Question

Part 1:

Copper (II) forms a complex with four molecules of ammonia according to the following:

Cu2+(aq) + NH3(aq) --> Cu(NH3)2+(aq)    K1 = 1.9*104

Cu(NH3)2+(aq) + NH3(aq) --> Cu(NH3)22+(aq) K2 = 3.9*103

Cu(NH3)22+(aq) + NH3(aq) --> Cu(NH3)32+(aq) K3 = 1.0*103

Cu(NH3)32+(aq) + NH3(aq) --> Cu(NH3)42+(aq) K4 = 1.5*102

As such, the overall reaction for this process is the following:

Cu2+(aq) + 4NH3(aq) --> Cu(NH3)42+(aq) Kf = ?

What is the overall K for this reaction?

Part 2:

An experiment is created where 0.169 moles of Cu2+ and 0.761 moles of NH3 are combined in water to form 1.00 L of solution. These react according to the process in part 1. Create an ICE table and react the two substances COMPLETELY. After, determine which molecule is in excess, and enter the molarity of this substance below. Hint: what is the stoichiometry of the overall reaction?

Part 3:

Create another ICE table, this time beginning with the end conditions of ALL substances in the reaction in part 2. Allow this reaction to reach equilibrium, and determine the equilibrium concentration of the substance that was the limiting reactant in part 2. Enter this concentration below.

Hints: once more, pay special attention to the stoichiometry of the reaction. Also, which substances will you include in this K expression?

Answer 1

1. K = K1*K2*K3*K4

K = 1.9*10⁴ * 3.9*10³ * 1.0*10³ * 1.5*10²

   K = 1.11 * 10¹³

2. molarity of Cu²⁺ = 0.169 mol/1 L = 0.169 M

molarity of NH3 = 0.761 mol/ 1L = 0.761 M

[Cu]²⁺(aq) + 4 NH₃(aq) <----------> [Cu(NH₃)₄]²⁺(aq)

I 0.169 0.761 0

C -0.169 -4*0.169 +0.169

E 0 0.676 +0.169

3. [Cu(NH₃)₄]²⁺(aq) <-----------> Cu²⁺(aq) + 4NH₃(aq)

I 0.169 0 0.676

C -x +x +4x

E 0.169-x +x 0.676 + 4x

Cu²⁺ is limiting reagent, because its number of moles less than NH3.

K = [Cu²⁺] [NH₃]⁴ /  [Cu(NH₃)₄²⁺]

     = (x)(0.676+4x)⁴ / 0.169-x

  K = 1.11 * 10¹³ =  (x)(0.676)⁴ / 0.169-x

1.11 * 10¹³ =  (x) 0.21 / 0.169

x = 1.11 * 10¹³ * 0.169 / 0.21

x = 8.93 * 10¹² = [Cu²⁺]