Combining Ksp and Kf?

Can someone please EXPLAIN how I would do this?

Formation of a complex ion can often be used as a way to dissolve an insoluble material. The insoluble salt cadmium phosphate has a Ksp = 2.53 x 10^-33. The free metal cation, Cd^2+, will form a complex ion with 4 ions of CN^-. The Kf for forming the [Cd(CN)4]^2- complex ion = 6.00 x 10^18.

What is the overall equilibrium constant for this combined equation?

K=(Ksp)(Kf) so I have been multiplying 2.53 x 10^-33 by 6.00 x 10^18 but it's wrong.

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  • 9 years ago
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    When you combine two equilibrium reactions to an overall reaction, you cannot just multiply their equilibrium constants to obtain overall equilibrium constant. You have take into account that overall reaction must be balanced.

    The two reactions are

    (i) dissolution

    Cd₃(PO₄)₂(s) ⇄ 3 Cd²⁺(aq) + 2 PO₄³⁻(aq)

    Ksp = [Cd²⁺]³∙[PO₄³⁻]²

    (ii) complex formation

    Cd²⁺(aq) + 4 CN⁻(aq) ⇄ [Cd(CN)₄]²⁻(aq)

    Kf = [ [Cd(CN)₄]²⁻]² / ( [Cd²⁺] ∙ [CN⁻]⁴ )

    To get a balanced overall equation without Cd²⁺, you have to "add" three times of reaction (ii) to reaction (i):

    Cd₃(PO₄)₂(s) + 12 CN⁻(aq) ⇄ 3 [Cd(CN)₄]²⁻(aq) + 2 PO₄³⁻(aq)

    So the overall equilibrium equals Ksp time Kf cubed.

    (Note that Keq for a reaction multiplied by factor n equals Keq of the original reaction raised to the power of n)

    You can easily show this by examining the overall equilibrium constant:

    K = [[Cd(CN)₄]²⁻]³ ∙ [PO₄³⁻]² / [CN⁻]¹²

    = [Cd²⁺]³ ∙ [PO₄³⁻]² ∙ [[Cd(CN)₄]²⁻]³ / ( [Cd²⁺]³ ∙ [CN⁻]¹² )

    = { [Cd²⁺]³∙[PO₄³⁻]² } ∙ { [[Cd(CN)₄]²⁻] / ( [Cd²⁺] ∙ [CN⁻]⁴ ) }³

    = Ksp ∙ Kf³

    Hence

    K = 2.53×10⁻³³ ∙ (6.00×10¹² )³ = 5.46×10⁵

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