ESTIMATION OF DISSOLVED OXYGEN CONTENT OF WATER BY MODIFIED WINKLER’S METHOD EXPERIMENTALLY

AIM : ESTIMATION OF DISSOLVED OXYGEN CONTENT OF WATER BY MODIFIED WINKLER'S METHOD.

https://www.youtube.com/watch?v=frtln5ZoeNQ

THEORY: 

The term Dissolved Oxygen (DO) is used to describe the amount of oxygen dissolved in a unit volume of water. Dissolved oxygen is used as an indicator of the health of a water body. Aerobic bacteria thrive when oxygen is available in plenty. Thus, higher DO values are correlated with high productivity and little pollution.

The Winkler Method (devised by Winkler in 1888) is a technique used to measure dissolved oxygen in freshwater systems. This test is performed on-site, as delays between sample collection and testing may result in an alteration in oxygen content.

In a healthy water body, such as a lake, river or stream, the DO ranges between 5-8 mg/L or parts per million (ppm). The minimum DO level of 4-5 mg/L or ppm is desirable for survival of aquatic life. Higher temperature, biological impurities, Ammonia, Nitrates, Ferrous ions, chemicals such as Hydrogen sulphide and organic matter reduce DO values.

 Environmental significance 

Dissolved oxygen analysis can be used to determine:

·         the health or cleanliness of a lake or stream and keep a check on stream pollution

·         the amount and type of biomass a freshwater system can support

·         the amount of decomposition occurring in the lake or stream

It forms the basis for Biogeochemical Oxygen demand (BOD) test which is an important parameter to evaluate organic pollution potential of a waste.

It is an important step in water pollution control and waste water treatment process control.

 

PRINCIPLE

Two methods are commonly used to determine DO concentration: (1) The iodometric method which is a titration-based method and depends on oxidizing property of DO and (2) The membrane electrode procedure, which works based on the rate of diffusion of molecular oxygen across a membrane.

In the Iodometric method, divalent manganese (Mn²⁺) solution is added to the water sample in a glass-stopper bottle, followed by addition of strong alkali (OH^--). The divalent manganese salt in solution is rapidly precipitated by the strong alkali to divalent manganese hydroxide Mn(OH₂). Dissolved oxygen present in the water sample rapidly oxidize Mn(OH)₂ to form trivalent manganese hydroxide MnO(OH)₂, which appears as a brown precipitate.  

The second part of the Winkler test reduces (acidifies) the solution. In the presence of iodide ions (I^-) in an acidic solution, the oxidized manganese reverts to the divalent state, with the liberation of iodine (I₂) equivalent of the original DO content of the sample. This step of conversion of Iodide ions to iodine is known as fixation of O₂. The amount of iodine liberated is then titrated with a standard solution of thiosulfate (S₂O₃^2-). The titration end point can be detected visually with a starch indicator.

Interferences: Some oxidizing and reducing agents present in solution can interfere with the iodometric method. Oxidizing agents (e.g. nitrites) liberate iodine from iodides (positive interference) and some reducing agents reduce iodine to iodide (negative interference). Also, organic matter present in solution can be oxidized partially in the presence of oxidized manganese precipitate, thus causing negative errors. Hence, a modified Winkler method is used these days, making use of sodium azide to overcome the effect of nitrite interference.

When interference from nitrites is present, it is impossible to obtain a permanent end point. As soon as the blue color of the starch indicator has been discharged, the nitrites formed by the reaction reacts with more iodide ions to produce I₂ and the blue color of the starch indicator will return. The nitrite interference is easily overcome with use of sodium azide (NaN₃), which is incorporated in the alkali-KI reagent. When sulfuric acid is added, following reactions happen:

NaN₃ + H⁺ à HN₃ +Na⁺

HN₃ + NO₂^- + H⁺ à N₂ + N₂O + H₂O

 

Hence this Winkler’s method is known as Alsterberg modified Winkler’s method.

CHEMICAL REACTIONS INVOLVED

REQUIREMENTS:

Apparatus required

300 ml glass stoppered BOD bottles

500 ml conical/ Erlenmeyer flasks

250 ml graduated cylinders

Burette with burette stand

Pipettes with elongated tips

Wash bottle

Chemicals required:

1.      Manganese (II) sulphate solution (MnSO₄): add 40g MnSO₄ to 50 ml deionized water. Dissolve and make the final volume 100 ml.

