Question

In the presence of CN-, Fe3 forms the complex ion Fe(CN)63-. The equilibrium concentrations of Fe3 and Fe(CN)63- are 8.5 x 10-40 M and 1.5 x 10-3 M, respectively, in a 0.11 M KCN solution. Calculate the value for the overall formation constant Kf of Fe(CN)63-.

Answer 1

Answer:

Kf = 9.96x10⁴¹

Explanation:

When Fe³⁺ and CN⁻ are in water, complex Fe(CN)₆³⁻ is formed, thus:

Fe³⁺ + 6CN⁻ ⇄ Fe(CN)₆³⁻

Kf is defined as:

Kf = [Fe(CN)₆³⁻] / [Fe³⁺] [CN⁻]⁶

Equilibrium concentration of ions is:

[Fe(CN)₆³⁻] = 1.5x10⁻³M

[Fe³⁺] = 8.5x10⁻⁴⁰

[CN⁻] = [KCN] = 0.11M

Replacing in Kf expression:

Kf = [1.5x10⁻³M] / [8.5x10⁻⁴⁰] [ 0.11M]⁶

Kf = 9.96x10⁴¹

Answer 2

The value for the overall formation constant K-f of F-e(C-N)63  is K-f = 9.96x10⁴¹.

Calculation of the value of overall formation constant K-f:

At the time When F-e³⁺ and C-N⁻ are in the water, complex F-e(C-N)₆³⁻ is formed,

So, the following reaction should be

F-e-³⁺ + 6-C-N⁻ ⇄ F-e(C-N)₆³⁻

K-f should be defined as:

K-f = [F-e(C-N)₆³⁻] / [F-e³⁺] [C = N⁻]⁶

Now

The equilibrium concentration of ions is:

[F-e(C-N)₆³⁻] = 1.5x10⁻³ M

[F-e³⁺] = 8.5x10⁻⁴⁰

[C-N⁻] = [K-C-N] = 0.11 M

Now we have to Replace in Kf expression:

So,

K-f = [1.5 x 10⁻³ M] / [8.5 x 10⁻⁴⁰] [ 0.11 M]⁶

K-f = 9.96 x 10⁴¹

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