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Discuse common ion effect on the solubility of an ionic salt. |
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Answer» Solution :(1) Purification of common salt : In a saturated solution of common salt, the following equilibrium exists `NaClhArrNa_(aq)^(+)+Cl_(aq)^(-)` `K_(sp)=[Na^(+)][CL^(-)]` = a constant at constant temperature. If HCl gas is passed into the saturated solution, it ionises producing plenty of `Cl^(-)` ions `HCl toH^(+)+Cl^(-),Q_(sp)=[Na^(+)][Cl^(-)]` increases. According to Le Chatelier's principle, the reverse reaction gets favoured until `Q_(sp)` again becomes equal to `K_(sp)`. More NaCl(s) precipitates. Thus, pure NaCl crystals are obtained. Impurities present in common salt remain in the solution. (ii) Salting out of soap: Soap consists of sodium salts of fatty acids like `C_(17)H_(35)COONa`. A saturated solution of `C_(17)H_(35)COONa` formed during SAPONIFICATION contains plenty of dissolved `C_(17)H_(35)COONa`. It is recovered as follows : `C_(17)H_(35)COONa(s)hArrC_(17)H_(35)COO_(aq)+Na_(aq)^(+)` In a saturated solution, `K_(sp)=[C_(17)H_(35)COO^(-)][Na^(+)]` = a constant at constant temperature. If plenty of common salt is added to the above saturated solution, `[Na^(+)]` increases because NaCl ionises as `NaCl TONA^(+)+Cl^(-)` `Q_(sp)=[C_(17)H_(35)COO^(-)][Na^(+)]` increases. Now the backward reaction gets favoured until `Q_(sp)` becomes equal to `K_(sp).C_(17)H_(35)COONa(s)` precipitates. |
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