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Copper(I) ions in aqueous solution react with NH3 according to

Cu^+ + 2NH3 --> Cu(NH3)2^+ Kf=6.3x10^10

Calculate the solubility (in g/L) of CuBr (Ksp=6.3x10^-9) in 0.81 M NH3.

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  1. CuBr ==> Cu^+ + Br^-
    Cu^+ + 2NH3 ==> Cu(NH3)2^+
    ----------------------------
    add equn 1 to eqn 2.
    ...CuBr(s) + 2NH3 ==>Cu(NH3)2^+ + Br^-
    I.............0.81M...0............0
    C............-2x......x............x
    E...........0.81-2x...x............x

    Keq for the rxn shown is Ksp*Kf
    Substitute from the ICE chart and solve for x = [Cu(NH3)2] = (Br^-) = solubility of CuBr in M. Then convert to g/L.

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