   # The weak base methylamine, CH 3 NH 2 , has K b = 4 .2 × 10 − 4 . It reacts with water according to the equation. CH 3 NH 2 (aq) + H 2 O( l ) ⇌ CH 3 NH 3 + (aq) + OH − ( aq ) Calculate the equilibrium hydroxide ion concentration in a 0.25 M solution of the base. What are the pH and pOH 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 55PS
Textbook Problem
323 views

## The weak base methylamine, CH 3 NH 2 , has K b = 4 .2 × 10 − 4 . It reacts with water according to the equation. CH 3 NH 2 (aq) + H 2 O( l )  ⇌ CH 3 NH 3 + (aq) + OH − ( aq ) Calculate the equilibrium hydroxide ion concentration in a 0.25 M solution of the base. What are the pH and pOH of the solution?

Interpretation Introduction

Interpretation:

The equilibrium hydroxide ion concentration in a 0.25 M solution of the base has to be calculated. pH and pOH 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

Given chemical reaction:

CH3NH2(aq) + H2O(l) CH3NH3+(aq) + OH(aq)

The equilibrium expression:

kb[OH-][CH3NH3+][CH3NH2]

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.

CH3NH2(aq) + H2O(l) CH3NH3+(aq) + OH(aq)I         0.25                --                --                   --C          -x                  --              + x                  +xE     (0.25 - x)           --                 x                    x

Kb of CH3NH2 is 4.2 × 10-4

At equilibrium, x = [OH-] [CH3NH3+]

kb[OH-][CH3NH3+][CH3NH2]

4

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