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Calculate the equilibrium concentrations of NH 3 , Cu 2+ , Cu(NH 3 ) 2+ , Cu(NH 3 ) 2 2+ , Cu(NH 3 ) 3 2+ , and Cu(NH 3 ) 4 2+ 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(NO 3 ) 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

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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Chapter
Section
BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 15, Problem 102CP
Textbook Problem
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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

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.

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

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Chapter 15 Solutions

Chemistry: An Atoms First Approach
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