   Chapter 17, Problem 91GQ

Chapter
Section
Textbook Problem

What is the equilibrium constant for the following reaction? A g C l ( s ) + I − ( a q ) ⇄ A g I ( s ) + C l − ( a q ) Does the equilibrium lie predominantly to the left or to the right? Will AgI form if iodide ion. I−, is added to a saturated solution of AgCl?

Interpretation Introduction

Interpretation:

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

AgCl(s)+I1(aq)AgI(s)+Cl1(aq)

Also whether the reaction will be product favoured or reactant favoured. Will AgI form or not has to be predicted.

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

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 AgCl is 1.8×1010 .

The value of solubility product constant, Ksp for AgI is 8.5×1017 .

AgCl dissociates as follows in water,

AgCl(s)KspAg+(aq)+Cl(aq) (5)

Ag+ ions present in solution react with the iodide ions added to the solution to form AgI.

Ag+(aq)+I1(aq)KfAgI(s) (6)

Kf for this reaction is equal to reciprocal of the Ksp value for AgI.

Kf=1(8.5×1017)=1.17×1016

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

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