All questions and answers about the DO measurement method for water
Dissolved Oxygen (DO) of Water and Its Measurement Method
DO of Water:
The amount of dissolved oxygen present in a water sample to make it saturated with oxygen is called the DO of that water. At 15°C, the DO value in oxygen-saturated water is 10 mg/L or 10 ppm [ppm = parts per million; incidentally mentioned, 1 picometer = 10-12 m]. At a temperature of 20°C, the DO value in oxygen-saturated water becomes 9.2 ppm.
Determination of DO Value:
The DO value can be directly measured by dipping an oxygen sensor-equipped probe or electrode into the water and reading the meter. Besides this, the DO of water is determined in the laboratory using Winkler’s iodometric method. In normal surface water, the DO value should be around 5 mg/L or higher, and in river estuaries, the DO value remains greater than 6 mg/L.
Principle and Working Procedure of DO Determination
Principle of DO Determination:
An estimation of water purity can be obtained by determining its dissolved oxygen. Dissolved oxygen is the basis for measuring Biochemical Oxygen Demand (BOD), which is used to determine the pollution level of wastewater. Measuring dissolved oxygen is essential to maintain the aerobic characteristics of recipient water, sewage, and industrial wastewater. Dissolved oxygen can be measured via iodometry using the modified Winkler’s method (Winkler’s method).
Procedure (Method):
i. Manganese sulfate is added to a water sample containing alkaline potassium iodide. This produces manganese hydroxide and potassium sulfate.
MnSO4 + 2KOH → Mn(OH)2↓ + K2SO4
ii. Mn(OH)2 is oxidized by the dissolved oxygen in the water to produce alkaline manganese oxide (or manganese oxyhydroxide).
2Mn(OH)2 + O2 (dissolved oxygen) → 2MnO(OH)2↓ [alkaline manganese oxide]
iii. Adding H2SO4 to alkaline manganese oxide produces manganese sulfate:
MnO(OH)2 + 2H2SO4 → Mn(SO4)2 + 3H2O
iv. The produced Mn(SO4)2 liberates iodine from potassium iodide.
Mn(SO4)2 + 2KI → MnSO4 + K2SO4 + I2
v. The liberated or free iodine is titrated with standard sodium thiosulfate solution using starch as an indicator.
2Na2S2O3 + I2 → Na2S4O6 (sodium tetrathionate) + 2NaI
Incidentally, the amount of liberated iodine is chemically equivalent to the amount of oxygen present in the sample. Dissolved oxygen is usually expressed in mg/L (ppm) units.
Calculation of DO:
According to the above chemical equations, the equivalence relationship is as follows—
2 mol Na2S2O3 ≡ 1 mol I2 ≡ 1 mol Mn(SO4)2 ≡ 1 mol MnO(OH)2 ≡ ½ mol O2
Therefore, 2 mol Na2S2O3 ≡ ½ mol O2
Or, 1 mol Na2S2O3 ≡ ¼ mol O2 ≡ ¼ mol DO
Therefore, 2 mol Na2S2O3 ≡ ½ mol O2
Or, 1 mol Na2S2O3 ≡ ¼ mol O2 ≡ ¼ mol DO
Importance of DO in Water and Its Environmental Impact
(1) The decomposition of organic matter in river and pond water decreases its DO; as a result, aerobic (aerobic) aquatic animals and fishes die due to the lack of oxygen. However, anaerobic (anaerobic) aquatic plants and certain harmful bacteria grow rapidly in that water, and fishes die from bacterial infections.
(2) If the DO level in water is low, incomplete oxidation of biodegradable organic matter occurs. Consequently, harmful compounds like methane (CH4), hydrogen sulfide (H2S), phosphine (PH3), and amines are produced in the water. Then that polluted water spreads a severe foul odor in the surroundings. Sewage, feces, industrial liquid waste, and biodegradable wastes primarily reduce the DO level of water.
(2) If the DO level in water is low, incomplete oxidation of biodegradable organic matter occurs. Consequently, harmful compounds like methane (CH4), hydrogen sulfide (H2S), phosphine (PH3), and amines are produced in the water. Then that polluted water spreads a severe foul odor in the surroundings. Sewage, feces, industrial liquid waste, and biodegradable wastes primarily reduce the DO level of water.
* Special Warning: If the DO level of surface water falls below 5 ppm, aerobic aquatic animals will decrease, and anaerobic plants and harmful microorganisms will increase. The net result of this is that the natural balance of the aquatic environment will be completely destroyed, and toxic gases and unhealthy odors will spread all around.
Q&A Bank on Dissolved Oxygen (DO) of Water
A) Knowledge-Based Questions & Answers
Question : What is meant by DO of water?
Answer : DO refers to the total amount of dissolved oxygen present in a water sample to make it saturated.
Question : What is the international unit of DO?
Answer : The common unit for measuring DO is mg/L or ppm (parts per million).
Question : Which method is used to determine DO in the laboratory via iodometry?
Answer : The modified Winkler’s method is used to determine the DO of water through iodometry.
Question : What should be the ideal DO value in surface water at a river estuary?
Answer : The DO value in water at a river estuary should ideally be greater than 6 mg/L.
Question : What is meant by 1 ppm?
Answer : 1 ppm means one part of solute by weight in one million parts of solution by weight. Mathematically, 1 ppm = 1 mg/L.
B) Analytical Questions & Answers
Question : How does the DO value change with temperature and why?
Answer : As temperature increases, the solubility of oxygen in water decreases, resulting in a lower DO value. For example, at 15°C, the DO in oxygen-saturated water is 10 ppm, but at 20°C, it drops to 9.2 ppm.
Question : What happens if the DO value in water falls below 5 ppm?
Answer : If the DO value falls below 5 ppm, the natural balance of the aquatic ecosystem is disrupted. Aerobic aquatic animals and fish die due to oxygen deficiency, while anaerobic microorganisms multiply, making the water toxic and foul-smelling.
Question : Explain the cause of foul odor in polluted water.
Answer : In water heavily contaminated with organic waste and low DO, anaerobic decomposition occurs. This produces toxic gases like methane (CH4), hydrogen sulfide (H2S), phosphine (PH3), and amine compounds, which cause a severe foul odor.
Question : Show the reaction between iodine and sodium thiosulfate in Winkler’s method with an equation.
Answer : In Winkler’s method, free iodine is titrated with standard sodium thiosulfate solution using starch as an indicator. The reaction is:
2Na2S2O3 + I2 → Na2S4O6 (sodium tetrathionate) + 2NaI
C) Multiple Choice Questions (MCQ)
1. What is the DO value of oxygen-saturated water at 15°C?
Answer: d) 10.0 ppm
2. What is the amount of dissolved oxygen (DO) at 20°C?
Answer: b) 9.2 mg/L
3. What should be the minimum DO value in normal surface water?
Answer: c) 5 ppm
4. Which indicator is used in Winkler’s method for titration?
Answer: c) Starch solution
5. What amount of Na₂S₂O₃ is equivalent to 1 mole of O₂ gas?
Answer: c) 4 mol
6. Which is the modern device for direct DO measurement?
Answer: b) Oxygen sensor probe
7. Which gas is produced in anaerobic decomposition causing a foul odor?
Answer: b) H₂S
