Your task is not to seek for love, but merely to seek and find all the barriers within yourself that
Idea Transcript
Module 5: Water technology 2015 Problems 1) In a COD experiment 30 cm3 of an effluent sample required 9.8 cm3 of 0.001 M potassium dichromate for oxidation. Calculate the COD of the sample. i) 1000 cm3 of 1M K2Cr2O7 solution contains 294 g of K2Cr2O7
9.8 cm3 of 0.001 M K2Cr2O7 solution contains 294 X 9.8 X 0.001 1000 X 1 = 2.8812 mg ii) K2Cr2O7 + H2SO4 K2SO4 + Cr2 (SO4)3 + 4H2O + 3[O] 294 g of K2Cr2O7 = 48 g of oxygen. 2.8812 mg of K2Cr2O7 =
48 X 2.8812 = 0.4704 mg of oxygen 294 iii) COD of the effluent sample = 0.4704 X 1000 30 = 15.68 mg of oxygen / dm3 2) In a COD experiment, 25 cm3 of an effluent sample required 8.3 cm3 of 0.001 M potassium dichromate for oxidation. Calculate the COD of the sample. i) 1000 cm3 of 1M K2Cr2O7 solution contains 294 g of K2Cr2O7 8.3 cm3 of 0.001 M K2Cr2O7 solution contains 294 X 8.3 X 0.001 1000 X 1 = 0.3983 mg
ii) K2Cr2O7 + H2SO4
H2SO4 + Cr2 (SO4)3 + 4H2O + 3[O]
294 g of K2Cr2O7 = 48 g of oxygen. 0.3983 mg of K2Cr2O7 =
48 X 0.3983 = 0.4704 mg of oxygen 294 iii) COD of the effluent sample = 0.4704 X 1000 25 = 15.93 mg of oxygen /dm3
Module 5: Water technology 2015 3.
Calculate the COD of the effluent sample when 25 ml of an effluent requires 8.3 ml of 0.001M K2Cr2O7 for oxidation. [Given molecular mass of K2Cr2O7 =294).
Solution: Given Concentration of K2 Cr2O7 =0.001M Molecular mass of K2Cr2O7 =294 Volume of the effluent sample =25 ml Volume of the K2 Cr2 O7 consumed by the effluent =8.3ml i)_1000ml of 1M K2Cr2O7 =294 g 8.3 ml of 0.001M K2Cr2O7 = (294×8.3×0.001)/1000 Amount of K2Cr2O7 present =2.4402mg (ii)_1mol of K2Cr 2O7 = 6 equivalents of oxygen i.e., 294 mg of K2Cr2O7 6×8 mg of oxygen 6 8 2.4402 2.4402 mg of K2Cr2O7 =0.3984 mg 294 (iii)_ COD in 25 ml of water =0.3984 mg 1000ml of water =398.4/25=15.92 mg COD of water=15.92 mg/dm3