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-.

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

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i  

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