After an algal bloom in the Hawkesbury River, the key question is not just “what grew?” but “how much oxygen is left?” Chemists answer that by measuring dissolved oxygen directly and by asking how much oxygen microbes will consume as they break down the organic load.
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A river sample is taken after a bloom. The water looks green and murky, and dead fish are reported downstream. A scientist says, “The dissolved oxygen reading matters now, but the BOD value may matter even more over the next few days.”
📚 Core Content
Dissolved oxygen is one of the most urgent water-quality parameters because aquatic ecosystems can fail quickly when it falls too low.
Dissolved oxygen (DO) is the concentration of oxygen gas dissolved in water. Fish, many invertebrates and aerobic microbes depend on it. If DO falls too low, aquatic organisms experience stress or die.
DO can decrease because of higher temperature, reduced mixing, or increased biological consumption of oxygen. This is why a low DO reading is often a warning sign that the system is under chemical or biological stress.
Wrong: Heavy metals are dangerous because they are radioactive.
Right: Heavy metals are toxic because they bioaccumulate and interfere with enzyme function, not because of radioactivity.
A DO meter gives speed. The Winkler method gives stoichiometric chemistry. Both aim to answer the same question: how much oxygen is dissolved in the sample right now?
A DO meter is a convenient field instrument that provides a rapid dissolved oxygen reading. The Winkler titration is a classical chemical method that converts the dissolved oxygen in the sample into a titre value through a chain of redox reactions.
The Winkler method is especially useful in teaching because it shows that water-quality monitoring is not only about sensors. It is also about chemical transformations that can be followed quantitatively.
The Winkler method does not titrate dissolved oxygen directly. Oxygen is first trapped chemically, then converted into iodine, and finally linked to a thiosulfate titre through stoichiometry.
The Winkler method works because dissolved oxygen is converted into a measurable amount of iodine, and that iodine is then titrated with sodium thiosulfate.
Dissolved oxygen gives a snapshot. BOD gives a forecast of oxygen stress caused by biodegradable organic matter.
Biochemical Oxygen Demand, usually measured as BOD5, is the amount of dissolved oxygen consumed by microorganisms as they decompose organic matter over five days at 20°C in the dark.
The danger of eutrophication is not only the algal bloom itself, but what happens after that bloom begins to die.
This is why BOD and DO together are so useful: one shows how much oxygen is available, and the other shows how strongly the system is likely to consume it.
📊 Data Interpretation
| Sample | Initial DO / mg L-1 | Final DO after 5 days / mg L-1 | BOD5 / mg L-1 |
|---|---|---|---|
| Site A | 8.8 | 7.4 | 1.4 |
| Site B | 7.2 | 3.6 | 3.6 |
| Site C | 6.5 | − | 9.1 |
Site A suggests relatively clean water. Site B shows moderate pollution pressure. Site C has a BOD above 8 mg L-1, indicating heavy pollution and serious oxygen demand.
✏️ Worked Examples
Given: A 100.0 mL water sample requires 8.00 mL of 0.0100 mol L-1 Na2S2O3(aq) in the Winkler titration.
Find: Dissolved oxygen concentration in mg L-1.
Method: First find moles of thiosulfate.
n(Na2S2O3) = cV = 0.0100 × 0.00800 = 8.00 × 10-5 molUse the Winkler ratio 1 mol O2 : 4 mol Na2S2O3.
n(O2) = (8.00 × 10-5) / 4 = 2.00 × 10-5 molConvert oxygen moles to mass in the 100.0 mL sample.
m(O2) = nM = 2.00 × 10-5 × 32.00 = 6.40 × 10-4 g = 0.640 mg in 100.0 mLScale to 1.00 L.
DO = 0.640 × 10 = 6.40 mg L-1Answer: The dissolved oxygen concentration is 6.40 mg L-1.
Given: Initial DO = 8.4 mg L-1; final DO after 5 days at 20°C in the dark = 4.9 mg L-1.
Find: BOD5 and pollution interpretation.
A BOD of 3.5 mg L-1 lies in the moderate-pollution range.
Answer: BOD5 = 3.5 mg L-1, indicating moderate pollution.
🧠 Activities
1 A 100.0 mL sample requires 6.50 mL of 0.0200 mol L-1 Na2S2O3(aq). Calculate the dissolved oxygen concentration.
2 Explain why the thiosulfate titre can be used to determine oxygen even though thiosulfate never reacts directly with dissolved oxygen in the final step.
