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In aqueous solution the fe 2 ion forms a complex with six cyanide anions. Write the formation constant expression for the equilibrium between the hydrated metal ion and the aqueous complex. Under that, write the balanced chemical equation for the last step in the formation of the complex.

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Answer:

Kf = [[Fe(CN)₆]⁻⁴] / [Fe²⁺] [CN⁻]⁶

Last step: [Fe(CN)₅]³⁻(aq)  + CN⁻(aq) ⇄  [Fe(CN)₆]⁻⁴(aq)

Explanation:

When the Fe²⁺ ion is in presence of cyanide anions, CN⁻, in an aqueous solution, the hexahydrated complex is formed, thus:

Fe²⁺(aq) + CN⁻(aq) ⇄ [Fe(CN)]⁺(aq)                    K₁

[Fe(CN)]⁺(aq)  + CN⁻(aq) ⇄  [Fe(CN)₂](aq)          K₂

[Fe(CN)₂](aq)  + CN⁻(aq) ⇄  [Fe(CN)₃]⁻(aq)         K₃

[Fe(CN)₃]⁻(aq)  + CN⁻(aq) ⇄  [Fe(CN)₄]²⁻(aq)      K₄

[Fe(CN)₄]²⁻(aq)  + CN⁻(aq) ⇄  [Fe(CN)₅]³⁻(aq)     K₅

[Fe(CN)₅]³⁻(aq)  + CN⁻(aq) ⇄  [Fe(CN)₆]⁻⁴(aq)   K₄ → Last step.

Global reaction:

Fe²⁺(aq) + 6CN⁻(aq) ⇄ [Fe(CN)₆]⁻⁴(aq)

And Kf is defined as:

Kf = [[Fe(CN)₆]⁻⁴] / [Fe²⁺] [CN⁻]⁶

The balanced chemical equation should be

K-f = [[F-e(C-N)₆]⁻⁴] / [F-e²⁺] [C-N⁻]⁶

And,

[F-e(C-N)₅]³⁻(aq)  + C-N⁻(aq) ⇄  [F-e(C-N)₆]⁻⁴(a-q)

Balanced chemical equation:

At the time When the F-e²⁺ ion is in presence of cyanide anions, C-N⁻, should be an aqueous solution, the hexahydrated complex should be created

So,

F-e²⁺(-aq) + C-N⁻(a-q) ⇄ [F-e(C-N)]⁺(a-q)                    K₁

[F-e(C-N)]⁺(a-q)  + C-N⁻(a-q) ⇄  [F-e(C-N)₂](a-q)          K₂

[F-e(C-N)₂](a-q)  + C-N⁻(a-q) ⇄  [F-e(C-N)₃]⁻(a-q)         K₃

[F-e(C-N)₃]⁻(a-q)  + C-N⁻(a-q) ⇄  [F-e(C-N)₄]²⁻(a-q)      K₄

[F-e(C-N)₄]²⁻(a-q)  + C-N⁻(a-q) ⇄  [F-e(C-N)₅]³⁻(a-q)     K₅

[F-e(C-N)₅]³⁻(a-q)  + C-N⁻(a-q) ⇄  [F-e(C-N)₆]⁻⁴(a-q)   K₄ → Last step.

Now

Global reaction:

F-e²⁺(a-q) + 6-C-N⁻(a-q) ⇄ [F-e(C-N)₆]⁻⁴(a-q)

And K-f should be K-f = [[F-e(C-N)₆]⁻⁴] / [F-e²⁺] [C-N⁻]⁶

Learn more about equation here: https://brainly.com/question/19119593

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