Chemical Oxygen D emand. O 2 1 Litre H 2O . . FAS (Ferrous Ammonium Sulphate). The excess Potassium Dichromate is back titrated against a Std.PRINCIPLE Std. Potassium dichromate acidified with Sulphuric acid is added in excess to a known volume of effluent(waste water) sample. ions) .25 N K2Cr2O7 30 cm3 of 6 N H2SO4 1g of Ag2SO4 (Catalyzes the oxidation of Organic matter) 1g of HgSO4 (Mercuric Sulphate prevents the interference of Cl.PROCEDURE V ml of waste water sample 10 cm3 of 0. . FAS using Ferroin as an indicator (Titre value = V1 ml).• Excess of K2Cr2O7 is titrated against std. • End point: Bluish Green to Reddish brown • Blank titration is performed taking water (NOT the waste water)( Titre value = V2 ml). Sample titre value. a .Normality of FAS.Blank titre value.CALCULATION Volume of K2Cr2O7 required for the sample = V2 .V1 ml COD of sample = N x (b – a) x 8000 mg L-1 V N . b . . V – Volume of Waste water sample. COD of sample = 0. 28.PROBLEMS 1.0) x 8000 mg L-1 25 = 225.6 mg dm-3 .05 N a = 14.05 x (28. Data Given: b = 28. The volume of test sample sample used is 25 cm3.1 – 14.In a COD test.1 cm3 N = 0.1 cm3 and 14.0 cm3 V = 25 cm3 SOLUTION: COD of sample = N x (b – a) x 8000 mg L-1 V Substituting the values. Calculate the COD of the sample solution.05 N FAS solution were required for blank and sample titration respectively.0 cm3 of 0. 93 mg of O2 .2) 25 cm3 of an effluent sample requires for oxidation 8.3 cm3 of 0.398 mg of O2 b)Calculation of COD 25 cm3 of effluent = 0.3 cm3 (0.001M solution of K2Cr2O7 = 48 x 0. COD of the sample = 15.398 mg x 1000 cm3 = 15. Data given: V = 25 cm3 K2Cr2O7 = 8.001 M ) Solution: a)To find the Oxygen equivalent 1000 cm3 of 1M K2Cr2O7 48g of O2 8.3 cm3 1000 cm3 =0.3 cm3 of 0.001 x 8.001 M K2Cr2O7. Calculate the COD of the effluent samples.398 mg of O2 1000 cm3 of effluent = 0.93 mg of O2 25 cm3 ≅ Therefore. 25 x 45 = 11.25N 25 Weight of Oxalic Acid in 1 dm3 = 0.5 N) Eq mass of Oxalic Acid = 45 Solution: (N x V) of effluent = (N x V) of K2Cr2O7 N of effluent = (N x V) of K2Cr2O7 V of effluent = 0. calculate the amount of oxalic acid present in 1 dm3(Given equivalent mass of Oxalic acid = 45) Data Given: V=25 cm3 K2Cr2O7 = 12.5 = 0.5 x 12.3) 25 cm3 of an industrial effluent requires 12.5 cm3 0. Calculate the COD of the sample.25 g . Assuming that the effluent contains only oxalic acid.5 cm3 (0.5N K2Cr2O7 for complete oxidation. carried out to remove heavy metal ions. remaining organic compounds and colloidal impurities .phosphates.SEWAGE TREATMENT Three stages of the treatment Primary Treatment – involves physical and chemical methods Secondary Treatment – involves biological methods Tertiary Treatment. . Thank You .