Topic outline

  • Winkler method

    Measuring dissolved oxygen (DO) in seawater is a basic requirement in oceanography.  This section describes the measurement of DO using the Winkler method.


    (1) Collect seawater in oxygen bottles and make oxygen in the seawater to combine with manganese hydroxide precipitate.

    (2) Dissolve the manganese hydroxide oxide formed by the dissolved oxygen with hydrochloric acid.  Iodine is generated in the sample water in proportion to the amount of bound oxygen.

    (3) Titrate the amount of iodine with sodium thiosulfate solution. Convert the amount of iodine into the amount of dissolved oxygen.

    The following is a more detailed explanation of the procedure and principle of DO measurement by the Winkler method.

    (1) Collect seawater in an oxygen bottle and add Fixing Solution I (an acidic solution of manganese chloride) and Fixing Solution II (a strongly alkaline mixture of sodium hydroxide and potassium iodide) before putting the lid on.

    (2) Put the lid on the oxygen bottle and invert it up and down 30 times.

    The left side of the picture below is immediately after adding Fixing Solution I and Fixing Solution II and putting the lid on. The acidic manganese chloride solution (with Mn2+) becomes white precipitates of manganese hydroxide (Mn(OH)2) when it becomes strongly alkaline with solution II.

    This white precipitate is mixed well by tipping the oxygen bottle up and down 30 times, then dissolved oxygen (O2) in seawater combines with manganese hydroxide (Mn(OH)2) to form brown manganese hydroxide (MnO(OH)2).  These are a mixture of both white and brown precipitates.  If the seawater contains more dissolved oxygen, the precipitates are darker brown (top of the lower panel).  If seawater contains little or no oxygen, the precipitates will remain white (bottom).

     After a few hours of standing in this state, the precipitates will accumulate at the bottom of the bottle.

    Winkler method

    海水中に酸素(O2)がある場合: Seawater with oxygen

    海水中に酸素(O2)がない場合: Seawater without oxygen

    The below figure shows the acid-base reaction in which manganese ions (Mn2+) precipitate white manganese hydroxide under alkaline conditions, and the redox reaction in which manganese hydroxide and oxygen combine to produce brown manganese hydroxide oxide.

    acid-bace reaction and redox reaction of Mn

    酸化数: Oxidation number  酸塩基反応: Acid-base reaction  酸化還元反応:  Redox reaction  コロイド状白色沈殿: Colloidal white precipitates  褐色沈殿:  Brown precipitates

    When the precipitates accumulate at the bottom of the bottle (right side of the above picture), open the lid of the bottle and add 6 mol/L of hydrochloric acid.  Then the precipitates of manganese hydroxide (white) and manganese hydroxide oxide (brown) that have accumulated in the oxygen bottle will dissolve to produce manganese ions (Mn2+).  At this time, the following acid-base and redox reactions occur.

    Above equation:  The oxidation number of manganese in manganese hydroxide (Mn(OH)2) is +2. Even if this dissolves to produce Mn2+, the oxidation number of manganese remains +2. Since the oxidation number does not change (no electron transfer), this is an acid-base reaction.

    Below equation:  The oxidation number of manganese in manganese hydroxide oxide (MnO(OH)2) is +4. When this dissolves to produce Mn2+, the oxidation number of manganese is +2. Since the oxidation number changes from +4 to +2 (transfer of electrons), this is a redox reaction.

    Reductants to supply electrons (e-) to manganese in this redox reaction (the other half reaction) are needed. Iodide ion (I-), in the fixing solution I,  is responsible. The iodide ion releases electrons (e-) (oxidized) to form iodine molecules (I2).

    Reaction around Mn

    酸塩基反応: Acid-base reaction  酸化還元反応:  Redox reaction  ヨウ化物が電子を供給: iodide supply electrons  ヨウ素発生量を測定: measure amount of iodide(I2) generation  塩酸: hydrochloric acid  滴定分析のときに電子(e-)を供給するヨウ化物(I)を、I液に予め含ませておく: Add Iodide (I-), which provides electrons (e-) during titration analysis, in fixing solution I beforehand.

