1. Titration method and why
| Step | Why |
|---|---|
| Rinse the pipette with the solution it will measure | Stops water or other residue diluting the aliquot |
| Rinse the burette with the titrant | Stops water diluting the titrant |
| Rinse the conical flask with distilled water only | Water adds no moles of reagent; rinsing with a reagent would add an unmeasured amount |
| Fill the burette, including the jet below the tap, and remove the funnel | Air in the jet or drips from the funnel make the volume readings wrong |
| Use a pipette filler; let the pipette drain as calibrated (don't blow out the last drop) | The pipette is calibrated to deliver its volume this way |
| Add a few drops of indicator | Indicators are weak acids or weak bases, so too much can react and affect the titre |
| Do a rough titration, then accurate ones, adding dropwise near the end point while swirling | Gives a precise end point |
| Use a white tile; read the meniscus at eye level | Makes the colour change and the reading clear |
Diagram placeholder
Titration apparatus
Labels to include:
- Burette clamped vertically, with tap and jet
- Meniscus read at eye level
- Conical flask with aliquot + indicator
- White tile
- Volumetric pipette with pipette filler
The titrant in the burette is added to a measured aliquot in the conical flask, which stands on a white tile, until the indicator changes colour permanently.
2. Indicators and end points
Choose an indicator whose colour-change range lies within the steep pH change near the equivalence point, and describe the change in the direction of the titration:
- Phenolphthalein: colourless in acid, pink in alkali. Acid added to alkali: pink → colourless. Alkali added to acid: colourless → first permanent pale pink.
- Methyl orange: red in acid, yellow in alkali. Acid added to alkali: yellow → orange.
Either indicator works for a strong acid with a strong alkali. Which one to use depends on the acid and base, not on a rule like "NaOH always uses phenolphthalein".
3. Recording results and concordance
Record the initial and final burette readings to 2 d.p. (to the nearest 0.05 cm3 on a burette graduated in 0.1 cm3). Concordant titres usually agree within 0.10 cm3, unless the question says otherwise. Calculate the mean using concordant titres only, leaving out the rough titration and any outlier.
| Rough | 1 | 2 | 3 | |
|---|---|---|---|---|
| Final / cm³ | 25.30 | 24.60 | 25.15 | 24.40 |
| Initial / cm³ | 0.50 | 0.50 | 0.70 | 0.25 |
| Titre / cm³ | 24.80 | 24.10 | 24.45 | 24.15 |
Titres 1 and 3 (24.10 and 24.15) are concordant; titre 2 is an outlier. Mean = (24.10 + 24.15) ÷ 2 = 24.125 = 24.13 cm3.
Concordant results show repeatability (precision). They do not prove the answer is accurate: a systematic error such as a wrongly made standard solution affects every titre equally.
4. Direct titration calculations
5. Using an aliquot factor
If you make a solution up to 250.0 cm3 and titrate 25.0 cm3 portions, the moles in the whole flask are 10 times the moles in one aliquot. Multiply moles by this factor, not concentration: the concentration of the aliquot and the flask are the same.
6. Back titration
Use a back titration when the substance is insoluble or reacts slowly, so it can't be titrated directly. React it with a known excess of acid, then titrate the leftover acid.
- Initial moles of acid added.
- Moles of leftover acid from the titration (× aliquot factor if needed).
- Moles of acid reacted = initial − leftover.
- Use the equation ratio to find moles of the substance, then its mass or purity.
7. Measurement uncertainty
Absolute uncertainty is the ± value. Percentage uncertainty compares it with the measurement. Percentage error compares your result with an accepted value. Uncertainty values depend on the instrument; use the ones in the question.
How many readings?
- Burette titre: two readings (initial and final). If each is ±0.05 cm3, the titre is ±0.10 cm3. If the question gives an overall uncertainty for the titre (e.g. ±0.15 cm3), use it as given; don't double it again.
- Volumetric flask or pipette: a single calibrated volume, so use its uncertainty once.
