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