COD

hemicalxygenemand

WHAT is COD

2O

Organic and In

organic waste

1 Litre

H2O

PRINCIPLE Std. Potassium dichromate acidified with Sulphuric acid is added in excess to a known volume of effluent(waste water) sample.

The excess Potassium Dichromate is back titrated against a Std. FAS (Ferrous Ammonium Sulphate).

PROCEDURE V ml of waste water sample

10 cm3 of 0.25 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- ions)

• Excess of K2Cr2O7 is titrated against std. FAS using Ferroin as an indicator (Titre value = V1

ml).

• End point: Bluish Green to Reddish brown

• Blank titration is performed taking water (NOT the waste water)( Titre value = V2 ml).

CALCULATIONVolume of K2Cr2O7 required for the sample = V2 - V1 ml

COD of sample = N x (b – a) x 8000 mg L-1

V

N - Normality of FAS.b - Blank titre value.a - Sample titre value.V – Volume of Waste water sample.

PROBLEMS1.In a COD test, 28.1 cm3 and 14.0 cm3 of 0.05 N FAS solution were required for blank and sample titration respectively. The volume of test sample sample used is 25 cm3. Calculate the COD of the sample solution.

Data Given:b = 28.1 cm3 a = 14.0 cm3 N = 0.05 N V = 25 cm3

SOLUTION:COD of sample = N x (b – a) x 8000 mg L-1

VSubstituting the values, COD of sample = 0.05 x (28.1 – 14.0) x 8000 mg L-1 25 = 225.6 mg dm-3

2) 25 cm3 of an effluent sample requires for oxidation 8.3 cm3 of 0.001 M K2Cr2O7. Calculate the COD of the effluent samples.

Data given:V = 25 cm3

K2Cr2O7 = 8.3 cm3 (0.001 M )

Solution:•To find the Oxygen equivalent

1000 cm3 of 1M K2Cr2O7 ≅ 48g of O2

8.3 cm3 of 0.001M solution of K2Cr2O7 = 48 x 0.001 x 8.3 cm3

1000 cm3

=0.398 mg of O2

•Calculation of COD 25 cm3 of effluent = 0.398 mg of O2

1000 cm3 of effluent = 0.398 mg x 1000 cm3 = 15.93 mg of O2

25 cm3

Therefore, COD of the sample = 15.93 mg of O2

3) 25 cm3 of an industrial effluent requires 12.5 cm3 0.5N K2Cr2O7 for complete oxidation. Calculate the COD of the sample. Assuming that the effluent contains only oxalic acid, calculate the amount of oxalic acid present in 1 dm3(Given equivalent mass of Oxalic acid = 45)

Data Given:V=25 cm3K2Cr2O7 = 12.5 cm3 (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 = 0.5 x 12.5 = 0.25N

25

Weight of Oxalic Acid in 1 dm3 = 0.25 x 45 = 11.25 g

SEWAGE TREATMENT

Three stages of the treatment

Primary Treatment – involves physical and chemical methods

Secondary Treatment – involves biological methods

Tertiary Treatment- carried out to remove heavy metal ions,phosphates, remaining organic compounds and colloidal impurities

Thank You