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Solublity detemination

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Information about Solublity detemination

Published on March 3, 2014

Author: ShmmonAhmad

Source: slideshare.net

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Compiled by-Shamon Ahmad,, M.Pharma (Q.A) Chandigarh Group of Colleges, Landran, Mohali(Punjab India)email-shmmon@gmail.com(Date-15/02/13) Solubility Determination How is the solubility of a substance determined? A known amount of the solvent--for example, 100 mL--is put in a container. Then the substance whose solubility is to be determined is added until, even after vigorous and prolonged stirring, some of that substance does not dissolve. Such a solution is said to be saturated because it contains as much solute as possible at that temperature. In this saturated solution, the amount of solute is the solubility of that substance at that temperature in that solvent. Doing this experiment with water as the solvent and sodium chloride as the solute, we find that, at 20°C, 35.7 g of the salt dissolve in 100 mL water. The solubility of sodium chloride is, then, 35.7 g/100 mL water at 20°C. Sodium chloride is a moderately soluble salt. The solubility of sodium nitrate is 92.1 g/100 mL water at 20°C; sodium nitrate is a very soluble salt. At the opposite end of the scale is barium sulfate, which has a solubility of 2.3 X 10 -4 g/100 mL water at 20°C. Barium sulfate is an insoluble salt. Determination of solubility : The solubility of salts is generally determined by gravimetric method. First of all a saturated solution is prepared. Some part of this saturated solution is weighed out in a porcelain dish. The solution is evaporated slowly to dryness on a sand bath. The dish is cooled and weighed again. The observations are recorded as follows: solubility of salts by gravimetric method 1. Mass of empty dish = w gram 2. Mass of dish + solution = w1 g 3. So Mass of solution = (w1 – w) g 4. Mass of dish + residue = w2 g So Mass of residue = (w2 – w) g = x g

and Mass of solvent = (mass of solution – mass of residue) = (w1 – w) – (w2 – w) = (w1 – w2) = y g Thus, the solubility of salt = x/y × 100 g per 100 g of solvent. Example: 50 g of a saturated aqueous solution of potassium chloride at 30°C is evaporated to dryness, when 13.2 g of dry KCl was obtained. Calculate the solubility of KCl in water at 30°C. Solution: Mass of water in solution = (50 – 13.2) = 36.8 g Solubility of KCl = Mass of KCl/Mass of water × 100 = 13.2/36.8 × 100

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