Winkler test for dissolved oxygen

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The Winkler test is used to determine the level of dissolved oxygen in water samples. The test was first developed by Lajos Winkler while working on his doctoral dissertation in 1888. The amount of dissolved oxygen is the biological activity of the water masses. Phytoplankton and macroalgae present in the water mass produce oxygen by the way of the photosynthesis. At the same time bacteria and all other organisms (zooplankton, algae, fish) consume the oxygen by respiration. The result of this two mechanisms give the concentration in dissolved oxygen, it indicates the growth production (of biomass). The difference between the physical concentration of oxygen in the water (theoretical concentration if there was no living organisms) and the actual concentration of oxygen is called the Biological Demand in Oxygen.

Contents

What follows is a set of obsolete instructions on how the test was once performed. Note that this method is no longer performed as described below, although the principle remains the same. For example, current testing typically uses a 300 mL sample.

Materials

Method Collect and label each water sample in a 25 ml stoppered volumetric flask, making sure no air bubbles are trapped within the sample. (Two samples per location are required for Biological Oxygen Demand testing.)
Immediatey after sampling, add 0.1 ml of manganese(II) sulfate solution, and mix carefully, without letting in air. Add 0.2 ml of alkaline potassium iodide, and again mix without letting in air. A pinkish-brown precipitate should appear.
At this point the sample may be stored for later analysis in the laboratory.

In the alkaline solution, dissolved oxygen will oxidize manganese(II) ions to the tetravalent state.

2 Mn(OH)2(s) + O2(aq) ───→ 2 MnO(OH)2(s)

MnO(OH)2 appears as a brown precipitate.

There is confusion about whether the oxidised manganese is tetravalent or trivalent. Some sources claim that Mn(OH))3 is the brown precipitate, but hydrated MnO2 is more likely to give the brown colour.

Materials

Method Add 0.3 ml sulfuric acid to each sample, and mix. Allow the sample to stand for two minutes. If the precipitate does not dissolve into iodine solution, add a further 0.1 ml acid.

The Mn(SO4)2 formed by the acid converts the iodide ions into iodine with itself being reduced back to manganese(II) ions in an acidic medium.

Mn(SO4)2 + 2 I-(aq) ───→ Mn2+(aq) + I2(aq) + 2 SO42-(aq)

Fill the burette with thiosulfate solution and note the burette reading. Transfer 10 ml of the sample to a conical flask, and add a few drops of starch solution. The subsample should turn blue. Titrate the subsample with thiosulfate until it turns clear. (You may find the endpoint easier to see if the conical flask is stood on a sheet of filter paper.) Record and repeat the titration.

2 S2O32-(aq) + I2 ───→ S4O62-(aq) + 2 I-(aq)

From the above stoichiometric equations, we can find that:

1 mole of O2─→ 4 moles of Mn(OH)3 ─→ 2 moles of I2

Therefore, after determining the number of moles of iodine produced, we can work out the number of moles of oxygen molecules present in the original water sample. The oxygen content is usually presented as mg dm-3.

In the above method, each milliliter of thiosulfate titer is equivalent to 0.1 mg of oxygen in the 10 ml subsample. Thus 1 ml of thiosulfate is equivalent to 1 mg oxygen per 100 ml fresh water.

Success of the method is critically dependent upon the manner in which the sample is manipulated. At all stages, every method must be made to assure that oxygen is neither introduced to nor lost from the sample. Furthermore, the sample must be free of any solutes that will oxidize iodide or reduce iodine.

Instrumental methods for measurement of Dissolved Oxygen have widely supplanted the routine use of the Winkler test, although it is still used to check instrument calibration.

To determine five-day Biological Oxygen Demand (BOD5), several dilutions of a sample are analyzed for dissolved oxygen before and after a five-day incubation at 20 degrees Celsius (68 degrees Fahrenheit) in the dark. In some cases the sample must be provided a source of oxygen-using bacteria, called "seed", in order to obtain results. The difference in DO and the dilution factor are used to calculated BOD5. The resulting number (usually reported in parts per million or milligrams per Liter) is useful in determining the relative organic strength of sewage or other polluted waters.

The BOD5 test is an example of analysis that determines classes of materials in a sample.

  • Winkler, 1888 (Ber. Deutsch Chem. Gos., 21, 2843).
  • Moran, Joseph M.; Morgan, Michael D., & Wiersma, James H. (1980). Introduction to Environmental Science (2nd ed.). W.H. Freeman and Company, New York, NY ISBN 0-7167-1020-X
  • Standard Methods for the Examination of Water and Wastewater - 20th Edition ISBN 0-87553-235-7. This is also available on CD-ROM and online by subscription

Good overview of technique

Manganese (III) consistently claimed

NB: Gives unbalanced equation for formation of MnO(OH)2 Claims manganese (III)

Gives manganese (IV) consistently

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