Chemistry · Year 12 · Module 8 · Lesson 7
HSC Exam Practice
Monitoring Dissolved Oxygen & BOD
Short answer
1.Short-answer questions
Define dissolved oxygen and state the unit in which it is typically reported in water-quality monitoring.
Define biochemical oxygen demand (BOD5) and state the standard conditions under which it is measured.
Distinguish between dissolved oxygen and BOD5 as indicators of river health.
Describe the sequence of reactions in the Winkler titration that links dissolved oxygen to the volume of sodium thiosulfate used.
Explain why dissolved oxygen saturation decreases as water temperature increases, using particle theory and solubility principles.
Outline the eutrophication process and explain why it causes dissolved oxygen to fall even after an algal bloom initially appears to have stabilised.
Data response
2.Data response — DO vs temperature (E2)
The figure below shows dissolved oxygen saturation (mg L−1) versus temperature (°C) for freshwater at standard pressure.
- Describe the relationship between temperature and dissolved oxygen shown in the graph. Refer to at least one specific data value. (2 marks)
- Estimate the dissolved oxygen saturation at 20°C. (1 mark)
- Identify from the graph above what temperature the dissolved oxygen drops below the 6 mg L−1 healthy-river threshold. Explain the chemical reason for this behaviour. (3 marks)
3.Multi-step calculation — Winkler titration (E4)
A 150.0 mL water sample collected from the Cooks River, Sydney, is analysed by the Winkler method. After the manganese and iodide steps are performed, the released iodine requires 9.60 mL of 0.0150 mol L−1 sodium thiosulfate to reach the colourless endpoint. A second portion of the same sample is incubated for 5 days at 20°C in the dark; the final dissolved oxygen is 4.2 mg L−1. (M(O2) = 32.00 g mol−1; ratio 1 mol O2 : 4 mol Na2S2O3)
- Calculate the moles of sodium thiosulfate used. (1 mark)
- Calculate the moles and mass of dissolved oxygen in the sample. (2 marks)
- Calculate the dissolved oxygen concentration in mg L−1. (2 marks)
- Calculate BOD5 for this sample and classify the water quality. (2 marks)
- State one limitation of using BOD5 alone to assess the ecological risk of this site. (1 mark)
Extended response
4.Extended response (E3)
Evaluate the use of dissolved oxygen (DO) measurement and BOD5 as complementary tools for assessing water quality in Australian river systems. In your response, refer to at least one named Australian example, the Winkler titration method, and the limitations of each measurement in isolation.
Chemistry · Year 12 · Module 8 · Lesson 7
Answer Key & Marking Guidelines
Section 1 · Short answer · 2 marks · Band 3
Sample response. Dissolved oxygen is the concentration of molecular oxygen (O2) dissolved in water; it is the form of oxygen available for respiration by aquatic organisms. It is reported in milligrams per litre (mg L−1) or as percentage saturation.
Marking notes. 1 mark for identifying dissolved O2 as oxygen gas dissolved in water and available for aquatic life. 1 mark for stating mg L−1 (accept % saturation).
Section 1 · Short answer · 2 marks · Band 3
Sample response. BOD5 is the amount of dissolved oxygen consumed by microorganisms decomposing organic matter in a water sample. It is measured over 5 days at 20°C in the dark using the formula BOD5 = initial DO − final DO.
Marking notes. 1 mark for “O2 consumed by microorganisms decomposing organic matter.” 1 mark for both standard conditions (5 days AND 20°C AND dark; accept 2 of 3).
Section 1 · Short answer · 3 marks · Band 3–4
Sample response. Dissolved oxygen measures how much oxygen is currently available in the water; it is a snapshot of the immediate ecological state. BOD5 measures how much oxygen will be consumed by microbial decomposition over five days; it is a forecast of oxygen demand from organic pollution. A river may have a currently acceptable DO but a dangerously high BOD, indicating that oxygen will fall in coming days as decomposition continues.
Marking notes. 1 mark for DO defined as current O2 concentration. 1 mark for BOD5 defined as O2 consumed by decomposition. 1 mark for explaining why both are needed (DO = snapshot; BOD = forecast/predictive).
Section 1 · Short answer · 4 marks · Band 4
Sample response. Step 1: Dissolved O2 oxidises Mn2+ to MnO2, trapping the oxygen chemically in the sample. Step 2: In acid solution, MnO2 oxidises iodide ions (I−) to iodine (I2); the amount of iodine produced is stoichiometrically proportional to the original dissolved oxygen. Step 3: The iodine is titrated with sodium thiosulfate (Na2S2O3) to the colourless endpoint. The stoichiometric ratio 1 mol O2 : 4 mol Na2S2O3 allows the original dissolved oxygen to be calculated from the titre volume.
Marking notes. 1 mark per step (3 steps) plus 1 mark for the 1:4 ratio or explicit link of titre to DO. Max 4.
Section 1 · Short answer · 3 marks · Band 4
Sample response. As temperature increases, water molecules have greater average kinetic energy. Gas molecules in solution can overcome the intermolecular forces holding them in the liquid phase and escape to the gas phase more readily. According to Henry’s Law, the solubility of a gas in a liquid decreases with increasing temperature. Therefore, at higher temperatures, fewer O2 molecules remain dissolved and DO saturation falls.
Marking notes. 1 mark for referencing increased kinetic energy at higher temperature. 1 mark for gas molecules escaping solution more readily / decreased solubility. 1 mark for applying to DO specifically (or invoking Henry’s Law).
