Lesson 5 of 8
Titration and quantitative analysis
Make the measured volume correspond to the reacting amount.
A-Level 9476 (2026-2027)
I can use apparatus and make detailed measurements and observations.Syllabus
Syllabus 9476, MMO. Use apparatus and make detailed measurements and observations
I can explain titration and standard solutions.Syllabus
Syllabus 9476, T1. Titration and standard solutions
Make a titre correspond to the reacting amounts
The endpoint, rinsing and complete transfer determine what the number means.
- Prepare the measured solutions
Rinse the burette with its solution and the pipette with the solution it will measure. Fill the burette tip, remove the funnel and record the initial reading. Use a pipette filler.
- Transfer a fixed aliquot
Deliver the pipetted solution into a clean conical flask. Allow the pipette to drain as designed; do not blow out a residual tip drop unless it is a blow-out pipette.
- Locate the endpoint
Add a small consistent amount of a suitable indicator. Swirl during addition, use a white background and add titrant dropwise near the endpoint. Rinse flask walls with distilled water if needed.
- Repeat and calculate
Record the final reading and titre. Obtain agreeing precise titres, select a justified mean, then use concentration, volume and the balanced mole ratio.
| Place | Suitable final rinse | Reason |
|---|---|---|
| Burette | Solution to be placed in it | Residual water would dilute the titrant |
| Volumetric pipette | Solution being measured | Residual water would reduce the amount in its fixed volume |
| Conical flask | Distilled water | Extra water changes concentration but not the already measured reacting amount |
| Volumetric flask before preparation | Distilled water | Water is part of the final solvent; the solution will be made to the calibration mark |
The equivalence point is the stoichiometric completion of reaction. The endpoint is the observed indicator change. Choose an indicator whose transition lies in the steep pH change near equivalence; universal indicator gives a broad colour change and is unsuitable for precise titration. The required final colour depends on the indicator and which solution is added.
| Indicator | Lower-pH side | Higher-pH side |
|---|---|---|
| Methyl orange | Red; passes through orange | Yellow |
| Screened methyl orange | Violet; passes through grey | Green |
| Thymolphthalein | Colourless | Blue |
With acid added to an alkali, methyl orange changes from yellow towards orange; screened methyl orange changes from green towards grey. With thymolphthalein, adding acid removes the blue colour, whereas adding alkali gives the first persistent pale blue. Add dropwise near the specified endpoint and swirl: a temporary colour where a drop lands is not the final well-mixed colour.
H2 also uses thymol blue. For its alkaline transition (about pH 8.0-9.6), the lower-pH colour is yellow and the higher-pH colour is blue, with green between. It has a separate red-to-yellow change at much lower pH. Use the transition appropriate to the supplied titration curve; an intermediate indicator colour does not automatically mean pH 7.
Worked exampleConnect a titre to an unknown
25.0 cm3 of an alkali requires 20.20 cm3 of 0.100 mol dm-3 HCl. The reaction is HCl + NaOH -> NaCl + H2O.
Try it first. Then show the worked stepsHide the worked steps
- n(HCl) = 0.100 × 20.20 / 1000 = 0.002020 mol.
- The 1:1 ratio gives n(NaOH) = 0.002020 mol in the aliquot.
- c(NaOH) = 0.002020 / 0.0250 = 0.0808 mol dm-3.
0.0808 mol dm-3. A different balanced equation would require its own mole ratio.
Check your understanding
After pipetting the alkali into the conical flask, a student adds a little distilled water. How does this affect the titre?
Show the answer
No effect, because the amount of alkali in the flask is unchanged.
Water dilutes the alkali but does not change the moles present, so the same volume of acid neutralises it. This assumes nothing is lost and the end-point is still clear.
Prepare a known concentration and interpret indirect titrations
Track the whole solution, the aliquot and any reagent left over.
A standard solution has a reliably known concentration. For a suitable pure, stable solid, weigh accurately, dissolve it completely in a beaker, transfer quantitatively into a volumetric flask, and wash the beaker, rod and funnel into the flask. Cool to the appropriate temperature before making up to the mark. Add the final water dropwise with the meniscus at eye level, stopper and invert repeatedly to mix. A volumetric flask contains its stated volume; it is not a device for delivering arbitrary measured portions.
Worked examplePrepare a carbonate standard
What mass of anhydrous Na2CO3, Mr = 106.0, is needed for 250.0 cm3 of a 0.0400 mol dm-3 solution?
Try it first. Then show the worked stepsHide the worked steps
- n = cV = 0.0400 × 0.2500 = 0.0100 mol.
- m = nM = 0.0100 × 106.0 = 1.06 g. Use the actual measured mass to calculate the actual concentration.
- A 25.00 cm3 aliquot contains one tenth of the flask amount. Making a dilute solution does not change the amount transferred from the solid.
Target mass 1.06 g. Complete transfer and making to volume are both necessary.
In a back titration, first react the sample with a known excess of reagent. Measure the leftover reagent with a second titration, then subtract it from the starting amount. If only an aliquot of the leftover solution is titrated, scale that result to the whole flask before subtracting.
Worked exampleBack-titrate a carbonate sample
A 0.250 g sample is treated with 50.00 cm3 of 0.100 mol dm-3 HCl. The whole remaining acid needs 10.00 cm3 of 0.100 mol dm-3 NaOH. Only CaCO3 in the sample reacts with the acid.
Try it first. Then show the worked stepsHide the worked steps
- Starting HCl = 0.005000 mol; leftover HCl = n(NaOH) = 0.001000 mol.
- HCl consumed by sample = 0.004000 mol. CaCO3 + 2HCl -> CaCl2 + CO2 + H2O, so n(CaCO3) = 0.002000 mol.
- Using M(CaCO3) = 100.1 g mol-1, mass = 0.2002 g; percentage by mass = 0.2002 / 0.250 × 100.
80.1% CaCO3 by mass, assuming reaction is complete and other components do not consume acid.
| Method | What the signal means | Reasoning to retain |
|---|---|---|
| Acidified manganate(VII) | A faint persistent excess manganate colour marks the endpoint in the usual reduction titration | Use the specified acid and the balanced electron ratio; the titrant may be its own indicator |
| Direct iodine titration | A first persistent iodine/starch colour can mark a small excess of iodine titrant | Iodine oxidises the reducing analyte; use the stated reaction and endpoint instructions |
| Iodine-thiosulfate | Blue starch-iodine colour disappears as the final iodine is reduced | I2 + 2S2O32- -> 2I- + S4O62-; add starch near the endpoint in the usual procedure |
| Indirect iodine determination | An oxidant first liberates iodine from excess iodide | Link the oxidant:iodine ratio to iodine:thiosulfate; do not assume either stage is 1:1 |
Check your understanding
A 25.0 cm3 portion is taken from a 250.0 cm3 flask, and it contains 0.00120 mol of leftover acid. How much leftover acid is in the whole flask?
Show the answer
0.0120 mol, because the portion is one tenth of the flask
250.0 / 25.0 = 10, so multiply by ten. Use the whole-flask amount when you subtract in a back titration.