Question

A 1.00 L solution contains 2.75x10^-4 M Cu(NO3)2 and 1.80x10^-3 M ethylenediamine (en) . The Kf for Cu(en)_2 ^2+ is 1x20^20 . what is the concentrationof Cu^2+ (aq) in the solution?

Answer 1

A 0.000275 M of Cu(NO3)2 and a 0.00180 M of ethlenediamine(en) forms a solution

Cu²⁺ + 2 en -------> Cu(en)₂²⁺

From the formuale given Cu(en)₂²⁺, we will get that Cu(NO3)2 reacts with twice as much mole of en

So a 0.000275 M of Cu(NO3)2 reacts with= 0.00055 M (en)

Now the unreacted en = 0.00180 M of en - 0.00055 M of reacted (en)

So the unreacted en = 0.00125 M of en

And the 0.000275 M Cu(NO3)2 reacts to form 0.000275 M Cu(en)₂²⁺

Now for the given Kf Cu(en)₂²⁺ is 1x10²⁰

To calculate for Cu²⁺(aq), the Kf = Product / Reactant

Kf = [Cu(en)₂ 2+] / [Cu²⁺] [en]²

1 X 10²⁰ = [ 0.000275] /[ Cu²⁺] [ 0.00125]²

Cu²⁺ = 0.000275 / (1 X 10²⁰) (0.00125)²

Cu²⁺ = 1.76 X 10^-18 M

Note - You have given the Kf value of Cu(en)₂²⁺ as 1 X 20²⁰ which i think is a typing mistake & should be 1 X 10²⁰