    When hydrochloric acid is added to the DO bottle to dissolve the precipitate, iodine molecules (I2) are generated in proportion to the amount of dissolved oxygen.

    The amount of these iodine molecules (I2) is examined by titrating with a sodium thiosulfate solution of known concentration.

    When iodine molecules are present in the solution (*), it displays brown to yellow in color.

    Titrating thiosulfate to the iodine molecule, the iodine molecule (I2) is reduced to I-. The endpoint (the point at which the drops of thiosulfate necessary to eliminate I2 are completed) is the point where the color disappears.

    ある量のI2を全て還元: Reducing all I2 of a certain amount  それに要するS2O32−を滴下:  Titrating S2O32- required for it

    [note] It is important to note that the iodine molecule (I2) is an insoluble black solid. As shown in the figure below, I2 can combine with any surrounding I- and dissolve as triiodide ion (I3-). Since I2 and triiodide ion (I3-) are in equilibrium, I2 disappears as soon as triiodide ion (I3-) loses its color.

    In manual analysis, add starch to the iodine solution (sample water), which has lost its brown color, and light purple color will be displayed due to the iodine-starch reaction. It is easier to determine the endpoint because the color shade is more pronounced.

    An automatic device is also available to titrate sample water with thiosulfuric acid by potentiometric titration. The following is an overview of potentiometric titration.

    In a potentiometric titration, monitor the redox potential in the sample water, titrating with c. The redox potential is different before the endpoint (when there are iodine molecules in the sample water) and after the endpoint (when thiosulfuric acid accumulates). The endpoint is the inflection point of the potential curve. By reading the inflection point, measure the amount of thiosulfate solution dropped to the endpoint.

    電位差滴定の説明: Explanation of potentiometric titration  コニカルビーカーの液が褐色のとき、徐々にチオ硫酸を滴下: Add thiosulfate acid steadily until solution in the conical beaker is brown  電子を受けたい物質が残存: Remaining substances that are able to receive electrons  コニカルビーカーの液が無色になった: Solution in the conical beaker turns colorless  電子を与えたい物質が貯まる: Accumulating substances that are able to give electrons  褐色: brown  無色透明: clear and colorless  酸化還元電位: Oxidation−Reduction Potential  高: high  低: low

  • Reaction equation of Winkler method (Summery)

    Reprinted below is a description of the Winkler method written in the student experiment.

    Principle of Winkler method

    Add manganese chloride and sodium hydroxide solution to a certain amount of samples to precipitate manganese hydroxide.

    Mn2+ + 2OH  → Mn(OH)2↓    (colloidal white precipitation)                         (1)

    At this time, the oxygen dissolved in the water oxidizes some of the manganese hydroxide (divalent Mn) to trivalent Mn(OH)3. (or to manganese hydroxide MnO(OH)2. Mn(OH)3 is used in the above equation because it is easier to perform thermodynamic calculations.)

    2Mn(OH)2 + 1/2O2 + H2→ 2Mn(OH)3↓    (brown precipitation)                         (2)

    Add potassium iodide and hydrochloric acid to it, then the oxidized manganese ions are reduced by potassium iodide in acidic conditions, and iodine liberates.

    2Mn(OH)3 + 2I ̶  + 6H+  →  2Mn2+ + I2 +6H2O                         (3)

    Titrate liberated-iodine with sodium thiosulfate solution whose concentration is known and indirectly determines the amount of oxygen.

    I2 + 2S2O32 ̶  → 2I ̶  + S4O62 ̶                          (4)

    After all, four molecules of sodium thiosulfate (or four moles of thiosulfate ions) are equivalent to one molecule of oxygen (O2) (or one mole of O2 molecules).

    adding hydrochloric acid to the mixed precipitate of Mn(OH)2 and Mn(OH)3, the precipitate of Mn(OH)2 will also dissolve and become Mn2+. Because the Mn oxidation number of Mn(OH)2 remains unchanged at +2, it is not in the redox reaction (without I2 generation).

