0.11 M KCN Fe^3+ Fe(CN)6^3- 8.5×10^(-40) M 1.5×10^(-3) M

0.11 M KCN Fe^3+ Fe(CN)6^3- 8.5×10^(-40) M 1.5×10^(-3) M

 

 

In the presence of CN^-, Fe^3+ forms the complex ion Fe(CN)6^3-.

The equilibrium concentrations of Fe^3+ and Fe(CN)6^3- are

8.5×10^(-40) M and 1.5×10^(-3) M, respectively, in a 0.11 M KCN solution.

Calculate the value for the overall formation constant of Fe(CN)6^3-.

Fe^3+(aq) + 6CN^-(aq) ⇌ Fe(CN)6^3-(aq) ... K_overall = ?

 

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Fe^3+(aq) + 6CN^-(aq) ⇌ Fe(CN)6^3-(aq)

 

Kf = [Fe(CN)6^3-] / [Fe^3+] [CN^-]^6

 

 

 

 

평형에서,

[Fe^3+] = 8.5×10^(-40) M

[Fe(CN)6^3-] = 1.5×10^(-3) M

[CN^-] = 0.11 M = [KCN]

 

 

 

Kf = (1.5×10^(-3)) / [(8.5×10^(-40)) (0.11)^6]

= 9.96×10^41