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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