Question

Would copper carbonate precipitate from aqueous solution by treatment of calcium carbonate by copper(II) chloride?

Answer 1

Hint: This question can be solved by considering the $ {K_{sp}} $ (solubility product constant) for the given product. If the value of the ionic product is greater than the value of $ {K_{sp}} $ , then a precipitate is formed. If the ionic product is smaller than the $ {K_{sp}} $ then no precipitation occurs. The precipitation stops when the Ionic product is equal to the $ {K_{sp}} $ .

Complete answer:
We will start solving the question by writing the chemical equation for the reaction occurring. The reaction can be given as:
$ CuC{l_2}(aq) + CaC{O_3}(s) \to CaC{l_2}(aq) + CuC{O_3}(s) \downarrow $
The given reaction is a simple double displacement reaction in which the Chloride ion of Copper (II) chloride is replaced by carbonate anion in Copper Carbonate and carbonate in Calcium carbonate is replaced by chloride anion.
We would probably see the precipitation of copper carbonate. The sparingly soluble Calcium carbonate will pass into the solution to give calcium chloride. The colour of the solution will change from dark blue to lighter blue as the $ C{u^{ + 2}} $ ion is precipitated. The cupric chloride is a complex salt that is formed and may cause inconveniences in the reaction.
The best way to determine whether the precipitation will occur can be decided only experimentally. But if we are given the $ {K_{sp}} $ of the respective salts formed we can determine which will precipitate first. The $ {K_{sp}} $ of copper carbonate is $ 1.4 \times {10^{ - 10}} $ and that of calcium carbonate is $ 2.8 \times {10^{ - 9}} $ , hence we can say that copper carbonate is more insoluble than calcium carbonate. The salt with the least $ {K_{sp}} $ will precipitate first. Hence calcium carbonate will precipitate.

Note:
The fresh precipitate will contain $ N{a_2}S{O_4} $ . To get rid of it, wash the precipitate thoroughly with distilled water. We can check if $ N{a_2}S{O_4} $ has been removed by adding diluted hydrochloric acid and aqueous $ BaC{l_2} $ . When there is no longer formation of white ppt the $ N{a_2}S{O_4} $ is considered to be removed.