Posted by Erin on Monday, April 4, 2011 at 1:35am.
What the problem means is that the solubility is increased due to the formation of the Cu(NH3)4^+2 ion and in fact that concn is greater than that of the Cu^+2 by itself. Here is what I would do.
Cu(OH)2 ==> Cu^+2 + 2OH^-
Cu^+2 + 4NH3 ==> Cu(NH3)4^+2
Write the Ksp expression for Cu(OH)2.
Write the K for the complex. That is
K = [Cu(NH3)4^+2]/(Cu^+2)(NH3)^4 and I would include the NH3 equilibrium also.
That is NH3 + HOH ==> NH4^+ + OH^-
Write Kb for NH3.
I would do an ICE chart for Cu(OH)2 in which Cu(OH)2 = S (for solubility)
S = (Cu^+2)+[Cu(NH3)4^+2]
You also know that total NH3 = 0.52 and that is
0.52 = (NH4^+) + NH3 + [Cu(NH3)4^+2].
You solve these two equations together with the 3 above you wrote (Ksp, K, and Kb) for S. I worked through the problem and obtained about 0.0027 M for solubility if I didn't make an error somewhere.
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