Question

The copper(I) ion forms a chloride salt that has Ksp = 1.2 10-6. Copper(I) also forms a complex ion with Cl ‾.

Cu+(aq) + 2 Cl ‾(aq) reverse reaction arrow CuCl2‾(aq) K = 8.7 104

Calculate the solubility of copper(I) chloride in pure water. (Ignore CuCl2‾ formation)

Answer 1

CuCl --> Cu+ & Cl-

Ksp = [Cu+2] [Cl-]

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Cu+(aq) + 2 Cl ‾(aq) ==> CuCl2 ‾(aq)

K = [CuCl2 ‾] / [Cu+2] [Cl-]^2


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(K) (Ksp) = [Cu+2] [Cl-] times [CuCl2 ‾] / [Cu+2] [Cl-]^2

reduces to:
(K) (Ksp) = [CuCl2 ‾] / [Cl-]

(8.7 x 10^4) (1.2 x 10^-6. ) = [CuCl2 ‾] / [0.13]

(8.7 x 10^4) (1.2 x 10^-6. ) = [CuCl2 ‾] / [0.13]

1.044 e-1 = [CuCl2 ‾] / [0.13]

dissolved copper = 0.01357 Molar

your answer
with 2 sig figs is
0.014 Molar Copper in solution
I think this is right.

Answer 2

I think Dan's solution takes into account the formation of the CuCl2^- complex but the problem says to ignore that. If we ignore it, the only thing we have is
CuCl ==> Cu^+ + Cl^-
Ksp + (Cu^+)(Cl^-) = 1.2E-6
Substitute x = Cu^+ and x = Cl^- and solve for x.