A student finds the concentration of ethanoic acid, CH3COOH\mathrm{CH_{3}COOH}CH3COOH, in a bottle of vinegar.
The label on the bottle says: "Contains 6.0 g of ethanoic acid in every 100 cm3\mathrm{cm^{3}}cm3."
The vinegar is too concentrated to titrate directly. The student uses a pipette to put 25.0 cm3\mathrm{cm^{3}}cm3 of vinegar into a 250 cm3\mathrm{cm^{3}}cm3 volumetric flask and makes the solution up to the mark with distilled water.
The student then titrates 25.0 cm3\mathrm{cm^{3}}cm3 portions of the diluted vinegar with sodium hydroxide solution of concentration 0.100 mol/dm3\mathrm{mol/dm^{3}}mol/dm3. The student uses phenolphthalein indicator.
CH3COOH+NaOH→CH3COONa+H2O \mathrm{CH_{3}COOH} + \mathrm{NaOH} \rightarrow \mathrm{CH_{3}COONa} + \mathrm{H_{2}O} CH3COOH+NaOH→CH3COONa+H2OThe table shows the results.
| Titration | Titre (cm3\mathrm{cm^{3}}cm3) |
|---|---|
| 1 | 22.30 |
| 2 | 21.55 |
| 3 | 21.45 |
| 4 | 21.50 |
The sodium hydroxide is added from the burette to the diluted vinegar in the flask.
State the colour change of the phenolphthalein at the end point.
Calculate the mean titre.
Calculate the concentration of ethanoic acid in the original vinegar, in mol/dm3\mathrm{mol/dm^{3}}mol/dm3.
Concentration = ...... mol/dm3\mathrm{mol/dm^{3}}mol/dm3
Evaluate the claim on the label. Use your answer to (c).
Relative atomic mass (Ar): C = 12.0 H = 1.0 O = 16.0
57 exam-style questions on OCR GCSE Chemistry 5.1 Monitoring chemical reactions, covering 5.1.1 The technique of titration, 5.1.2 Theoretical mass, percentage yield and atom economy, 5.1.3 Concentration of a solution in mol/dm3, 5.1.4 Concentration calculations and titration results, 5.1.5 Molar gas volume and volumes of gases in reactions, and 5.1.6 Choosing a reaction pathway. Each one has a worked solution and a mark scheme showing where the marks go.