Chemistry 102: Acid Equilibrium: Find concentration of Cu^2+ in the solution

Question

A 1.00-L solution contains 2.75×10^-4 M  Cu(NO₃)₂ and 1.80×10^-3 M  ethylenediamine (en). The K_f for Cu(en)₂²⁺ is 1.00 × 10²⁰.

Cu2+(aq)+2en(aq)Cu(en)2+2​(aq)

Kf​=[Cu(en)2+2​]/[Cu2+][en]2​

What is the concentration of Cu²⁺(aq) in the solution?

Answer 1

Given Data:

Initial Concentration of Cu²⁺ = 2.75 x 10^-4 M.

Initial Concentration of en = 1.80 x 10^-3 M.

Recall that Molarity= Moles of solute / Liters of Solution

Therefore, in a 1.00 L solution we have;

(2.75 x 10^-4 M) (1.00 L) = 0.000275 moles of Cu²⁺

(1.80 x 10^-3 M) (1.00 L) = 0.00180 moles of en

According to the reaction equilibrium, Cu2+(aq)+2en(aq)Cu(en)2+2​(aq)

1 mole of Cu²⁺ will react with 2 moles of en.

So 0.000275 moles of Cu²⁺ will react with 0.00055 moles of en.

At eq. there will be 0.00180 moles en - 0.00055 moles en = 0.00125 moles of en remaining as reactant.

(0.00125 moles) / (1.00 L) = 0.00125 M en.

Now we can use Kf​=[Cu(en)2+2​]/[Cu2+][en]2​ and The K_f for Cu(en)₂²⁺ is 1.00 × 10²⁰.

2.75 x 10^-4 M Cu(NO₃)₂ forms 2.75 x 10^-4 M Cu(en)₂²⁺

1.00 x 10²⁰ = [2.75 x 10^-4] / [Cu²⁺] x [0.00125]²

[Cu²⁺] = [2.75 x 10^-4] / 1.00 x 10²⁰ x [0.00125]²

[Cu²⁺] = 1.76 x 10^-18 M