Determination of DO & BOD in a water sample

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Determination of DO & BOD in a water sample

Objective
           Using volumetric analysis, the amount of dissolved oxygen (DO) in a water sample can be found. On standing another sample for 5 days, the difference between these two values is defined as biochemical oxygen demand in 5 days (BOD5). The aquatic life needs certain amount of DO for survival, whereas the value of BOD reflects the water quality.

Principle
           In alkaline solution, dissolved oxygen will oxidize manganese(II) into manganese(III):
                        4Mn
2+(aq)   +   8OH-(aq)   +   O2(aq)   +   2H2O(l)   →   4Mn(OH)3(s)
              The amount of DO can be found by titrating the iodine produced from potassium iodide by manganese(III) with sodium thiosulphate:
                       2Mn(OH)
3(s)   +   2I-(aq)   +   6H+(aq)   →   2Mn2+(aq)   +   I2(aq)   +   6H2O(l)
                         I2(aq)   +   2S2O32-(aq)   →   2I-(aq)   +   S4O62-(aq)

Chemicals
         water sample, MnSO4, conc.H2SO4, alkaline KI, standard Na2S2O3, starch solution

Apparatus
         titration apparatus, pipette, 250cm3 volumetric flask x2, beaker, droppers, magnetic stirrer, white tile

Procedure
         1.> Collect a water sample with two 250cm3 volumetric flasks. Remember to fill the flasks completely with water without trapping any air bubbles.
         2.> To one of the volumetric flask, add about 1 cm
3 of MnSO4 solution well below the surface with a dropper.
         3.> Similarly, introduce about 1 cm
3 of alkaline KI solution to the same flask. Be sure that no air becomes entrapped and then invert the stopped
               flask to distribute the precipitate uniformly.
         4.> When the precipitates has settled at least 2 cm below the stopper, introduce about 1 cm
3 of conc.H2SO4 well below the surface. Stop the flask and
               using a magnetic stirrer to mix the solution until the precipitate dissolves.
         5.> Measure 200 cm
3 of the acidified sample into a 500 cm3 conical flask, titrate with standard Na2S2O3 solution. Until the solution becomes pale
               yellow, add a few drops of starch solution and complete the titration.
         6.> While another volumetric flask containing the sample is stopped, without adding any chemicals, and keep in a cool, dark place (avoid the
               photosynthesis of phytoplankton) for 5 days.
         7.> After 5 days, investigates the sample by repeating the steps 1-5. The difference between 2 DO is defined as BOD
5.

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Data Analysis
        Concentration of standard Na2S2O3 at the first day:        0.0149       M
        Volume of 0.0149M Na
2S2O3 used = 40.00 - 26.75 =       13.25       cm3
        Amount of 0.0149M Na2S2O3 used = 13.25 x 0.0149 =     0.1974     mmol
        Amount of I
2 generated = 1/2 (0.1974) =      0.0987     mmol
        Amount of Mn(OH)
3 generated = 0.0987 x 2 =      0.1974      mmol
        Amount of O
2 present in 200 cm3 of aquarium water at room temperature
        = 1/4 (0.1974) = 0.04935 mmol = 0.04935 x 32 = 
    1.5792     mg
        Dissolved oxygen (DO
0) in aquarium water in the first day at room temperature
        = 1.5792
÷ (200x10-3) =     7.896     mgdm-3
        
       Concentration of standard Na2S2O3 after 5 days:      0.0127      M
        Volume of 0.0127M Na
2S2O3 used = 34.40 ...

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