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Determination of dissolved oxygen in water sample by Winkler’s method
Dr. Mausumi Adhya
HOD and Associate Professor
Supreme Knowledge Foundation, West Bengal, India

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Theory
Dissolved oxygen (DO) is essential for maintaining healthy water bodies. It is the amount of oxygen
dissolved in unit volume of water. The DO level of 4-5mg/L is desirable for survival of aquatic life.
The amount of DO present in natural water can be estimated by Winkler’s method.
Winkler’s method: In alkaline medium DO oxidizes manganese sulphate (MnSO4) to insoluble brown
precipitate of hydrated manganese dioxide (MnO2).
Mn2++ 2OH-+ ½ O2=MnO2 (hydrated brown precipitate) + H2O
Hydrated MnO2 dissolves in dilute H2SO4. Acidified solution of MnO2 oxidizes iodide to iodine
quantitatively.
MnO2+4H++2I-= Mn2++ I2+ 2 H2O
The liberated I2 is estimated by standard sodium thiosulphate (Na2 S2O3) solution using starch as
indicator.
I2+ 2e=2I- E0 = 0.54 V
S4O6
2-+ 2e=2S2O3
2- E0= 0.08 V
The reduction potential value of I2/ I- is higher than S4O6
2-/ S2O3
2-. Hence I2 is reduced to I- and S2O3
2-
is oxidized to S4O6
2-.
I2+ 2S2O3
2-=S4O6
2- + 2I-
Hence, 2Na2S2O3 ≡ I2 ≡ MnO2 ≡ 1/2 O2
2 equivalent of Na2S2O3 ≡ 1 equivalent O2
i.e. 2000 ml of (N) Na2S2O3≡ 8 g O2
Na2S2O3 is a secondary standard solution and is standardized by primary standard potassium
dichromate (K2Cr2O7) solution iodometrically using starch indicator.
Cr2O7
2- + 14H+ + 6e = 2Cr3+ + 7H2O E0=1.33 V
I2+ 2e=2I- E0 = 0.54 V
In acid medium, reduction potential value of Cr2O7
2-/ Cr3+ is greater than I2/ I-. Hence K2Cr2O7 is
reduced to Cr3+ and I- is oxidized to I2.
Cr2O7
2- + 14H+ + 6e = 2Cr3+ + 7H2O
3×(2I-- 2e= I2)
Overall reaction Cr2O7
2- + 14H+ + 6I-= 2Cr3+ +7H2O + 3I2
The liberated I2 is estimated by standard sodium thiosulphate (Na2 S2O3) using starch indicator.

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Materials
1. Apparatus: Burette with burette stand, pipette, conical flask, beaker, volumetric flask, reagent bottle,
measuring cylinder, dropper
2. Chemicals: K2Cr2O7, Na2S2O3, KI, Conc HCl, NaOH, sodium azide (NaN3), KF, MnSO4, H2SO4, starch
Procedure
1. Preparation of reagents
(I) Preparation of 250 ml N/20 K2Cr2O7
Cr2O7
-2 + 14H+ + 6e = 2Cr+3 + 7H2O
Gram-equivalent weight of K2Cr2O7 = g molecular weight/6 = [39×2+52×2+16×7]/6 g = 49 g
To prepare 1000 ml 1 N K2Cr2O7 gm equivalent required= 49 g
Therefore, to prepare 250 ml N/20 K2Cr2O7 gm equivalent required= (49×250)/(1000×20) =0.613 g
0.613 g K2Cr2O7 is accurately weight out in 250 ml volumetric flask and dissolved in distilled water
with continuous stirring. The water is added upto the mark. Finally the solution is made uniform by
shaking.
(II) Preparation of 1000 ml N/20 Na2S2O3solution
I2+ 2S2O3
2-=S4O6
2- + 2I-
2S2O3
2- - 2e= S4O6
2-
Therefore, gram-equivalent of Na2S2O3.5H2O =g molecular weight/1= [23×2+32×2+16×3+18×5]/1=248
g
To prepare1000 ml N/20 Na2S2O3, gm equivalent required=248/20=12.4 g
12.4 g Na2S2O3 is weight out in 1000 ml volumetric flask and dissolved in distilled water with
continuous stirring. The water is added upto the mark. Finally the solution is made uniform by
shaking.
(III) Preparation of 20 ml 4% KI solution
0.8 g KI is dissolved in 20 ml distilled water.
(IV) Preparation of 50 ml alkaline iodide-azide solution
2.5 gm NaOH, 1 gm KI and 0.25 gm NaN3 are dissolved in 50 ml distilled water.
(V) Preparation of 20ml 40% KF or NaF
8g KF or NaF is dissolved in 20 ml distilled water.
(VI) Preparation of 20ml 40% MnSO4
8g MnSO4 is dissolved in 20 ml distilled water.
(VII) Preparation of 1% starch solution