Section 1 · Short answer · 4 marks · Band 4–5
Sample response. Eutrophication begins when excess nutrients (nitrates, phosphates) enter the waterway, stimulating rapid algal growth (bloom). The dense algal layer reduces light penetration, killing submerged plants. As algae die, aerobic bacteria decompose the accumulated biomass. This decomposition consumes dissolved oxygen, causing BOD to rise. Even after the visible bloom appears to have cleared, the large amount of dead organic matter continues to sustain intense microbial respiration over the following days, progressively depleting DO until hypoxic conditions develop and fish kills occur.
Marking notes. 1 mark for nutrients → algal bloom. 1 mark for algal/plant death → decomposition by bacteria. 1 mark for decomposition → rising BOD/consuming DO. 1 mark for explaining the delayed effect after bloom “appears stable” — ongoing decomposition continues to deplete DO.
Section 2 · Data response · 6 marks · Band 4–5
(a) Trend — 2 marks. As temperature increases from 0°C to 35°C, dissolved oxygen saturation decreases continuously (inverse / negative relationship). For example, at 0°C DO is approximately 14.6 mg L−1; by 35°C it has fallen to approximately 7.0 mg L−1. The rate of decrease is steepest at lower temperatures. 1 mark: identifies inverse relationship. 1 mark: quotes at least one specific data point from the graph.
(b) Estimate at 20°C — 1 mark. Approximately 9.1 mg L−1 (accept 8.8–9.4).
(c) Threshold temperature + reason — 3 marks. From the graph, DO saturation drops below 6 mg L−1 above approximately 33–35°C. Reason: higher temperatures give water molecules greater kinetic energy, reducing the ability of intermolecular forces to retain dissolved O2 molecules in solution. By Henry’s Law, gas solubility decreases with increasing temperature. 1 mark: identifies threshold correctly from graph (accept 32–35°C). 1 mark: kinetic energy / Henry’s Law explanation. 1 mark: links specifically to O2 escaping solution.
Section 2 · Multi-step calculation · 8 marks · Band 4–5
(a) 1 mark. n(Na2S2O3) = cV = 0.0150 × 0.00960 = 1.44 × 10−4 mol.
(b) 2 marks. n(O2) = 1.44 × 10−4 ÷ 4 = 3.60 × 10−5 mol (1 mark). m(O2) = 3.60 × 10−5 × 32.00 = 1.152 × 10−3 g = 1.152 mg (1 mark).
(c) 2 marks. DO = 1.152 mg in 150.0 mL = 1.152 × (1000/150) = 7.68 mg L−1 (accept 7.7). 1 mark for correct scaling method; 1 mark for correct answer.
(d) 2 marks. BOD5 = 7.68 − 4.2 = 3.48 mg L−1 (accept 3.5). 1 mark calculation; 1 mark classification: moderate pollution (2–8 mg L−1).
(e) 1 mark. Accept any one valid limitation: BOD5 does not show the current (initial) DO level at the time of sampling; it does not account for chemical oxygen demand from non-biodegradable toxins; it takes 5 days to obtain a result, making rapid real-time decisions difficult; it does not reveal the source of organic pollution.
Section 3 · Extended response · 8 marks · Band 5–6
- 1 mark Defines dissolved oxygen accurately (O2 dissolved in water; measured in mg L−1; available to aquatic life).
- 1 mark Defines BOD5 accurately (O2 consumed by microorganisms decomposing organic matter; 5 days, 20°C, dark).
- 1 mark Explains the Winkler titration as the chemical basis for DO measurement, including the 1 mol O2 : 4 mol Na2S2O3 ratio or the redox chain (Mn2+ → MnO2 → I2 → thiosulfate titre).
- 1 mark Uses a named Australian example correctly (e.g. Darling River / Murray–Darling fish kills 2018–2019; Hawkesbury River algal blooms; Sydney Water Cooks River monitoring; GBR AIMS turbidity monitoring).
- 1 mark Explains why DO alone is insufficient — it is a snapshot and does not predict future oxygen depletion from organic load.
- 1 mark Explains why BOD5 alone is insufficient — it takes 5 days, does not show current DO, and does not detect non-biodegradable chemical pollution.
- 1 mark Explains how the two measurements are complementary — DO gives immediate ecological status; BOD5 forecasts ongoing demand; together they reveal both present and future risk.
- 1 mark Reaches an evaluative judgement: neither measurement alone is sufficient for comprehensive water quality assessment; both are required for a complete picture, and ideally would be combined with other indicators (e.g. turbidity, nutrient levels, temperature).
Sample response (high-band). Dissolved oxygen (DO) is the concentration of O2 dissolved in water and available to aquatic organisms, measured in mg L−1. It is determined by the Winkler titration, in which dissolved O2 is trapped as MnO2, then converted to I2, which is titrated with sodium thiosulfate using the ratio 1 mol O2 : 4 mol Na2S2O3. A healthy freshwater river should maintain DO above 6 mg L−1; values below this indicate ecological stress. BOD5 measures the O2 consumed by microorganisms decomposing organic matter over 5 days at 20°C in the dark. Clean water has BOD5 below 2 mg L−1; values above 8 mg L−1 indicate heavy pollution. In the Murray–Darling fish kills of 2018–2019, initial DO readings were close to threshold (around 7–8 mg L−1) but BOD was high due to decomposing cyanobacterial biomass; subsequent DO plummeted to below 2 mg L−1, causing mass deaths of Murray cod and golden perch. This demonstrates the limitation of DO alone: a single snapshot can appear acceptable even when conditions will deteriorate sharply. Conversely, BOD5 alone does not reveal the current oxygen status, takes five days to determine and fails to detect chemical pollutants. Together, however, the two measurements are highly complementary: DO provides immediate status and BOD5 provides a forecast of oxygen demand. Used in combination with temperature and nutrient data, they give Australian river managers the information needed to anticipate and prevent ecological disaster.