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chemistry

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calculate Cu2+ in a 0.15M CuSO2 (aq) solution that is also 6.5M in free NH3.

Cu^2+(aq)+ 4NH3(aq)<=> [Cu(NH3)4]^2+ kf=1.1*10^13

  • chemistry - ,

    Do you mean CuSO4?
    Kf = 1.1E13 = [Cu(NH3)4]^2+/(Cu^2+)(NH3)^4.

    Do this ICE chart in two steps. Here is the first one. With Kf so large the reaction goes to completion (almost) like this.
    ...Cu^2+ + NH3 ==>Cu(NH3)2+4
    I..0.15.....6.5.......0
    C..-0.15...-0.6.....+0.15
    E....0......5.9......0.15

    Now redo the ICE chart BUT showing the small amount of reverse reaction like this.
    ....Cu^2+ + 4NH3 ==> Cucomplex^2+
    I...0.......5.9.......0.15
    C...+x......+4x........-x
    E...x......5.9+4x.....0.15-x

    Now substitute this last E line into the kf expression and solve for x = (Cu^2+)
    Post your work if you get stuck.

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