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A paste of 1 g starch is prepared by distilled water. This paste is poured in 100 ml boiling distilled
water in a beaker with constant stirring by a glass rod until clear solution is obtained. The solution is
transferred into a stoppered bottle. The aqueous solution of starch is less stable. Hence it is better to
use freshly prepared starch solution.
2. Standardization of Na2S2O3 by standard N/20 K2Cr2O7
10 ml K2Cr2O7 is pipetted out in a 250 ml conical flask. , 4 ml 4% KI solution and 2 ml concentrated
HCl are added to it. The mixture is shaken and allowed to stand in dark for 5 min covering with watch
glass. Therefore, the watch glass and the side of flask are washed with 50 ml distilled water. The
solution is titrated rapidly with Na2S2O3 running from burette with continuous stirring until the colour
of the solution is straw yellow. Then freshly prepared 1% starch solution is added and colour of the
solution is changed to blue. Therefore the solution is titrated by Na2S2O3 until the blue colour is
disappeared leaving bright green colour. Experiment is repeated three times.
3. Estimation of dissolved O2 in water sample
50 ml of sample water is taken in a 250 ml conical flask and shaken with 1 ml 40% KF solution. Then,
1ml 40% MnSO4 solution and 1 ml alkaline-iodide-azide solution are added to it. The mixture is
shaken for few minutes. Therefore, 5 ml 1:4 H2SO4 is added to it and the solution is shaken to dissolve
MnO2 completely. The mouth of the flask is covered by a watch glass and kept the conical in dark for 5
min. Liberated iodine is titrated like previously. Experiment is repeated three times.
Results
Table 1: Standardization of Na2S2O3 by standard K2Cr2O7
Number of
observation
Volume of
K2Cr2O7
(ml)
Burette reading Mean
volume of
Na2S2O3 (ml)
Strength of
K2Cr2O7 ( N)
Strength of
Na2S2O3
(N)
Initial Final Actual
1
10 X 1/20 S
2
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Strength of Na2S2O3 = (10×1/20)/X =S (N)
Table 2: Estimation of dissolved O2 in water sample
Number of
observation
Volume of
sample water
(ml)
Burette reading Mean volume
of Na2S2O3 (ml)
Strength of
Na2S2O3 (N)
Amount of
DO in g/L
Initial Final Actual
1
50 Y S
2
3

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Calculations
Reaction occurs during titration of DO are given below
Mn+2+ 2OH-+ ½ O2=MnO2+ H2O
MnO2+4H++2I-= Mn+2+ I2+ 2 H2O
I2+ 2S2O3
2-=S4O6
2- + 2I-
Hence, 2Na2S2O3 ≡ I2≡ MnO2≡1/2 O2
i.e. 2000 ml of (N) Na2S2O3≡ 8 g O2
1000 ml of (N) Na2S2O3≡ 4 g O2
Y ml S (N) Na2S2O3= [4×Y×S]/1000 g O2
Hence 50 ml sample water contains [4×Y×S]/1000 g O2
Therefore, amount of DO present in 1 L sample water = [4× Y × S]/50 g
Conclusion
The amount of oxygen is present in water sample is…………………….. gm /L
Precautions
1. For preparing all solutions for this experiment, the distilled water should be boiled to remove DO.
2. Starch should be added close to end point i.e. in straw yellow colour.
Viva Voce
1. What is DO? What is its unit?
2. How the amount of DO present in water is measured? Explain with reactions.
3. Why Na2S2O3 solution is secondary standard? How its exact strength can be determined?
Answer. Na2S2O3 is a secondary standard solution as its aqueous solution is unstable. On standing the
aqueous solution turns turbid due to deposition of colloidal sulphur.
2S2O3
2- + H+= HSO3
- + S
The decomposition may be caused by presence of bright sunlight, bacteria or carbondioxide.
4. Why azide solution is used for estimation of DO in water?
Amount of DO in sample water
= [Volume of Na2S2O3 required for titration of sample water × strength of Na2S2O3 ×4/ volume of sample water
taken] g/L

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Answer. In water sample nitrate salts may be present and it can interfere in titration. Hence to remove
interference azide solution is used.
5. Why KF or NaF solution is used for estimation of DO in water?
Answer. KF or NaF is used to remove interference Fe+3 which may present in water.
6. Why should starch be added when the solution is straw yellow in colour?
Starch is a polymer of amylose and amylopectin. Amylose reacts with iodine in presence of iodide to
form an intense blue coloured complex. The complex is visible at very low concentration of iodine. So
starch is added close to end point i.e. in straw yellow colour.