   # K f for the complex ion Ag(NH 3 ) 2 + is 1.7 × 10 7 . K sp for AgCl is 1.6 × 10 −10 . Calculate the molar solubility of AgCl in 1.0 M NH 3 . ### Chemistry: An Atoms First Approach

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

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 15, Problem 73E
Textbook Problem
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## Kf for the complex ion Ag(NH3)2+ is 1.7 × 107. Ksp for AgCl is 1.6 × 10−10. Calculate the molar solubility of AgCl in 1.0 M NH3.

Interpretation Introduction

Interpretation: The molar solubility of AgCl in 1.0M NH3 is to be calculated.

Concept introduction: Concept introduction: Solubility is defined as the maximum amount of solute that can dissolve at a certain amount of solvent at certain temperature. The solubility product, Ksp is the equilibrium constant that is applied when salt partially dissolve in a solvent. The solubility product of dissociation of AxBy is calculated by the formula,

Ksp=[A]x[B]y

### Explanation of Solution

Explanation

To determine: The molar solubility of AgCl in 1.0M NH3 .

The molar solubility of AgCl in 1.0M NH3 is 4.7×10-2mol/L_ .

Given

The molarity of NH3 is 1.0M .

The overall formation constant for Ag(NH3)2+ is 1.7×107 .

The stated reactions are,

AgClAg++ClKsp(AgI)=1.6×1010Ag++2NH3Ag(NH3)2+Kf=1.7×107

Add the two equations. The final reaction is

AgCl(s)+2NH3Ag(NH3)2++Cl

The overall equilibrium constant for the reaction is calculated by the formula,

K=KspKf

Where,

• Ksp is the solubility product of AgI .
• Kf is the overall formation constant.

Substitute the value of Ksp and Kf in the above expression.

K=(1.6×1010)(1.7×107)=2.72×103

At equilibrium, the equilibrium constant expression is expressed by the formula,

K=ConcentrationofproductsConcentrationofreactants

Where,

• K is the equilibrium constant

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