Question

A 1.00-L solution contains 2.00×10^-4 M Cu(NO3)2 and 1.40×10^-3 M ethylenediamine (en). The Kf for Cu(en)22+ is 1.00 × 10^20.

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

Answer 1

Answer: Approximately 3.92 × 10^-15 M.

Step-by-step explanation:

The reaction that takes place is:

Cu2+ + 2en ↔ Cu(en)22+

Let's assume that x is the concentration of Cu2+ that reacts with en to form Cu(en)22+.

Therefore, the equilibrium concentrations are:

[Cu2+] = 2.00×10^-4 - x [en] = 1.40×10^-3 - 2x [Cu(en)22+] = x

The expression for Kf is:

Kf = [Cu(en)22+] / ([Cu2+][en]^2)

Substituting the equilibrium concentrations into the Kf expression gives:

1.00 × 10^20 = x / ((2.00×10^-4 - x)(1.40×10^-3 - 2x)^2)

Since Kf is extremely large, we can assume that x is very small compared to the initial concentrations of Cu2+ and en. Therefore, we can approximate the equation by ignoring x in the denominator:

1.00 × 10^20 ≈ x / ((2.00×10^-4)(1.40×10^-3)^2)

Solving for x gives:

x ≈ 1.00 × 10^20 * (2.00×10^-4)(1.40×10^-3)^2 x ≈ 3.92 × 10^-15 M

Thus, the concentration of Cu2+(aq) in the solution is approximately 3.92 × 10^-15 M.