Biochemical oxygen demand BOD Test Kit
Biochemical oxygen demand Test kits for measuring Biological Oxygen Demand in water and wastewater
Complete BOD Test Kit manufacturer and supplier. Dissolved oxygen. Biochemical oxygen demand Test kits for measuring Biological Oxygen Demand in water and wastewater for applications ranged 2 - 3000 mg/l O2.
Reagents:
Manganous sulphate solution Alkali-iodine reagent
Starch indicator
0.00025 No Sodium thiosulphate solution
Estimation of dissolved oxygen in water:
Reagents: Manganous sulphate solution Alkali-iodine reagent
Starch indicator
0.00025 No Sodium thiosulphate solution
Procedure:
. 20 mL of sample water was taken in a dry breakers.
. Adding 3 drops of Manganous sulphate solution and 2 drops of alkali iodine solution to it.
. Mix well and allow the precipitate to settle down.
. Adding 2 drops of conc. H2 SO4 and mix well till the precipitate
dissolves.
.20 drops of the above solution was taken in a dry test tube.
. Adding 1 drop of starch to the tube. A blue colour of the solution appears.
. Adding Sodium thiosulphate solution drop wise with a calibrated pipette to the tube till the blue colour disappears.
. Counting the number of drops of Sodium thiosulphate solution used.
Calculation:
Volume of water = No. of drops x Drop value of the pipette = p mL Volume of Na2 S2 O3 = No. of drops x Drop value of the pipette = q mL Dissolved oxygen (ppm) = 2 x q / p
Result: The DO of the sample is ___________mg/L or ppm.
Estimation of BOD of water:
Reagents: Manganous sulphate solution
Alkail-iodine reagent
Starch indicator 0.00025 N Sodium thiosulphate solution Procedure:
. Three vials A, B and C was taken and filled with the water sample.
. Determine the DO of A immediately by adding 1 mL of MnSO4 + 2 mL of H2 SO4.
.ToBadding1mLofMnSO4 +2mLofH2 SO4andkeep for 5 days.
.ToCadding1mLofMnSO4 +2mLofH2 SO4 and2mL of sample water and keep for 5 days.
. After 5 days determine DO of B ( blank bottle) and C (sample bottle).
Calculation:
If the values of DO of the vials A, B and C be a, b and c then
BOD (mg /L) = (b-c) x 1000 / volume of sample in mL.
Result: The BOD of the sample is __________ mg /L.
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