Question 19N.2.SL.TZ0.2
| Date | November 2019 | Marks available | [Maximum mark: 7] | Reference code | 19N.2.SL.TZ0.2 |
| Level | SL | Paper | 2 | Time zone | TZ0 |
| Command term | Calculate, Determine, Suggest | Question number | 2 | Adapted from | N/A |
The biochemical oxygen demand of a water sample can be determined by the following series of reactions. The final step is titration of the sample with sodium thiosulfate solution, Na2S2O3 (aq).
2Mn2+ (aq) + O2 (aq) + 4OH− (aq) → 2MnO2 (s) + 2H2O (l)
MnO2 (s) + 2I− (aq) + 4H+ (aq) → Mn2+ (aq) + I2 (aq) + 2H2O (l)
2S2O32− (aq) + I2 (aq) → 2I− (aq) + S4O62− (aq)
A student analysed two 300.0 cm3 samples of water taken from the school pond: one immediately (day 0), and the other after leaving it sealed in a dark cupboard for five days (day 5). The following results were obtained for the titration of the samples with 0.0100 mol dm−3 Na2S2O3 (aq).
Determine the mole ratio of S2O32− to O2, using the balanced equations.
[1]
4 : 1 ✔
Calculate the number of moles of oxygen in the day 0 sample.
[2]
✔
« ✔
NOTE: Award [2] for correct final answer.
The day 5 sample contained 5.03 × 10−5 moles of oxygen.
Determine the 5-day biochemical oxygen demand of the pond, in mg dm−3 (“parts per million”, ppm).
[2]
«difference in moles per dm3 = (6.45 × 10−5 − 5.03 × 10−5) × =»
4.73 × 10−5 «mol dm−3» ✔
«convert to mg per dm3: 4.73 × 10−5 mol dm−3 × 32.00 g mol−1 × 1000 mg g–1 = » 1.51 «ppm/mg dm−3» ✔
NOTE: Award [2] for correct final answer.
Calculate the percentage uncertainty of the day 5 titre.
[1]
« «%»✔
Suggest a modification to the procedure that would make the results more reliable.
[1]
repetition / take several samples «and average» ✔