Question

Consider the following equilibrium: Cu2+ + 4NH3 Cu(NH3)42+ (Kf = 4.8e+12)

A solution is made by mixing 19.0 mL of 1.30 M CuSO4 and 1.50 L of 0.330 M NH3. Assume additive volumes to answer the following questions. Give all answers to three sig figs.

a) What is the concentration of NH3 in the resulting solution?

[NH3] = ______ M

b) What is the concentration of Cu(NH3)42+ in the resulting solution?

[Cu(NH3)42+] = ______M

c) What is the concentration of Cu2+ in the resulting solution?

[Cu2+] = ______M

Answer 1

Let us initial concentration

0.019 L x 1.3 M = 0.0247 mol CuSO4 or [Cu²⁺]

1.5 L x 0.33 M = 0.495 mol NH₃

Draw ICE Table

	Cu2+	NH3	Cu(NH3)42+
Initial	0.0247	0.495	0
Change	0.0247	0.0988	0.0247
Equi	0	0.3962	0.0247

Concentration of NH3

[NH3]/ total volume
0.3962 mol / 1.519L = 0.2608 M

Concentration of Cu(NH₃)₄²⁺

0.0247 / 1.519 = 0.016260 M

Kf = [products] / [reactants]
Kf = Cu(NH₃)₄²⁺ / Cu²⁺ x NH₃

Now, find Cu²⁺

Cu²⁺ = 0.016260 / 4.8 x 10¹² (0.2608)⁴

Cu²⁺ = 0.7322 x 10^-12 M