water - THE UNIVERSITY OF HONG KONG DEPARTMENT OF CIVIL...

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Unformatted text preview: THE UNIVERSITY OF HONG KONG DEPARTMENT OF CIVIL ENGINEERING PART 2 LAB REPORT ON WASTEWATER ANALYSIS NAME: HO CHING MAN ANITA UNIVERSITY NO: 2000055225 GROUP NO: 8A DATE OF SUBMISSION: 8-4-2002 Chemical Oxygen Demand Test (COD Test) Date of test: 18 th March, 2002 Objective: To determine the COD of given samples by Closed Reflux Method Reagents: 1. COD reagent (potassium dichromate: 0.04M) 2. Standard Ferrous ammonium sulfate, FAS Procedure: 1. Pipet 5 ml water sample and 10 ml COD reagent into culture tubes. 2. Tighten the tube cap and mix thoroughly. 3. Place the tube in preheated oven at 150 C for 2 hours. 4. After cooling to room temperature, determine COD. COD determination by titration: 1. Add about 2 drops of ferroin indicator 2. Titrate with standard FAS solution 3. The end point is color change from blue-green to reddish brown Molarity of FAS solution = (ml COD reagent / ml FAS used in titration) 3 0.03 COD as mg O 2 /L = (A-B) X M X 8000/ mL sample where A = mL FAS used for blank B = mL FAS for sample M = molarity of FAS Experimental result and Calculation Samples: Sample 1: Blank Sample 2: Treated wastewater Sample 3: Raw wastewater Standard dichromate solution Blank Treated wastewater Raw wastewater Volume of FAS used 3.7 mL 7.05 mL 6.7 mL 2.7 mL Molarity of FAS: M = (ml K 2 Cr 2 O 7 added to sample / ml FAS used in titration in standard) x 0.03 = 5 ml/ 3.7 ml x 0.03 = 0.0405 M COD of Treated wastewater: COD (mg O 2 /l) = (A-B) x (M / amount of waste water added) x 8000 mg O 2 /mole Fe 2+ = (7.05 ml 6.70 ml ) x (0.0405 M / 5 ml) x 8000 mg O 2 /mole Fe 2+ = 22.703 mg O 2 /l COD of Raw Wastewater: COD (mg O 2 /l) = (A-B) x (M / amount of waste water added) x 8000 mg O 2 /mole Fe 2+ = (7.05ml 2.70 ml ) x (0.0811 M / 5 ml) x 8000 mg O 2 /mole Fe 2+ = 282.162 mg O 2 /l Discussion: The chemical oxygen demand (COD) test is used to measure the pollution strength of domestic and industrial wastes. It allows measurement of waste in terms of the total quantity of oxygen required for oxidation to carbon dioxide and water in accordance with the following equation: 3 2 2 2 ) 2 3 2 ( ) 4 3 2 4 ( cNH O H c a nCO O c b a n N O H C c b a n +- + -- + + It is based upon the fact that most of the organic compounds can be oxidized by the action of strong oxidizing agents under acid conditions. Among all the oxidation of compounds in wastewater, three main types are involved. (A) Organics compounds- During oxidation, organic compounds change into carbon dioxide and water. Organics + H + CO 2 + H 2 O K 2 Cr 2 O 7 oxidize- Volatile organics like low-molecular weight fatty acid are present in the vapour space and do not come into contact with the oxidizing liquid. To oxidize them, a catalyst Ag + must be added to decompose them into CO 2 and H 2 O....
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water - THE UNIVERSITY OF HONG KONG DEPARTMENT OF CIVIL...

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