Determine Amount of Dissolved Oxygen

 EXPERIMENT NO. 05

 

OBJECT: To determine the amount of dissolved oxygen (D.O.) present in given sample of water with the help of Winkler’s Method.

 

OBSERVATION TABLE:

Titration between prepared water sample and N/200 hypo solution using starch as an indicator.

 

S.

No.

Volume of Water Sample (mL)

Burette Reading (mL)

Concordant Reading (Y) (mL)

Initial

Final

1

20

0.0

7.5ml

 

7.5ml

2

20

0.0

----

3

20

0.0

----

 

CALCULATION:

Volume of water sample taken =100 mL

Volume of reagent added = 10 mL MnSO4 +10 mL alkaline KI + 10 mL H2SO4 Total volume of prepared water sample = 100+10+10+10 = 130 mL

Volume of prepared solution taken for titration =20 mL

 

Volume correction:

130 mL prepared water sample contain =100 mL tap water

So, 20 mL prepared = (100 x 20)/130 = 15.38 mL tap water

 

According to the law of equivalence:

 

1 mole of O2 = 2 moles of I2 = 4 moles of Na2S2O3

Hence, 1 mole of Na2S2O3 = 1/4 of O2 = 8g of O2 = 1 mL of Na2S2O3 = 8mg of O2

 

N1V1 of Sample

= N2V2 of Na2S2O3

N1×15.38

= N/200×7.5

N1

= 7.5/200×1/15.38 = 0.0024

                                                                                                                             

Amount of dissolved oxygen (D.O.) = Normality×Eq.Weight of Oxygen

                                                            = 0.0024×8g/L

 

Strength of D.O. = 0.0024×8×1000mg/L or ppm = 19.2 ppm

 

RESULT: The dissolved oxygen content in given water sample is 19.2 ppm.

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