Experiment to determine Biological oxygen demand

Principle: A known volume of sewage water is diluted with a known amount of dilution water (containing nutrient for bacterial growth). The solution is kept in incubator for a period of 5 days at 20⁰ C. The DO of diluted water sample (D₁) and the solution taken out of the incubator (D₂) are determined. From these values of BOD is calculated.

Procedure:

A)   Preparation of two separate diluted water samples:

Pipette out a known volume of sewage sample (A ml) and dilute it to a known volume (B ml) by adding dilution water containing nutrient and Free O₂. Fill it in two separate BOD bottles.

B)   Blank titration:

In bottle 1, find DO immediately as follows.

·         Add 2ml of MnSO₄ and 3ml of alkaline KI. Stopper the bottle and mix well for 10- 15min.

·         Wait for 2 min. Then add 1ml 1 ml of conc. H₂SO₄ to dissolve MnO₂ ppt.

·         Pipette out known volume (V ml) of this solution and titrated against std. Na₂S₂O₃ using starch as indicator.

·         Note the volume of Na₂S₂O₃ consumed (V₁ml).

C)   Back titration:

·         Bottle 2 is incubated for 5 days at 20^o C.

·         After 5 days, add 2ml of MnSO₄ and 3ml of alkaline KI. Stopper the bottle and mix well for 10- 15min.

·         Wait for 2 min .Then add 1ml 1 ml of conc. H₂SO₄ to dissolve MnO₂ ppt.

·         Pipette out same known volume (V ml) of this solution and  titrated against same std. Na₂S₂O₃ using starch as indicator.

·         Note the volume of Na₂S₂O₃ consumed (V₂ml)

D) Calculation:

1)   DO in blank titration:

Volume of water sample taken= V ml

         Volume of sodium thiosulphate consumed = V₁ ml

         Normality of sodium thiosulphate used = N₁

         Normality of oxygen content in water =    V₁ x N₁

                                                                     V

                                                               = N₂

          Hence oxygen content in blank solution = N₂ x Equivalent weight of oxygen

                                             = N₂ x 8 x 1000 mg/dm³ = D₁ mg/dm³

2)   DO in back titration:

           Volume of water sample taken= V ml

           Volume of sodium thiosulphate consumed = V₂ ml

           Normality of sodium thiosulphate used = N₁

           Normality of oxygen content in water =    V₂ x N₁

                                                                         V

                                                               = N₃

           Hence oxygen content in back solution = N₃ x Equivalent weight of oxygen

                                               = N₃ x 8 x 1000 mg/dm³ = D₂ mg/dm³

                 Therefore BOD = (D1-D2) x B   mg/dm³

                                                      A