1 A river sample has initial DO 9.0 mg L-1 and final DO 7.6 mg L-1 after 5 days. Calculate BOD5 and classify the water.
2 A second sample has BOD5 = 9.5 mg L-1. What does this suggest about pollution level and likely ecological stress?
3 Explain eutrophication as a sequence of linked events starting from excess nutrients and ending in fish kill.
1. What does dissolved oxygen measure?
What is NOT does dissolved oxygen measure?
2. In the Winkler method, sodium thiosulfate titrates:
3. A water sample has initial DO 8.0 mg L-1 and final DO after 5 days 5.5 mg L-1. What is the BOD5?
4. What does a BOD5 value above 8 mg L-1 indicate?
What is NOT does a BOD 5 value above 8 mg L -1 indicate?
5. Which sequence best describes eutrophication?
1. Explain the sequence of reactions in the Winkler titration and how the final thiosulfate titre is linked to dissolved oxygen. 4 marks
2. Explain the difference between dissolved oxygen and BOD5, and why both are useful when assessing river health after an algal bloom. 4 marks
3. Evaluate the usefulness of BOD5 as an indicator of pollution in the Hawkesbury River after an algal bloom. In your answer, refer to what BOD5 reveals, one limitation of using it alone, and how it should be interpreted with other data. 5 marks
Return to the Hawkesbury River case and refine your first answer using proper analytical language.
n(thiosulfate) = 0.0100 × 0.01000 = 1.00 × 10-4 mol.
n(O2) = (1.00 × 10-4) / 4 = 2.50 × 10-5 mol.
m(O2) = 2.50 × 10-5 × 32.00 = 8.00 × 10-4 g = 0.800 mg in 100.0 mL.
DO = 8.00 mg L-1.
1. n(thiosulfate) = 0.0200 × 0.00650 = 1.30 × 10-4 mol. n(O2) = 1.30 × 10-4 / 4 = 3.25 × 10-5 mol. m(O2) = 3.25 × 10-5 × 32.00 = 1.04 × 10-3 g = 1.04 mg in 100.0 mL. Therefore DO = 10.4 mg L-1.
2. The thiosulfate is linked to oxygen because it titrates iodine, and the amount of iodine produced is stoichiometrically related to the original dissolved oxygen through the Winkler redox chain.
1. BOD5 = 9.0 − 7.6 = 1.4 mg L-1, indicating clean water.
2. A BOD5 of 9.5 mg L-1 suggests heavy pollution and strong oxygen demand, meaning serious ecological stress is likely.
3. Excess nutrients promote algal bloom growth. As algae and plants die, microbes decompose the organic matter, raising oxygen demand and lowering dissolved oxygen. This can lead to hypoxia and fish kill.
1. A — dissolved oxygen measures oxygen available in the water.
2. D — thiosulfate titrates iodine, not oxygen directly.
3. B — BOD5 = 8.0 − 5.5 = 2.5 mg L-1.
4. C — values above 8 mg L-1 indicate heavy pollution.
5. A — that is the correct eutrophication sequence.
Q1 (4 marks): In the Winkler titration, dissolved oxygen first oxidises Mn2+ to MnO2. The oxidised manganese species then oxidises iodide ions to iodine. That iodine is titrated with sodium thiosulfate. Because the amount of iodine produced depends on the oxygen originally present, and the thiosulfate titre measures the iodine, the final titre can be converted stoichiometrically to dissolved oxygen. The key HSC ratio is 1 mol O2 to 4 mol Na2S2O3.
Q2 (4 marks): Dissolved oxygen measures how much oxygen is currently dissolved in the water and available for aquatic life. BOD5 measures how much oxygen microorganisms consume over five days while decomposing organic matter. After an algal bloom, DO shows the immediate oxygen status of the river, while BOD5 indicates how strongly the system is likely to keep consuming oxygen. Together, they give both a snapshot and a forecast of oxygen stress.
Q3 (5 marks): BOD5 is useful because it indicates how much biodegradable organic matter is present by measuring the oxygen consumed during microbial decomposition. After an algal bloom, this is valuable because dead algae and plant matter can drive strong oxygen demand and threaten aquatic life. A limitation is that BOD5 alone does not show the current dissolved oxygen level at the time of sampling, so it cannot by itself describe immediate river condition. It should therefore be interpreted alongside DO and other water-quality data. Overall, BOD5 is a strong pollution indicator, but it is most useful as part of a broader assessment rather than as a stand-alone measure.
Sprint through questions on monitoring dissolved oxygen and biochemical oxygen demand. Pool: lessons 1–7.
Tick when you've finished the activities and checked your answers.