[hellowurld]

I am confused as to how to start this problem.  I'm trying to find the solubility of CuCl in a .35 M solution of CaCl.  Do I add the two together?  But when I do, won't the product be the same thing?  I know the solubility equilibrium, and I can figure it out once I get started so...how do I start that?

[Borek]

Common ion effect. Depending on the CuCl Kso you may probably safely assume CaCl₂ is the only source of Cl^- in the solution, although in general Cl^- comes from two sources. Just one of them can be neglected.

[Donaldson Tan]

Are you dealing with Copper(I) Chloride or Copper(II) Chloride?

I am definitely sure that CaCl actually refers to CaCl₂ because Calcium only exhibits one oxidation state when it is a compound. Unfortunately, this isn't true for Copper. However, Copper(I) ions tend to disproportionate in the presence of water to form a mixture of Copper(II) ions and Copper metal

2Cu⁺ (aq) -> Cu²⁺ (aq) + Cu (s) (aq).

[woelen]

In the presence of excess chloride ion, copper (I) tends to form the CuCl2(-) complex ion.

I have some solid CuCl. It does not dissolve in water. When some NaCl is added, then it does dissolve fairly easily, especially if a lot of NaCl is added. The liquid then quickly becomes very dark, almost black, due to aerial oxidation of the CuCl2(-) complex and the subsequent formation of very dark mixed-oxidation state complexes of copper (I) and copper (II).