Chemistry
Chemistry
9th Edition
ISBN: 9781133611097
Author: Steven S. Zumdahl
Publisher: Cengage Learning
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Chapter 16, Problem 102CP

Calculate the equilibrium concentrations of NH3, Cu2+, Cu(NH3)2+, Cu(NH3)22+, Cu(NH3)32+, and Cu(NH3)42+ in a solution prepared by mixing 500.0 mL of 3.00 M NH3 with 500.0 mL of 2.00 × 10−3 M Cu(NO3)2. The stepwise equilibria are

Cu 2 + ( a q ) + NH 3 ( a q ) CuNH 3 2 + ( a q ) K 1 = 1.86 × 10 4 CuNH 3 2 + ( a q ) + NH 3 ( a q ) Cu ( NH 3 ) 2 2 + ( a q ) K 2 = 3.88 × 10 3 Cu ( NH 3 ) 2 2 + ( a q ) + NH 3 ( a q ) Cu ( NH 3 ) 3 2 + ( a q ) K 3 = 1.00 × 10 3 Cu ( NH 3 ) 2 2 + ( a q ) + NH 3 ( a q ) Cu ( NH 3 ) 4 2 + ( a q ) K 4 = 1.55 × 10 2

Expert Solution & Answer
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Interpretation Introduction

Interpretation: The equilibrium concentrations of NH3 , Cu2+ , Cu(NH3)2+ , Cu(NH3)22+ , Cu(NH3)32+ and Cu(NH3)42+ in a solution prepared by mixing 500mL of 3MNH3 with 500mL of 2×103MCu(NO3)2 is to be calculated.

Concept introduction: The solubility product is the mathematical product of a substance’s dissolved ion concentration raised to its power of its stoichiometric coefficients. When sparingly soluble ionic compound releases ions in the solution, it gives relevant solubility product. The solvent is generally water.

Answer to Problem 102CP

The concentration of NH3 is 1.50M_ .

The concentration of Cu2+ is 1.77×10-17M_ .

The concentration of Cu(NH3)2+ is 4.93×10-13M_ .

The concentration of Cu(NH3)22+ is 2.866×10-9M_ .

The concentration of Cu(NH3)32+ is 4.30×10-6M_ .

The concentration of Cu(NH3)42+ is 1×10-3M_ .

Explanation of Solution

Explanation

To determine: The equilibrium concentrations of NH3 , Cu2+ , Cu(NH3)2+ , Cu(NH3)22+ , Cu(NH3)32+ and Cu(NH3)42+ in a solution prepared by mixing 500mL of 3MNH3 with 500mL of 2×103MCu(NO3)2 .

The initial concentration of Cu2+ is 1×10-3M_ and the initial concentration of NH3 is 1.50M_ .

The concentration of Cu2+ is calculated by the formula,

[Cu2+]initial=(VolumeofCu(NO3)2solution)×(ConcentrationofCu(NO3)2solution)(VolumeofCu(NO3)2solution+VolumeofNH3solution)

Where,

  • [Cu2+]initial is the initial concentration of Cu2+ .

The volume of Cu(NO3)2 solution is 500mL .

The volume of NH3 solution is 500mL .

The concentration of Cu(NO3)2 solution is 2×103M .

Substitute the values of volume of solutions and concentration of Cu(NO3)2 solution in the above formula.

[Cu2+]initial=(500mL)×(2×103M)(500mL+500mL)=1×10-3M_

The concentration of NH3 is calculated by the formula,

[NH3]initial=(VolumeofNH3solution)×(ConcentrationofNH3solution)(VolumeofCu(NO3)2solution+VolumeofNH3solution)

Where,

  • [NH3]initial is the equilibrium concentration of NH3 .

The volume of Cu(NO3)2 solution is 500mL .

The volume of NH3 solution is 500mL .

The concentration of NH3 solution is 3M .

Substitute the values of volume of solutions and concentration of NH3 solution in the above formula.

