With the displacement reaction of the chloride ions and sulphate ions in excess by carbonate ions, how would there be any copper left in solution?
There should be not (edit: Ksp of the carbonate is in the 10-10
range, so with an excess of CO32-
I would expect something like nM solution, negligible) - but if the solution is blue, probably some are left. Most likely in some complexed form (otherwise it should precipitate), which suggests to me solution contained something more that just pure copper chloride and sulfate.
Is there just copper ions in solution once the copper carbonate precipitate comes out?
In general - assuming solution contained only ions mentioned earlier - you remove Cu, but everything else is there. Cl-
, excess CO32-
(not to mention HCO3-
from the hydrolysis, plus OH-
; there will be also traces of other ions - like H+
- but their amounts will be orders of magnitude lower). Charge balance says something like
] + [H+
] = [Cl-
] + 2×[SO42-
] + [HSO4-
] + 2×[CO32-
] + [HCO3-
] + [OH-
(not that it matters much in this particular case, but is a general and important property of such solutions, technically traces of copper should be also included on the left, together with several forms of complexed copper ions - all in trace quantities).
By washing the precipitate out I get that I remove the soluble ions from the filtrate which would be Na+, Cl- and SO42- ions, as well as some Cu2+ ions given the solution is still blue right?
More or less yes.
Note that what you precipitate is not a simple copper carbonate, it is kind of a mixture of basic copper carbonates with typically a slightly variable composition.