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Calculate the molar concentration of uncomplexed Zn2+ (aq) in a solution that contains 0.22 mol of Zn(NH3)4 2+ per liter and 0.3109 M NH3 at equilibrium. Kf for Zn(NH3)4 2+ is
2.9 X 10^9

I started this way...

(0.3109 + 4)2/0.22-x)= 2.9 X 10^9

  • Chemistry -

    I think it helps to draw a picture.
    ......Zn^2+ + 4NH3 ==> Zn(NH3)4 +

    The problem gives the equilibrium conditions; therefore, you need not worry about a reaction.

    Kf = [Zn(NH3)4}^2+/(Zn^2+)(NH3)^4
    (Zn^2+) = x
    [Zn(NH3)4]^2+ = 0.22M
    (NH3) = 0.3109
    Solve for (Zn^2+) which is the only unknown in the equation.

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