[NH3]initial=(500mL)×(3M)(500mL+500mL)=1.50M_

The concentration of Cu(NH3)42+ is 1×10-3M_ .

Explanation

The given stepwise equilibria for the formation of complex ion is as follows,

Cu2+(aq)+NH3(aq)CuNH32+(aq)K1=1.86×104 (1)

CuNH32+(aq)+NH3(aq)Cu(NH3)22+(aq)K2=3.88×103 (2)

Cu(NH3)22+(aq)+NH3(aq)Cu(NH3)32+(aq)K3=1.00×103 (3)

Cu(NH3)32+(aq)+NH3(aq)Cu(NH3)42+(aq)K4=1.55×102 (4)

Since the equilibrium constants for every reaction is high and there is an excess of NH3 , this means that the reaction goes to completion and the net reaction in solution is,

Cu2+(aq)+4NH3(aq)Cu(NH3)42+(aq)

The concentration of the species in the solution is calculated by the table given below.

Cu2+(aq)+4NH3(aq)Cu(NH3)42+(aq)Beforereaction1×103M1.50M0Afterreaction01.50M4(1×103)M1.50M1×103M

The entire Cu2+ will form Cu(NH3)42+ because reaction goes to completion. Hence the

concentration of Cu(NH3)42+ is 1×10-3M_ .

The concentration of Cu(NH3)32+ is 4.30×10-6M_ .

The equilibrium constant for equation (4) is calculated by the formula,

K4=[Cu(NH3)42+][Cu(NH3)32+][NH3]

Where,

  • K4 is the equilibrium constant of equation (4).

Substitute the value of K4 and the concentration of Cu(NH3)42+ and NH3 in the above formula.

1.55×102=[1×103][Cu(NH3)32+][1.50][Cu(NH3)32+]=4.30×10-6M_

Therefore, the concentration of Cu(NH3)32+ is 4.30×10-6M_ .

The concentration of Cu(NH3)22+ is 2.866×10-9M_ .

The equilibrium constant for equation (3) is calculated by the formula,

K3=[Cu(NH3)32+][Cu(NH3)22+][NH3]

Where,

  • K3 is the equilibrium constant of equation (3).

Substitute the value of K3 and the concentration of Cu(NH3)32+ and NH3 in the above formula.

1×103=[4.30×106][Cu(NH3)22+][1.50][Cu(NH3)22+]=2.866×10-9M_

Therefore, the concentration of Cu(NH3)22+ is 2.866×10-9M_ .

The concentration of CuNH32+ is 4.93×10-13M_ .

The equilibrium constant for equation (2) is calculated by the formula,

K2=[Cu(NH3)22+][CuNH32+][NH3]

Where,

  • K2 is the equilibrium constant of equation (2).

Substitute the value of K2 and the concentration of Cu(NH3)22+ and NH3 in the above formula.

3.88×103=[2.87×109][CuNH32+][1.50][CuNH32+]=4.93×10-13M_

Therefore, the concentration of CuNH32+ is 4.93×10-13M_ .

The concentration of Cu2+ is 1.77×10-17M_ .

The equilibrium constant for equation (1) is calculated by the formula,

K1=[CuNH32+][Cu2+][NH3]

Where,

  • K1 is the equilibrium constant of equation (1).

Substitute the value of K2 and the concentration of Cu(NH3)22+ and NH3 in the above formula.

1.86×104=[4.93×1013][Cu2+][1.50][Cu2+]=1.77×10-17M_

Therefore, the concentration of Cu2+ is 1.77×10-17M_ .

Conclusion

Conclusion

The concentration of NH3 is 1.50M_ .

The concentration of Cu2+ is 1.77×10-17M_ .

The concentration of Cu(NH3)2+ is 4.93×10-13M_ .

The concentration of Cu(NH3)22+ is 2.866×10-9M_ .

The concentration of Cu(NH3)32+ is 4.30×10-6M_ .

The concentration of Cu(NH3)42+ is 1×10-3M_ .

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Chapter 16 Solutions

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