   # Calculate the hydronium ion concentration and pH in a 0.20 M solution of ammonium chloride NH 4 Cl . ### Chemistry & Chemical Reactivity

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

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 16, Problem 59PS
Textbook Problem
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## Calculate the hydronium ion concentration and pH in a 0.20 M solution of ammonium chloride NH 4 Cl .

Interpretation Introduction

Interpretation:

The hydronium ion concentration and pH in a 0.20 M solution of ammonium chloride (NH4Cl) has to be calculated.

Concept introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka is acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb is base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

Ion product constant for wter  Kw= [H3O+][OH-]       =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]Reltion between pH and pOH pH + pOH =14

### Explanation of Solution

Ammonium chloride dissociates into ammonium ion and chloride ions.The symbols of these ions are NH4+ and Cl- respectively.

Chloride ions does n’t affect for the pH of the solution But the pH of the solution is completely depends upon the ammonium ions (NH4+) conjugated acid of weak base is (NH3).

Ammonium ion reacts with the water, the chemical equilibrium reaction is as follows.

NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)

The equilibrium expression for the above reaction.

Ka[NH3][H3O+][NH4+]

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.

NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)   I         0.20               --              --                   --C         -x                  --            +x                  +x    E      (0.20-x)           --              x                    x

Ka of ammonia is 5

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