 |
| Coefficient (mole ratio) | 1 | 1 |
Moles in the full flask = moles in aliquot × flask volume/aliquot volume = 0.00200 × 250.0/25.00 = 0.0200 mol.
Original concentration = total moles/original sample volume = 0.0200/0.01000 = 2.00 mol dm⁻³.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
A 1.100 g impure NaOH sample contains an impurity that does not react with sulfuric acid. The sample is made up to 250.0 cm3. A 25.00 cm3 aliquot of this solution requires 12.50 cm3 of 0.1000 mol dm−3 H2SO4 solution for complete neutralisation. H2SO4 + 2NaOH → Na2SO4 + 2H2O. M(NaOH) = 40.0 g mol−1. Calculate the percentage purity by mass of NaOH in the sample. (5 marks)
Total: 5 marks
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H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.
| Quantity | H₂SO₄ | NaOH aliquot |
|---|---|---|
| Concentration, c / mol dm⁻³ | 0.1000 | Not needed for purity |
| Volume, V / dm³ | 12.50/1000 = 0.01250 | 25.00/1000 = 0.02500 |
| Amount, n / mol | n = cV 0.1000 × 0.01250 = 0.001250 [1] | 0.001250 × 2 = 0.002500 [1] |
| Coefficient (mole ratio) | 1 | 2 |
Moles in full flask = moles in aliquot × flask volume/aliquot volume = 0.002500 × 250.0/25.00 = 0.02500 mol. [1]
Pure mass = n × M = 0.02500 × 40.0 = 1.00 g. [1] Percentage purity = pure mass/sample mass × 100 = 1.00/1.100 × 100 = 90.9% (3 significant figures). [1]
5.00 cm3 of stock H2SO4 is diluted to 250.0 cm3. A 25.00 cm3 aliquot of the diluted sulfuric acid solution requires 22.60 cm3 of 0.1500 mol dm−3 NaOH solution for complete neutralisation. H2SO4 + 2NaOH → Na2SO4 + 2H2O. M(H2SO4) = 98.1 g mol−1. Calculate the original acid concentration in mol dm−3 and g dm−3. (5 marks)
Total: 5 marks
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H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.
| Quantity | H₂SO₄ (diluted) | NaOH |
|---|---|---|
| Concentration, c / mol dm⁻³ | c = n/V 0.001695/0.02500 = 0.06780 [1] | 0.1500 |
| Volume, V / dm³ | 25.00/1000 = 0.02500 | 22.60/1000 = 0.02260 |
| Amount, n / mol | 0.003390/2 = 0.001695 [1] | n = cV 0.1500 × 0.02260 = 0.003390 [1] |
| Coefficient (mole ratio) | 1 | 2 |
Original concentration = diluted concentration × final dilution volume/original sample volume = 0.06780 × 250.0/5.00 = 3.39 mol dm⁻³. [1]
Mass concentration = c × M = 3.39 × 98.1 = 333 g dm⁻³ (3 significant figures). [1]
Back titration and purity
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A back titration measures what is left over. You add a known excess of reagent to the sample, then titrate the unreacted excess.
- Calculate the amount in moles of reagent originally added.
- Use the titration and its equation ratio to find the amount in moles of that reagent left over.
- Subtract: amount that reacted with the sample = initial amount - amount remaining.
- Use the sample-reaction equation to find the amount in moles of sample substance, then its mass or purity.
The two reactions may have different coefficient ratios. Apply each ratio at the stage where it belongs.
Purity of limestone
0.250 g limestone reacts with 25.00 cm3 of 0.200 mol dm−3 HCl. All the hydrochloric acid remaining after the reaction is completely neutralised by 10.00 cm3 of 0.100 mol dm−3 NaOH solution. M(CaCO3) = 100.1 g mol−1.
Calculate the percentage by mass of CaCO₃ in the limestone. Assume the impurities do not react. The reactions are CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ and HCl + NaOH → NaCl + H₂O.
Worked answer
The NaOH titration measures the remaining HCl. HCl + NaOH → NaCl + H₂O.
| Quantity | HCl added | NaOH titre | HCl left |
|---|---|---|---|
| Concentration, c / mol dm⁻³ | 0.200 | 0.100 | Not needed |
| Volume, V / dm³ | 25.00/1000 = 0.02500 | 10.00/1000 = 0.01000 | Not needed |
| Amount, n / mol | n = cV 0.200 × 0.02500 = 0.00500 | n = cV 0.100 × 0.01000 = 0.00100 | Same moles as NaOH 0.00100 |
| Coefficient (mole ratio) | Not compared | 1 | 1 |
Moles of HCl consumed = initial − remaining = 0.00500 − 0.00100 = 0.00400 mol.
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O. Moles of CaCO₃ = 0.00400/2 = 0.00200 mol (1 : 2 ratio).
Pure mass = n × M = 0.00200 × 100.1 = 0.2002 g. Percentage purity = pure mass/sample mass × 100 = 0.2002/0.250 × 100 = 80.08%, or 80.1%.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
A 3.000 g mineral contains CaCO3 and material that does not react with nitric acid. It is treated with 50.00 cm3 of 1.200 mol dm−3 HNO3. After the reaction, all the excess nitric acid is completely neutralised by 18.50 cm3 of 0.4000 mol dm−3 NaOH solution. CaCO3 + 2HNO3 → Ca(NO3)2 + CO2 + H2O. M(CaCO3) = 100.1 g mol−1. Calculate the percentage by mass of CaCO3. (6 marks)
Total: 6 marks
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HNO₃ + NaOH → NaNO₃ + H₂O: the back titration has a 1 : 1 ratio.
| Quantity | HNO₃ added | NaOH titre | HNO₃ left |
|---|---|---|---|
| Concentration, c / mol dm⁻³ | 1.200 | 0.4000 | Not needed |
| Volume, V / dm³ | 50.00/1000 = 0.05000 | 18.50/1000 = 0.01850 | Not needed |
| Amount, n / mol | n = cV 1.200 × 0.05000 = 0.06000 [1] | n = cV 0.4000 × 0.01850 = 0.007400 | Same moles as NaOH 0.007400 [1] |
| Coefficient (mole ratio) | Not compared | 1 | 1 |
Moles of HNO₃ consumed = initial − remaining = 0.06000 − 0.007400 = 0.05260 mol. [1] CaCO₃ : HNO₃ = 1 : 2, so moles of CaCO₃ = 0.05260/2 = 0.02630 mol. [1]
Pure mass = n × M = 0.02630 × 100.1 = 2.63263 g. [1] Percentage purity = pure mass/sample mass × 100 = 2.63263/3.000 × 100 = 87.75% (4 significant figures). [1]