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.