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Date November 2021 Marks available 1 Reference code 21N.2.sl.TZ0.6
Level SL Paper 2 Time zone TZ0
Command term Deduce Question number 6 Adapted from N/A

Question

Biochemical oxygen demand (BOD) can be determined by the Winkler Method.

A 25.00 cm3 sample of water was treated according to the Winkler Method.

Step I:   2Mn2+ (aq) + O2 (g) + 4OH− (aq) → 2MnO2 (s) + 2H2O (l)

Step II:  MnO2 (s) + 2I− (aq) + 4H+ (aq) → Mn2+ (aq) + I2 (aq) + 2H2O (l)

Step III: 2S2O32− (aq) + I2 (aq) → 2I− (aq) + S4O62− (aq)

The iodine produced was titrated with 37.50 cm3 of 5.000 × 10−4 mol dm−3 Na2S2O3.

Outline what is measured by BOD.

[1]
a.

A student dissolved 0.1240 ± 0.0001 g of Na2S2O3 to make 1000.0 ± 0.4 cm3 of solution to use in the Winkler Method.

Determine the percentage uncertainty in the molar concentration.

[2]
b.

Calculate the amount, in moles of Na2S2O3 used in the titration.

[1]
c(i).

Deduce the mole ratio of O2 consumed in step I to S2O32− used in step III.

[1]
c(ii).

Calculate the concentration of dissolved oxygen, in mol dm−3, in the sample.

[2]
c(iii).

The three steps of the Winkler Method are redox reactions.

Deduce the reduction half-equation for step II.

[1]
c(iv).

Markscheme

«amount of» oxygen used to decompose the organic matter in water ✔

a.

«0.0001 g0.1240 g×100%=» 0.08 «%»
OR
«0.4 cm31000.0 cm3×100%=» 0.04 «%» ✔


«0.08 % + 0.04 % =» 0.12/0.1 «%» ✔


Award [2] for correct final answer.

Accept fractional uncertainties for M1, i.e., 0.0008 OR 0.0004.

b.

«37.50 cm31000× 5.000 × 10−4 mol dm−3 =» 1.875 × 10−5 «mol» ✔

c(i).

1:4 ✔

Accept “4 mol S2O32– :1 mol O2“, but not just 4:1.

c(ii).

«1.875×10-5mol×14=» 4.688 × 10−6 «mol» ✔

«4.688×10-6 mol25.00 cm31000=» 1.875 × 10−4 «mol dm−3» ✔


Award [2] for correct final answer.

c(iii).

MnO2 (s) + 2e + 4H+ (aq) → Mn2+ (aq) + 2H2O (l) ✔

c(iv).

Examiners report

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

Syllabus sections

Core » Topic 9: Redox processes » 9.1 Oxidation and reduction
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