Question 1. A sample of river water was acquired as an attempt to characterize t

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Question 1. A sample of river water was acquired as an attempt to characterize the health of the water source. Part of this characterization was to determine the biological oxygen demand (BOD) of the river water with precision. The measurement was repeated four time using four different dilutions to determine an accurate BODs of the river water. All four BODs had a total volume of 300 mL and had the same initial dissolved oxygen (DO) concentration of 8.2 mg O2/L The dilution water was non-seeded distilled water. No seed was added as it was assumed there would be an adequate microbial population in the sample water. (a) Based on the measurement results below, determine which of the tests enable one to determine an accurate BODs measurement of the river: Sample A-300 mL of sample water was used and a final DO concentration of 0.06 mg O2/L was measured; Sample B-10 mL of sample water was used and a final DO concentration of 7.6 mg /L was measured; Sample C-100 mL of sample water was used and a final DO concentration of 3.1 mg O2/L was measured; Sample D - 150 mL of sample water was used and a final DO concentration of 0.74 mg Oa/L was measured Blank-300 mL of dilution water was used and a final DO concentration of 8.2 mg Oa/L was measured. (b) Determine the BODs of the river water. (c) What are two storage based factors that could invalidate the BOD test. Why do they invalidate the test?

Explanation / Answer

For question a

Initial D.O of blank or distilled water is 8.2 Mg/l

Also it is a general practise to incubate at least two bottle containing sample and diluted to 300 ml total volume and a blank to calculate bod ofsample. BOD residual of 1 Mg/ l is required for accurate measurement and Dilution factor greater than equal to 3 is required thus only two samples that qualify the criteria are:

So sample B and C clubbed with blank would give an accurate bod 5 of the river water.

Question b

BOD 5 of river water = ( DO initial - DO final) * Dilution factor

Dilution factor = Final volume of diluted sample/ volume of sample

For 10 ml sample water diluted to 300 ml Dilution factor = 30

Thus BOD 5 = (8.2-7.6) * 30 = 18 Mg/l

For 100 ml sample water diluted to 300 ml Dilution factor = 3

Thus BOD 5 = ( 8.2-3.1)* 3 = 15.3 Mg/l

Average BOD 5 = (18+15.3)/2 = 16.65 Mg/l which is BOD 5 of river water.

For question C

Sample bottles containing sample shall be stored at temperature of 4 degree Celsius to ensure that no biological activity takes place which might change the initial dissolved oxygen level of the sample prior to testing thereby affecting the accuracy of test.

Also the sample bottles should be airtight so that no alteration in dissolved oxygen level of the sample is allowed due to atmospheric exposure.

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