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# Determination of DO &amp; BOD in a water sample

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Introduction

Determination of DO & BOD in a water sample Objective Using volumetric analysis, the amount of dissolved oxygen (DO) in a water sample can be found. On standing another sample for 5 days, the difference between these two values is defined as biochemical oxygen demand in 5 days (BOD5). The aquatic life needs certain amount of DO for survival, whereas the value of BOD reflects the water quality. Principle In alkaline solution, dissolved oxygen will oxidize manganese(II) into manganese(III): 4Mn2+(aq) + 8OH-(aq) + O2(aq) + 2H2O(l) � 4Mn(OH)3(s) The amount of DO can be found by titrating the iodine produced from potassium iodide by manganese(III) with sodium thiosulphate: 2Mn(OH)3(s) + 2I-(aq) + 6H+(aq) � 2Mn2+(aq) + I2(aq) + 6H2O(l) I2(aq) + 2S2O32-(aq) � 2I-(aq) + S4O62-(aq) Chemicals water sample, MnSO4, conc.H2SO4, alkaline KI, standard Na2S2O3, starch solution Apparatus titration apparatus, pipette, 250cm3 volumetric flask x2, beaker, droppers, magnetic stirrer, white tile Procedure 1.> Collect a water sample with two 250cm3 volumetric flasks. ...read more.

Middle

= 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 (c) KCl http://hk.geocities.com/fatherofchemistry Precaution 1.> At all stages, every method should be made to assure that oxygen is neither introduced to nor lost from the water sample. ...read more.

Conclusion

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. ...read more.

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