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Date	November 2019	Marks available	1	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).

${{\text{n}}_{{{\text{s}}_2}{{\text{o}}_3}^{2 - }}} = «0.0258{\text{ d}}{{\text{m}}^3} \times 0.010{\text{ mol d}}{{\text{m}}^{ - 3}} =» 2.58 \times {10^{ - 4}}{\text{«mol»}}$ ✔