# Calculate the equilibrium constant for the following reaction. Z n ( O H ) 2 ( s ) + 2 C N − ( a q ) ⇄ Z n ( C N ) 2 ( s ) + 2 O H − ( a q ) Does the equilibrium lie predominantly to the left or to the right?

### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

Chapter
Section

### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 17, Problem 92GQ
Textbook Problem
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## Calculate the equilibrium constant for the following reaction. Z n ( O H ) 2 ( s ) + 2   C N − ( a q ) ⇄ Z n ( C N ) 2 ( s ) + 2   O H − ( a q ) Does the equilibrium lie predominantly to the left or to the right?

Interpretation Introduction

Interpretation:

The value of the equilibrium constant Knet has to be calculated for the given reaction.

Zn(OH)2(s)+2CN1(aq)Zn(CN)2(s)+2OH1(aq)

Also predict whether the reaction will be product favoured or reactant favoured.

Concept introduction:

Some metal ions when present in an aqueous solution containing anions or neutral species called Lewis base or ligands having a tendency to donate electron pairs to metal ions then complex ion formation will take place.

Example of metal ions that form complex ions includes,Cd2+,Fe2+,Zn2+,Ni2+ etc.

Example of Lewis bases includes,NH3,OH etc.

The complex ion remains in equilibrium with the metal ion and the ligand called complex ion formation equilibrium and the equilibrium constant is called as formation constant Kf.

A larger value of Kf implies that the complex ion formed is more stable. Kf is the measure of the strength of the interaction between the metal ions and the Lewis base to form the complex ion.

For example for general complex ion formation reaction,

xM+yL[MxLy]

Kf can be given as

Kf=[MxLy][M]x[L]y

Here,

• [MxLy] is the equilibrium concentration of complex ion.
• [M] is the equilibrium concentration of metal ion.
• [L] is the equilibrium concentration of the ligand.
• x and y are the coefficients of metal ion and ligand respectively.

Complex ions are stable and thus formation of these increase the solubility of the salt containing the metal ions same as in complex ions. Effect of complex ion formation on the solubility of salt can be explained as below,

AgBr when dissolved in water does not dissolve completely and dissociates as follows,

AgBr(s)KspAg+(aq)+Br(aq) (1)

For this reaction Ksp expression is,

Ksp=[Ag+][Br]

Ag+ ions are capable of forming complex with ligand so when aqueous ammonia solution (strong ligand) is added to the saturated solution of AgBr, Ag+ ions present in the solution form complex with NH3.

Ag+(aq)+2NH3(aq)Kf[Ag(NH3)2]+(aq) (2)

The expression for formation constant Kf is given as,

Kf=[[Ag(NH3)2]+][Ag+][NH3]2

Complex formation leads to the decrease in the concentration of the Ag+ ions in the solution, as a result, according to Le Chatelier’s principle the equilibrium in equation (1) move in the forward direction producing more of the Ag+ ions and the solubility of the slightly soluble salt AgBr increases.

Adding two equilibrium equation (1) and (2) new overall equilibrium constant can be defined.

Net chemical equation:

Ag+(aq)+2NH3(aq)Knet[Ag(NH3)2]+(aq)+Br(aq) (3)

Net equilibrium constant can be given as,

Knet=(Ksp)(Kf) (4)

### Explanation of Solution

The value of equilibrium constant, Knet for the given reaction is calculated below.

Given:

Refer to the Appendix J in the textbook for the value of Ksp .

The value of solubility product constant, Ksp for Zn(OH)2 is 3.0×1017 .

The value of solubility product constant, Ksp for Zn(CN)2 is 8.0×1012 .

Zn(OH)2 dissociates as follows in water,

Zn(OH)2(s)KspZn2+(aq)+2OH(aq) (5)

Zn2+ ions present in solution react with the cyanide ions added to the solution to form Zn(CN)2.

Zn2+(aq)+2CN1(aq)KfZn(CN)2(s) (6)

Kf for this reaction is equal to reciprocal of the Ksp value for Zn(CN)2.

Kf=1(8.0×1012)=1.25×1011

Add equation (5) and (6) to find the net chemical equation

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