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Given a mixture of barium chloride, and potassium chloride, how would we separate one salt from the other?

Answer
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Hint: We are given two salts barium chloride and potassium chloride. The chemical formulas for the two salts are BaCl2&KCl respectively. There are various processes of separation used in chemistry including decantation, filtration, distillation, crystallization etc. The combination of methods to be used for the given mixture will be discussed ahead.

Complete Step By Step Answer:
The solubility product constant is given as Ksp . The Ksp is the equilibrium constant for substances when dissolved in aqueous solutions. It depicts the extent to which a salt/substance is soluble in the solution. The more soluble the substance the more will be the value of its Ksp .
The Ksp of barium sulphate was found to be: Ksp=1.08×1010
To separate the given mixture, we’ll form the compound; barium sulphate. Since it is insoluble in water, it can be easily filtered off. The reaction of Barium chloride with Potassium sulphate, gives us Barium Sulphate. It is a displacement reaction and can be given as:
 BaCl2(aq)+K2SO4(aq)BaSO4(s)+2KCl(aq)
The reaction gives us a precipitate of barium sulphate which is white in colour. The other substance to be separated (potassium chloride) is formed as a by-product from the reaction itself. The barium sulphate will be accumulated down and the supernatant solution will contain only KCl. The solution is filtered off to obtain a pure KCl solution. KCl(s) can be obtained from this solution very easily. We can dry it, but this method will take longer. The easier method would be adding ethyl alcohol. Ethyl alcohol will precipitate out the KCl. This precipitate is subjected to drying at an elevated temperature and is collected on a frit.

Note:
The difference is solubility is a common separation technique used to separate two miscible solids as well as liquids. The physical techniques involved in this separation are: Decantation, Filtration, evaporation, precipitation etc.