A student finds the concentration of a solution of sodium carbonate, Na2CO3\mathrm{Na_{2}CO_{3}}Na2CO3.
The student uses a pipette to put 25.0 cm3\mathrm{cm^{3}}cm3 of the sodium carbonate solution into a conical flask and adds methyl orange indicator.
The student then adds hydrochloric acid of concentration 0.100 mol/dm3\mathrm{mol/dm^{3}}mol/dm3 from a burette.
Na2CO3+2HCl→2NaCl+H2O+CO2 \mathrm{Na_{2}CO_{3}} + 2\mathrm{HCl} \rightarrow 2\mathrm{NaCl} + \mathrm{H_{2}O} + \mathrm{CO_{2}} Na2CO3+2HCl→2NaCl+H2O+CO2The table shows the student's results.
| Titration | Titre (cm3\mathrm{cm^{3}}cm3) |
|---|---|
| rough | 24.10 |
| 1 | 23.40 |
| 2 | 23.75 |
| 3 | 23.35 |
| 4 | 23.45 |
State the colour change of the methyl orange at the end point.
Calculate the mean titre the student should use.
Calculate the concentration of the sodium carbonate solution in mol/dm3\mathrm{mol/dm^{3}}mol/dm3.
Give your answer to 3 significant figures.
Concentration = ...... mol/dm3\mathrm{mol/dm^{3}}mol/dm3
Calculate the concentration of the sodium carbonate solution in g/dm3\mathrm{g/dm^{3}}g/dm3.
Relative atomic mass (Ar): C = 12.0 Na = 23.0 O = 16.0
Concentration = ...... g/dm3\mathrm{g/dm^{3}}g/dm3
Another student repeats the titration with the same sodium carbonate solution, but uses hydrochloric acid of concentration 0.200 mol/dm3\mathrm{mol/dm^{3}}mol/dm3.
Predict the mean titre. Explain your answer.
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.