A solution contains 0.20 M Cl- and 0.20 M SO4^2-.
a) identify a cation that could be added to the solution to give a precipitate with only one of these anions.
MY answer: Cu+
b) write the net ionic equation for the precipitation reaction in part a)
MY answer: Cl^- + Cu^+ ==> Cu^+ + Cl^-
thank you, Im not very sure how to do b...
To determine the cation that could be added to the solution to give a precipitate with only one of the anions, you need to identify a cation that forms an insoluble compound with one of the anions while remaining soluble with the other. In this case, you want to find a cation that can form an insoluble compound with either Cl- or SO4^2-.
a) The cation Cu+ (copper) can be added to the solution to give a precipitate with only one of the anions. Copper forms an insoluble compound with Cl- called copper(I) chloride (CuCl), while it remains soluble with SO4^2-.
b) To write the net ionic equation for the precipitation reaction in part a), you need to first write the molecular equation and then convert it to the net ionic equation by eliminating the spectator ions.
The molecular equation for the reaction of Cu+ with Cl- in the solution would be:
Cu+(aq) + 2Cl-(aq) → CuCl(s)
Now, let's write the net ionic equation by eliminating the spectator ions (ions that appear on both sides of the equation):
Cu+(aq) + 2Cl-(aq) → CuCl(s)
In this equation, both Cu+ and Cl- appear on both sides, so they are spectator ions. Therefore, the net ionic equation is:
No net ionic equation exists (since there are no ions that are directly involved in the reaction).
It's important to note that the net ionic equation is not applicable in this scenario because the reaction does not involve any ions directly. The precipitate formation is due to the solubility rules rather than an actual reaction with ions.