# 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.