Alcoholic Potassium Hydroxide Solution

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.6 PREPARATION OF REAGENTS Alcoholic potassium hydroxide solution 500 ml of ethyl alcohol (95%) is measured using a measuring cylinder and transferred into a 1 L volumetric flask. 3.3333 g of potassium hydroxide (KOH) is weighed using an analytic balance and transferred into the 1 L flask. The resulting solution is then boiled using a water bath under a reflux condenser for 30 to 60 minutes. The resulting solution is distilled to obtain pure alcohol. 13.3333 g of KOH is dissolved in 1 litre of the distilled alcohol, keeping the temperature below 15.5oC while the alkali is being dissolved. Resulting solution is clear American Society for Testing and Materials, 1962). 10% potassium iodide 10 ml of potassium iodide is pipetted into a 100 ml volumetric flask. 90 ml…show more content…
Add 0.1 ml of phenolphthalein solution. Titrate with the ethanolic potassium hydroxide solution until permanent pale-pink colour is produced. 1 ml of 0.5 M hydrochloric acid is equivalent to 0.02806 g of KOH. Saponification Value Test Procedure About 2 g of oil is transferred in to a round bottomed flask and dissolved in 5 mL of distilled ethyl alcohol. 25 mL of 0.5 M alcoholic KOH is added into it and mixed well. The flask is fitted with a reflux condenser and refluxed for about one hour (till the reaction is complete and the liquid becomes clear). A blank experiment is simultaneously conducted in the same way without taking oil. Both the flasks are cooled and washed down the inner side of the reflux condenser into the respective round bottomed flask with minimum quantity of water. 2 drops of phenolphthalein is added and titrated against a standard solution of 0.5 M hydrochloric acid until the pink colour disappeared. The difference between the test (A) and blank(B) gives the volume of 0.5 M HCl equivalent of KOH used in saponifying the 2 g of the oil. 56 is the equivalent mass of KOH. Saponification number of oil =

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