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Date November 2019 Marks available 2 Reference code 19N.2.sl.TZ0.2
Level SL Paper 2 Time zone TZ0
Command term Determine Question number 2 Adapted from N/A

Question

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]
a(i).

Calculate the number of moles of oxygen in the day 0 sample.

[2]
a(ii).

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]
a(iii).

Calculate the percentage uncertainty of the day 5 titre.

[1]
b(i).

Suggest a modification to the procedure that would make the results more reliable.

[1]
b(ii).

Markscheme

4 : 1 ✔

a(i).

ns2o32=«0.0258 dm3×0.010 mol dm3=»2.58×104«mol» ✔

«2.58×104mol4=»6.45×105«mol» ✔

NOTE: Award [2] for correct final answer.

a(ii).

«difference in moles per dm3 = (6.45 × 10−5 − 5.03 × 10−5) × 1000 300.0

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.

a(iii).

«100×0.1cm320.1cm3=»0.5 «%»✔

b(i).

repetition / take several samples «and average» ✔

b(ii).

Examiners report

[N/A]
a(i).
[N/A]
a(ii).
[N/A]
a(iii).
[N/A]
b(i).
[N/A]
b(ii).

Syllabus sections

Core » Topic 1: Stoichiometric relationships » 1.3 Reacting masses and volumes
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