- Weighing by difference: two balance readings, so double the per-reading uncertainty.
| Measurement | Uncertainty | Percentage |
|---|---|---|
| Titre 24.13 cm³ | ±0.05 per reading → ±0.10 | 0.10 ÷ 24.13 × 100 = 0.41% |
| Pipette 25.0 cm³ | ±0.06 | 0.24% |
| Mass 2.65 g by difference | ±0.005 per reading → ±0.010 | 0.38% |
| Volumetric flask 250.0 cm³ | ±0.20 | 0.08% |
A larger titre (within the burette's capacity) reduces the percentage uncertainty. To get a larger titre, dilute the titrant or use a more concentrated sample; adding water to the conical flask doesn't change the titre. Repeating titrations improves reliability but doesn't remove systematic errors.
8. Practical errors
| Error | Effect |
|---|---|
| Burette rinsed with water, not titrant | Titrant diluted, so a larger titre is needed: titre too high |
| Air bubble in the jet that fills during titration | Measured volume larger than the volume delivered: titre too high |
| Distilled water in the conical flask | No effect: the moles in the aliquot are unchanged |
| Conical flask rinsed with a reagent | Extra unmeasured moles: result biased |
| End point overshot | Titre too high |
9. Working safely
Wear eye protection when handling acids and alkalis. Use a pipette filler, never your mouth. Let crucibles cool before weighing. Follow your school's risk assessment.
10. Quick checks
Finesse practice: indicative answers to check your reasoning, not official mark allocations. Ar: O 16.0, Mg 24.3.
Q1. Titres: rough 22.35, then 22.05, 22.15, 22.10 cm³. Find the mean titre.Show answer
Exclude the rough. 22.05, 22.15 and 22.10 all lie within 0.10 cm3, so all three are concordant.
Mean = (22.05 + 22.15 + 22.10) ÷ 3 = 22.10 cm3
Q2. Each burette reading is ±0.05 cm³. Find the percentage uncertainty in a 22.10 cm³ titre.Show answer
Two readings: ±0.10 cm3
0.10 ÷ 22.10 × 100 = 0.45%
Q3. 25.0 cm³ of HCl needs 22.10 cm³ of 0.100 mol dm⁻³ NaOH. Find [HCl].Show answer
n(NaOH) = 0.100 × 0.02210 = 2.210 × 10−3 mol = n(HCl) (1 : 1)
c = 2.210 × 10−3 ÷ 0.0250 = 0.0884 mol dm−3
Q4. A student rinsed the burette with water before filling it with NaOH. Effect on the calculated [HCl]?Show answer
The NaOH is diluted, so a larger titre is needed. Using the stated NaOH concentration with a larger titre gives too many moles, so the calculated [HCl] is too high.
Q5. 0.500 g of impure MgO reacts with 25.0 cm³ of 1.00 mol dm⁻³ HCl. The excess acid needs 15.00 cm³ of 1.00 mol dm⁻³ NaOH. Assume the impurities do not react. Find the % MgO (MgO + 2HCl → MgCl₂ + H₂O).Show answer
Initial HCl = 0.0250 mol; excess = n(NaOH) = 0.0150 mol
Reacted HCl = 0.0100 mol; n(MgO) = 0.00500 mol
Mass = 0.00500 × 40.3 = 0.2015 g; 0.2015 ÷ 0.500 × 100 = 40.3%
11. Sources
Sources and examiner guidance (reviewed 1 October 2026)
- Chemrevise — AQA 1.2 Calculations (amount of substance) revision guide (N. Goalby) — primary content reference (titrations, back titration, uncertainty)
- AQA 7405 specification — 3.1.2 Amount of substance (incl. Required Practical 1) — 3.1.2.5; Required Practical 1 (titration)
- AQA 7404/1 mark scheme, June 2022 — Q02.3–02.5: faults and effects, concordant mean, given overall uncertainty
- AQA 7404/1 examiner report, June 2022 — Q02.3–02.4: explain fault effects; titres to 2 d.p., concordant only
- AQA 7404/1 mark scheme, June 2023 — Q02.6 back titration; Q03.2 flask uncertainty not doubled
- AQA 7404/1 examiner report, June 2023 — Q02.6, Q03.2, Q10/Q11: ratio/subtraction errors, doubling, water rinse
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