General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN: 9781305580343
Author: Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher: Cengage Learning
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Chapter 22, Problem 22.71QP
Interpretation Introduction

Interpretation:

The concentration of Cu2+(aq) , NH3(aq) and [Cu(NH3)4]2+(aq) at equilibrium has to be determined when 0.10 mol Cu2+(aq) and 0.40 mol NH3(aq) are made up to 1.00 L of solution.

Concept Introduction:

Consider a ligand ‘X’ co-ordinates to central metal ion ‘A’, and the reaction is represented as,

A + 6X [AX6]n±

Equilibrium constant Kc or dissociation constant, Kd for formation of a complex compound by complexation reaction is represented as,

Kc = Kd = [A][X]6[AX6]n±

Expert Solution & Answer
Check Mark

Answer to Problem 22.71QP

Concentration of Cu2+(aq) at equilibrium is calculated as 6.1×104 M.

Concentration of NH3(aq) at equilibrium is calculated as 2.4×10-3 M.

Concentration of [Cu(NH3)4]2+(aq) at equilibrium is calculated as 0.10 M.

Explanation of Solution

The reaction between Cu2+(aq) , NH3(aq) and [Cu(NH3)4]2+(aq) can be represented as,

Cu2+(aq)+4NH3(aq) [Cu(NH3)4]+(aq)

Given that dissociation constant for the above reaction is Kd = 2.1×1013 also Kc = Kf = 2.1×1013

The concentration of each reactant species and the product at equilibrium is given as,

Concentration(M)     Cu2+(aq)      NH3(aq)    [Cu(NH3)4]+(aq)  _Initial                           0.10             0.40                          0Change                          -x               -4x                          +xEquilibrium                 0.10 - x     0.40-4x                        x

  As we know,

Kc = Kd = 2.1×1013=[[Cu(NH3)4]2+][Cu2+][NH3]4

Substitute the concentration Cu2+(aq) , NH3(aq) and [Cu(NH3)4]2+(aq) values in the above equation,

Kc = 2.1×1013=(0.10x)(4×(0.10x))4x

To solve the above equation, let y=0.10x and then x=0.10y .  Therefore the above equation becomes,

2.1×1013 = y(4y)4(0.10y)

Numerical value of ‘y’ is negligible as it is lesser than 0.10 . Hence 0.10y 0.10 Rearranging the above equation and solving for y,

2.1×1013 = y(4y)4(0.10y)y5 = (2.1×1013)(0.10)44 = 8.203×1017y = 8.203×10175 = 6.064×104

Therefore, at equilibrium,

Concentration of Cu2+(aq) ,

[Cu2+] = (0.10 M - x)= 6.064×104 M = 6.1×104M   

Concentration of NH3(aq) ,

  [NH3] = 4(0.10x)            = 4y = 4×6.064×104M           = 2.4×103M

Concentration of [Cu(NH3)4]2+= x = (0.10y)=0.10M

Conclusion

Concentration of product and reactant species can be determined knowing the dissociation constant of the reaction.

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Chapter 22 Solutions

General Chemistry - Standalone book (MindTap Course List)

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