Determination of dissolved oxygen in a water sample

3.> Similarly, introduce about 1 cm3 of alkaline KI solution to the same flask. Be sure that no air becomes entrapped and then invert the stopped

flask to distribute the precipitate uniformly.

4.> When the precipitates has settled at least 2 cm below the stopper, introduce about 1 cm3 of conc.H2SO4 well below the surface. Stop the flask and

using a magnetic stirrer to mix the solution until the precipitate dissolves.

5.> Measure 200 cm3 of the acidified sample into a 500 cm3 conical flask, titrate with standard Na2S2O3 solution. Until the solution becomes pale

yellow, add a few drops of starch solution and complete the titration.

6.> While another volumetric flask containing the sample is stopped, without adding any chemicals, and keep in a cool, dark place (avoid the

photosynthesis of phytoplankton) for 5 days.

7.> After 5 days, investigates the sample by repeating the steps 1-5. The difference between 2 DO is defined as BOD5.

Data Analysis

Concentration of standard Na2S2O3 at the first day: 0.0149 M

Volume of 0.0149M Na2S2O3 used = 40.00 - 26.75 = 13.25 cm3

Amount of 0.0149M Na2S2O3 used = 13.25 x 0.0149 = 0.1974 mmol

Amount of I2 generated = 1/2 (0.1974) = 0.0987 mmol

Amount of Mn(OH)3 generated = 0.0987 x 2 = 0.1974 mmol

Amount of O2 present in 200 cm3 of aquarium water at room temperature

= 1/4 (0.1974) = 0.04935 mmol = 0.04935 x 32 = 1.5792 mg

Dissolved oxygen (DO0) in aquarium water in the first day at room temperature

= 1.5792 ÷ (200x10-3) = 7.896 mgdm-3

Concentration of standard Na2S2O3 after 5 days: 0.0127 M

Volume of 0.0127M Na2S2O3 used = 34.40 - 22.95 = 11.45 cm3

Amount of 0.0127M Na2S2O3 used = 11.45 x 0.0127 = 0.1454 mmol

Amount of I2 generated = 1/2 (0.1454) = 0.0727 mmol

Amount of Mn(OH)3 generated = 0.0727 x 2 = 0.1454 mmol

Amount of O2 present in 200 cm3 of aquarium water (stood for 5 days) at room temperature

= 1/4 (0.1454) = 0.03635 mmol = 0.03635 x 32 = 1.1632 mg

Dissolved oxygen (DO5) in aquarium water stood for 5 days at room temperature

= 1.1632 ÷ (200x10-3) = 5.816 mgdm-3

Biochemical oxygen demand for 5 days (BOD5) = DO5 - DO0 = 2.08mgdm-3

Precaution

1.> At all stages, every method should be made to assure that oxygen is neither introduced to nor lost from the water sample.

2.> Sodium thiosulphate needs to be standardized before use since it will be oxidized by atmospheric oxygen easily.

3.> Avoid exposure to overflow of the solution in volumetric flask, as the solution is quite alkaline.

4.> Starch solution should not be introduced too early, as the blue-black precipitate formed between starch and I2 at high concentration is hard to

decolourize.

5.> Handle conc.H2SO4 with great care, if possible, wear gloves.

6.> Keep the set-up in dark before measuring DO5 to avoid photosynthesis taken place by microscopic algae in water which will produce oxygen.

Possible Error

1.> Oxygen is introduced or lost from the sample, e.g. air bubbles inside volumetric flask.

2.> Chlorinated water sample will oxidize I- to I2, thus more I2 is formed than expected.

Discussion

1.> Oxygen dissolved in water is necessary for aquatic life. Dissolved oxygen can be used as an indicator of oxygen content in water. It's measured

in milligrams of oxygen per litre of water (mgdm-3 / ppm). A minimal amount of DO is needed for survival of aquatic life and reducing odour

smell due to anaerobic degradation of organic matter. Biological degradation of waste matter can cause depletion of oxygen whereas aeration can

replenish oxygen.

2.> In the experiment, DO0 is 7.90 mgdm-3. This value is higher than the minimum requirement for fishes (4-6 mgdm-3). As the water used is

aquarium water, oxygen is always replenished by air pump and the DO is enough for fishes.

3.> Biochemical oxygen demand (BOD) is the amount of oxygen required to break down the organic matter present in a water sample. It can be used

as an indicator of water pollution. If the water sample has a high BOD, it indicates that a lot of organic waste is present and a lot of oxygen is

needed to break down the waste. Natural clean water has a BOD of about 1-4 mgdm-3 only. A low BOD indicates only small amount of organic

matter is present but there is still a little organic pollution. If BOD is high and if the oxygen depleted is greater than the oxygen replenished, then

some fishes may die, organic debris accumulates and anaerobic microorganisms begin to multiply, producing unpleasant taste and smell.

4.> Since air pump is applied to the water sample continuously, the DO is slightly higher than other water samples.

Conclusion

The value of DO and BOD can be used as an indicator of water pollution and the quality of the living environment of aquatic life.