2.      Alkaline Potassium Iodide Azide solution: Add 70g KOH and 15 g KI to a 250 ml beaker. Add 30 ml water and dissolve. Separately dissolve 1g NaN₃ in 10 ml water in a small beaker. Mix both. Make up the final volume 100 ml with water.

3.      Concentrated sulphuric acid

4.      0.0125 N or N/80 sodium thiosulphate solution: 3.1025g sodium thiosulphate in 100 ml water. Make the final volume 1000 ml with water. Add 0.4g NaOH to prevent thiosulphate deterioration.

5.      Freshly prepared 2% starch solution.

 PROCEDURE

1.      Measure the temperature of the given water sample first of all.

2.      Carefully fill a 300-mL glass Biological Oxygen Demand (BOD) stoppered bottle brim-full with sample water by immersing in the water sample.

3.      Immediately add 2mL of manganese sulfate to the collection bottle by inserting the calibrated pipette just below the surface of the liquid. (If the reagent is added above the sample surface, you will introduce oxygen into the sample.) Squeeze the pipette slowly so no bubbles are introduced via the pipette. (FIXATION STEP)

4.      Add 2 mL of alkali-iodide-azide reagent in the same manner.

5.      Stopper the bottle with care to be sure no air is introduced. Mix the sample by inverting several times. Check for air bubbles; discard the sample and start over if any are seen. If oxygen is present, a brownish-orange cloud of precipitate or floc of Manganese hydroxide will appear. When this floc has settled to the bottom, mix the sample by turning it upside down several times and let it settle again.

Here, the intensity of brown color formed is directly proportional to the DO content of the water sample. At this point, the sample is "fixed" and can be stored for up to 8 hours if kept in a cool, dark place. As an added precaution, cap the bottle with aluminum foil and a rubber band during the storage period.

6.      Add 2 mL of concentrated sulfuric acid via a pipette held just above the surface of the sample. Carefully stopper and invert several times to dissolve the floc.

7.      In a glass flask, titrate 100 mL of the sample with sodium thiosulfate to a pale straw color. Titrate by slowly dropping titrant solution (sodium thiosulfate) from a calibrated pipette into the flask and continually stirring or swirling the sample water.

8.      Add 2 ml of starch solution so a blue color forms.

9.      Continue slowly titrating until the sample turns clear. As this experiment reaches the endpoint (blue to colourless), it will take only one drop of the titrant to eliminate the blue color. Be especially careful that each drop is fully mixed into the sample before adding the next. It is sometimes helpful to hold the flask up to a white sheet of paper to check for absence of blue colour.

10.  The concentration of dissolved oxygen in the sample is equivalent to the number of milliliters of titrant used. Each ml of sodium thiosulfate added in step7 plus 9 equals 1 mg/L dissolved oxygen.

 

Observation table: 

Sample 1: ___________________

Temperature at which the water sample was collected = ……’C

Reading number	Volume of sample taken (ml)	Burette reading (ml)		Volume of Titrant used (ml)
		Initial	Final
1
2
3

 Average volume of Titrant used = …. ml

Calculations:

 Volume of sodium thiosulphate used = …. ml

            Therefore, Dissolved Oxygen content = …. X 0.0125 X 8 X 1000   = …. mg/L or ppm

                                                                                                            100

 General Calculations: 

Let the average volume of sodium thiosulphate solution used in titration = V₁ ml

Normality of sodium thiosulphate solution (N₁) = N/80 or 0.00125N (given)

Volume of sample taken for titration (V₂) = 100 ml

Dissolved oxygen (in mg/L) = V₁ X N₁ X 1000 x 8 = x mg/L or ppm

                                                              V₂

 

From the above stoichiometric equations, we can also find that:

1 mole of O₂ → 2 moles of MnO(OH)₂ → 2 mole of I₂ → 4 mole of S₂O₃²⁻

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/L of water.

 Results and Inference:

The normal value of 4-5 ppm is good enough for a water sample to support aquatic life.

Dissolved oxygen content of the ______ water sample is ____ ppm or mg/L, which means the tested sample is in (healthy/polluted) condition and (fit/unfit) for aquatic life.

Dissolved oxygen content of the ______ water sample is ____ ppm or mg/L, which means the tested sample is ………… for aquatic life.

Dissolved oxygen content of the tap water sample is ____ ppm or mg/L, which means the tested sample is low is organic content and living matter.

Also, dissolved oxygen levels depend on temperature, salinity and partial pressure or elevation. Oxygen is less soluble in cold waters. Hence, water samples at lower temperatures have lower DO values.

Precautions: