Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: hellowurld on May 31, 2007, 11:23:10 PM
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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?
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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.
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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).
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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).