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Mg(OH)_2 is soluble in NH_4CI, but not in NaCl. Why? |
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Answer» Solution :Adding `NH_4 CI` to `Mg(OH)_2`, the interaction takes place to give magnesium chloride and ammonium hydroxide. `Mg(OH)_2 + 2NH_4 CI toMgCl_2 + 2NH_4 OH ` `NH_4 OH ` being a weak base, reduces the `[OH^(-)]` in solution. Product of `[Mg^(2+)]` and `[OH^(-)]` REMAINS lower than `K_(sp)`of `Mg(OH)_2` to give no precipitation of `Mg(OH)_2` . Adding NACL to `Mg(OH)_2`, the interaction LEADS to give magnesium chloride and sodium hydroxide. `Mg(OH)_2 +2NaCl toMgCl_2 + 2NaOH` NaOH being a strong base produces free `OH^(-)` ions in solution. Ionic product of `[Mg^(2+)] and [OH^(-) ]` EXCEEDS the solubility product, showing precipitation of magnesium hydroxide |
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