     To do this in practice, add (1) a manganese chloride solution (known as fixing solution I) and (2) a potassium iodide-sodium hydroxide mixture (known as fixing solution II) sequentially to sample water collected at the observation site. Dissolved oxygen is fixed as Mn(OH)3 (called oxygen fixation) and later acidified with hydrochloric acid to liberate iodine.  Titrate liberate-iodine with sodium thiosulfate solution.

    [Reagent preparation]

    ① Manganese chloride solution (Fixing solution Ⅰ)

    Dissolve 200 g of manganese chloride (MnCl2-4H2O) in 500 mL of deionized water and add 2 mL of pure concentrated hydrochloric acid.

    ② Potassium iodide-sodium hydroxide mixture (Fixing solution Ⅱ)

    Dissolve 180 g sodium hydroxide in 500 mL deionized water and 200 g pure potassium iodide*. Store in a plastic container (because this is highly alkaline and will cause the glass lid to stick). After use, the dispensing vessel should be washed with deionized water and acid.

    * Use new reagents within 1 year after opening the package, as old or exposed to strong light one will release I2 and cause color change, which may lead to errors.

    ③ Hydrochloric acid (6 mol/L) (200ml)

    Dilute pure concentrated hydrochloric acid (12 mol/L) 2 times. You can measure it with a measuring cylinder.

    Add 100 mL of water in a beaker (500 mL). Measure 100 mL of concentrated hydrochloric acid in a draft with a measuring cylinder. Gradually add the concentrated hydrochloric acid to the beaker in the draft. Put the hydrochloric acid solution into the reagent bottle in the draft. Immediately after adjustment, the reagent is warm, so do not put the lid on. When cooled to room temperature, put the lid on.

    Warning: Handle concentrated hydrochloric acid in a draft chamber wearing protective glasses because hydrochloric acid emits vapor. Add concentrated hydrochloric acid to water gradually. Wipe up spilled hydrochloric acid with a wet tissue and wash the tissue with tap water.

    ④ Starch solution

    Knead about 1 g of starch with a small amount of water to form an even, itchy consistency and add to 100 mL of hot water (deionized water). Continue to boil while heating slowly until clear. When cooled, put the mixture into reagent bottles. 

    [note]   For long-term storage, add benzoic acid at a rate of 0.1 g per 100 mL of starch solution or acetic acid at a rate of 5 mL to prevent spoilage. If the bottle is too old, or if the bottom of the bottle becomes stagnant, the blue coloring of the iodine solution will weaken. Then replace it with a new one. For long-term stable use, a glycerin solution of starch is better. While warming glycerin, dissolve 10-20% soluble starch in it.

    ⑤ Standard stock solution of potassium iodate (0.016669 mol/L)

    Dissolve 1.7835 g potassium iodate (KIO3) in deionized water to make the total volume exactly 500 mL. Mix well by tipping up and down 20 times. Put this undiluted solution (10 times the concentration of the actual solution used in the experiment) into a brown bottle and keep it in a cool, dark place as much as possible. Before use, dilute the solution to 1/10 of its original concentration.

    The exact concentration of the standard solution of potassium iodate can be calculated by recording the exact weight weighed on an electronic balance. Therefore, it is not necessary to adjust it to exactly 3.567 g. Consider performing the experimental manipulation in a dexterous manner. Wash reagent bottles and other containers together to maintain the adjusted concentration.

    ⑥ Sodium thiosulfate solution (0.02 mol/L)

    Dissolve about 3.2 g of anhydrous sodium thiosulfate (Na2S2O3) in deionized water to make the total volume 1 L. Mix well by tipping up and down 20 times. Store in a reagent bottle.

    Since we are titrating with this sodium thiosulfate solution, we need to know the exact concentration. Sodium thiosulfate easily absorbs water to form hydrates. Even if you weigh an exact amount of anhydrous sodium thiosulfate, you will not be able to adjust the exact concentration of the solution. Therefore, the exact concentration should be determined using potassium iodate standard solution. The method of calibration is explained in the next chapter.

