Topic 5 of 8
Titration and quantitative analysis
Make the measured volume correspond to the reacting amount.
O-Level 6092 (2026) / SEC G3 K324 (2027)
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.
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.
Worked example
Connect 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.
- 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.