Science of hygiene : a text-book of laboratory practice for public heatlh students / by Walter C.C. Pakes.

  • Pakes, Walter C. C.
Date:
[1912]
Catalogue details

Licence: In copyright

Credit: Science of hygiene : a text-book of laboratory practice for public heatlh students / by Walter C.C. Pakes. Source: Wellcome Collection.

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    2. Into each of these measure 10 c.c. of the standard potassium permanganate solution and 10 c.c. of the specially prepared, organically pure 25% H2S04. 3. Shake the bottles up and place in the water bath at 270 C. for four hours. 4. At the end of this time add a few drops of the KI solution to each bottle. The pink colour will disappear and be replaced by a yellow one. 5. Fill the burette with thiosulphate solution, and carefully read the height of the fluid. ' 6. Run the thiosulphate into the control bottle until the yellow colour is almost discharged and then add a few drops of starch solution. The colour will now turn blue. Add the thiosulphate cautiously until all the colour is discharged, and read off the height of the fluid in the burette. 7. Repeat this with the bottle containing the sample of water. Explanation The potassium permanganate in the presence of sulphuric acid oxidizes the organic matter in the water. K2Mn208 + 3H2S04 = K2S04 + 2MnS04 + 3H20 + 5O The amount of this organic matter is to a certain extent gauged by the quantity of permanganate used up. The control bottle, containing only distilled water and there- fore no organic matter, will use up no oxygen, and so will contain permanganate at the end of the four hours, the equivalent of i milligram of oxygen. The potassium iodide is decomposed by the permanganate, and iodine is liberated. K2Mn208 + 8H2S04 + 10KI = 6K2S04 + 2MnS04 + 8H20 + 5l2 The amount of iodine liberated will be in proportion to the amount of permanganate in solution (i.e. not used up). In this case it will be the equivalent of 1 milligram of oxygen. The amount of iodine liberated is now measured by the quantity of thiosulphate required to decolourize the solution, which it does according to the following equation:— 2Na2S203 +12 = 2NaI + Na2S406 The starch solution is added towards the end of the operation in order the more easily to determine the actual end point. It performs no other function.
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