   # What are the equilibrium concentrations of NH 3 , NH 4 + and OH ‒ in a 0.15M solution of ammonia? What is the pH of the solution? ### 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 53PS
Textbook Problem
30 views

## What are the equilibrium concentrations of NH 3 , NH 4 + and OH‒ in a 0.15M solution of ammonia? What is the pH of the solution?

Interpretation Introduction

Interpretation:

The equilibrium concentrations of NH3,NH4+ and OH- in a 0.15M solution of ammonia has to be calculated and the pH of the solution also 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]

### Explanation of Solution

The Ammonia reacts with water. The equilibrium reaction is as follows.

NH3(aq) + H2O (l) NH4+(aq) + OH-(aq)

The equilibrium expression,

Kb(NH3) = [NH4+][OH-][NH3]

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.

NH3 (aq) +  H2O (l)  NH4+ (aq) +  OH-(aq)I         0.15                --                --                   --C          -x                 --              + x                  +xE     (0.15 - x)          --                 x                    x

Kb of ammonia is 1.8 × 10-5

At equilibrium, x = [OH-] [NH4+]

Kb(NH3) = [NH4+][OH-][NH3]

1.8×105 = (x)(x)0.15- x1.8×10-5 = (x)20

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