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4mn(oh), 2mn(oh) – LaMotte Dissolved Oxygen Water Quality User Manual

Page 14

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Immediately upon formation of the precipitate, the oxygen in the water
oxidizes an equivalent amount of the manganous hydroxide to brown-colored
manganic hydroxide. For every molecule of oxygen in the water, four
molecules of manganous hydroxide are converted to manganic hydroxide.
Chemically, this reaction can be written as:

4Mn(OH)

2

+

O

2

+

2H

2

O

⎯ →

4Mn(OH)

3

Manganous

Hydroxide

+

Oxygen

+

Water

⎯ →

Manganic

Hydroxide

After the brown precipitate is formed, a strong acid, such as Sulfamic Acid
Powder (6286) or Sulfuric Acid, 1:1 (6141) is added to the sample. The acid
converts the manganic hydroxide to manganic sulfate. At this point the
sample is considered “fixed” and concern for additional oxygen being
introduced into the sample is reduced. Chemically, this reaction can be written
as:

2Mn(OH)

3

+ 3H

2

SO

4

⎯ →

Mn

2

(SO

4

)

3

+

6H

2

O

Manganic

Hydroxide

+ Sulfuric

Acid

⎯ →

Manganic

Sulfate

+

Water

Simultaneously, iodine from the potassium iodide in the Alkaline Potassium
Iodide Azide Solution is oxidized by manganic sulfate, releasing free iodine
into the water. Since the manganic sulfate for this reaction comes from the
reaction between the manganous hydroxide and oxygen, the amount of iodine
released is directly proportional to the amount of oxygen present in the
original sample. The release of free iodine is indicated by the sample turning a
yellow-brown color. Chemically, this reaction can be written as:

Mn

2

(SO

4

)

3

+

2KI

⎯ →

2MnSO

4

+

K

2

SO

4

+

I

2

Manganic

Sulfate

+ Potassium

Iodide

⎯ →

Manganous

Sulfate

+ Potassium

Sulfate

+ Iodine

The final stage in the Winkler titration is the addition of sodium thiosulfate.
The sodium thiosulfate reacts with the free iodine to produce sodium iodide.
When all of the iodine has been converted the sample changes from
yellow-brown to colorless. Often a starch indicator is added to enhance the
final endpoint. Chemically, this reaction can be written as:

2Na

2

S

2

O

3

+

I

2

⎯ →

Na

2

S

4

O

6

+

2NaI

Sodium

Thiosulfate

+

Iodine

⎯ →

Sodium

Tetrathionate

+

Sodium

Iodide

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