Question #b667a

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Question #b667a

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Answer 1 · 2017-09-02T02:33:10

No.

Explanation:

Equilibrium will favor the most stable outcome. One way to approach this is to assume that all species ARE in solution, and then see which reactions of combinations produce the most stable result.

Barium carbonate is also essentially insoluble in water, and only soluble in acids.

The solubility product of $C u C O_{3}$ $CuCO_3$ is $3.8 \times 10^{- 9}$ $3.8 xx 10^-9$

The solubility product of $B a C O_{3}$ $BaCO_3$ is $5.1 \times 10^{- 9}$ $5.1 xx 10^-9$
http://www.csudh.edu/oliver/chemdata/data-ksp.htm
SO, even IF we had both metals in solution and added carbonate ion, the $C u C O_{3}$ $CuCO_3$ would precipitate out more completely. THUS, there is no driving force to exchange the carbonate from the copper with the barium ion in solution.

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