    (7) Capacity test of oxygen bottles (wash and dry oxygen bottles on the second day of the experiment, and measure the dry weight on the third day)

    To test the capacity of the oxygen bottles, measure the weight of the water in the bottles. First,  wash oxygen bottles with water, and drain the water inside. They will be placed in a dryer, and once completely dry,  measure their tare weight. On the second day of the experiment, we will fill beakers with deionized water (about 200 mL), wrap them and leave them overnight to acclimate to room temperature; on the third day, we will measure the temperature of the deionized water and the room temperature. Fill the oxygen bottle with this deionized water, cover it, and wipe off any water droplets on the outside with tissue paper. Measure the weight of the oxygen bottle with water to determine the weight of the water and convert it to volume. Complete this volume test before beginning the measurement of dissolved oxygen concentration in seawater.

    Perform experimental manipulations (1) through (6). If there is extra time, do additional experiments.

    ① Evaluation of sodium thiosulfate concentration

    Since sodium thiosulfate powder reagent can absorb water, it is difficult to adjust an accurate concentration. Titrate a standard solution of iodic acid with sodium thiosulfate to determine its concentration. Do it three times and take the average. If there is a clear failure, start over.

    (1) Put about 0.02 mol/L sodium thiosulfate solution in a burette. Thiosulfate is corrosive, so be careful not to get it in your eyes. Remove the burette from the stand and hold it in your hand. Set the funnel on top of the burette, and slowly pour the solution into it. After pouring, remove the funnel.

    (2) Dilute the potassium iodate standard stock solution (stored in a reagent bottle) to 1/10 of its original concentration (using a whole pipette and volumetric flask for accuracy). Measure  10 mL of the diluted solution with a whole pipette accurately and transfer it to a conical beaker. Add 0.2-0.3 g of small crystals of potassium iodide and 1 mL of hydrochloric acid (6 mol /L) to liberate the iodine. Immediately add deionized water to make the total volume approximately 100 mL.

    (3)Titrate the solution in the conical beaker with sodium thiosulfate.  When the yellow color of iodine fades, add about 1 mL of starch solution as an indicator. Set the endpoint as the first moment when the resulting blue color disappears, and determine the exact concentration from the volume of sodium thiosulfate solution required until then. If the concentration of the sodium thiosulfate solution deviates significantly from the target concentration of 0.02 mol/L, it is likely that a serious error has been made in the experimental operation.

    [Explanation of the reaction equation]

    Iodide ion (I-) is oxidized to I2 under hydrochloric acidic conditions. At this time, iodic acid (IO3-) is reduced to I2.

    IO3 + 5I + 6H+ → 6H2O + 3I2                 (1)

    Potassium iodide crystals are over-fed, so all IO3- in the conical beaker is turned to I2. I2 and I- combine to form I3-.

    I2(aq) + I- ⇆ I3-                                             (2)

    I3− displays yellow in color.

    Reduce this I2 with thiosulfate.

    I2 (aq) + 2S2O32 ̶ → 2I ̶  + S4O62 ̶                   (3)

    When I2 is completely reduced, I3- is also completely reduced at the same time. The endpoint is the point where the color disappears.

    Note that I2 is black solid, but slightly dissolves to become I2 (aq).

    I2 (s)  I2 (aq)                                              (4)

    Additional experiment (if you have extra time)

    In experimental manipulation (2), place 10 mL of the stock solution of the iodic acid standard in a conical beaker (10 times the amount of iodic acid added in the initial experiment). Add only 0.2-0.3 g of small crystals of potassium iodide, 1 mL of hydrochloric acid, and about 50 mL of deionized water in it. Then, the IO3- in the conical beaker will react to all of the added I-. In other words, the equilibrium in the reaction equation (2) shifts to the left, and I2 (aq) is generated because there is no I- in the conical beaker.  Since I2 is hardly soluble, I2(s) precipitates by the reaction equation (4). Let us check this. How can we dissolve this black solid? Just add potassium iodide.

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