Question

Find the solubility of CuI in 0.48 M KCN solution. The Ksp of CuI is 1.1×10−12 and the Kf for the Cu(CN)2− complex ion is 1×1024.

Answer 1

        CuI    <---> Cu^+ + I^-   Ksp

Cu^+ + 2CN^- <---> Cu(CN)2^-      Kf
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   CuI + 2CN^- <---> Cu(CN)2^- + I^-   Kc
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          CuI + 2CN^- <---> Cu(CN)2^- + I^-

initial        0.48 M          0         0

Change          -2x             +x       +x

equil        0.48-2x            x         x

   Kc = [Cu(CN)2^-][I^-]/[CN^-]^2

   Kc = X^2/(0.48-2X)

   Kc = Ksp*Kf

   kC = (1.1*10^-12*1*10^24) = 1.1*10^12

1.1*10^12 = X^2/(0.48-2X)^2

X = 0.24

X= solubility of CuI = 0.24 M