   Chapter 14, Problem 119E

Chapter
Section
Textbook Problem

# Calculate the pH of each of the following solutions.a. 0.10 M CH3NH3Clb. 0.050 M NaCN

(a)

Interpretation Introduction

Interpretation: The pH value for each of the given solutions to be calculated.

Concept introduction: The pH of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The pOH of a solution is calculated by the formula, pOH=log[OH]

The sum, pH+pOH=14

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

Ka=ConcentrationofproductsConcentrationofreactants

The value of Kw is calculated by the formula,

Kw=KaKb

To determine: The pH value for each of the given solution of 0.10M CH3NH3Cl .

Explanation

Explanation

The equilibrium constant expression for the given reaction is, Ka=[CH3NH2][H+][CH3NH3+]

CH3NH3+ is a stronger acid than H2O .

The dominant equilibrium reaction is,

CH3NH3+(aq)CH3NH2(aq)+H+(aq)

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

Ka=ConcentrationofproductsConcentrationofreactants

Where,

• Ka is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

Ka=[CH3NH2][H+][CH3NH3+] (1)

The Ka value is 2.28×10-11M_ .

The value, Kw=KaKb

The value of Kb for methyl amine is 4.38×104 .

The value of Ka is calculated by the formula,

Ka=KwKb

Substitute the value of Kb in the above expression.

Ka=1.0×10144.38×104=2.28×10-11_

The [H+] is 1.50×10-6M_ .

The change in concentration of CH3NH3+ is assumed to be x .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

CH3NH3+(aq)CH3NH2(aq)+H+(aq)Inititialconcentration0

(b)

Interpretation Introduction

Interpretation: The pH value for each of the given solutions to be calculated.

Concept introduction: The pH of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The pOH of a solution is calculated by the formula, pOH=log[OH]

The sum, pH+pOH=14

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

Ka=ConcentrationofproductsConcentrationofreactants

The value of Kw is calculated by the formula,

Kw=KaKb

To determine: The pH value for each of the given solution of 0.050M NaCN .

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Find more solutions based on key concepts 