# Calculate the pH after 0.020 mole of HCl is added to 1.00 L of each of the four solutions in Exercise 21.

### Chemistry: An Atoms First Approach

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

Chapter
Section

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 14, Problem 25E
Textbook Problem
25 views

## Calculate the pH after 0.020 mole of HCl is added to 1.00 L of each of the four solutions in Exercise 21.

(a)

Interpretation Introduction

Interpretation:

The pH value after 0.020mole of HCl is added to 1.00L of each of the given four solutions in Exercise 21 is 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 pH of a solution is calculated by the formula, pH=log[H+]

The sum, pH+pOH=14

To determine: The pH value after 0.020mole of HCl is added to 1.00L of 0.100M propionic acid.

### Explanation of Solution

Explanation

To find the concentration of HCl

The given number of moles of HCl is 0.020mol .

The volume of the solution is 1.00L .

The concentration is calculated by the formula,

Concentration=NumberofmolesVolumeofthesolution

Substitute the value of the given number of moles of HCl and the volume of the solution in the above expression.

Concentration=0.020mol1.0L=0.020M_

To find the concentration of [H+]

The dominant equilibrium reaction that takes place in the given case is,

HC3H5O2(aq)C3H5O2(aq)+H+(aq)

The change in the concentration of HC3H5O2 is assumed to be x .

The ICE table is formed for the given reaction.

HC3H5O2(aq)H+(aq)+C3H5O2(aq)Initialconcentration0.1000.0200Changex+x+xEquilibriumconcentration0.100x0.020+xx

The equilibrium concentration of [HC3H5O2] is (0.100x)M .

The equilibrium concentration of [C3H5O2] is xM .

The equilibrium concentration of [H+] is (0.020+x)M .

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

Ka=ConcentrationofproductsConcentrationofreactants

Where,

• Ka is the acid dissociation constant

(b)

Interpretation Introduction

Interpretation:

The pH value after 0.020mole of HCl is added to 1.00L of each of the given four solutions in Exercise 21 is 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 pH of a solution is calculated by the formula, pH=log[H+]

The sum, pH+pOH=14

To determine: The pH value after 0.020mole of HCl is added to 1.00L of 0.100M sodium propionate

(c)

Interpretation Introduction

Interpretation:

The pH value after 0.020mole of HCl is added to 1.00L of each of the given four solutions in Exercise 21 is 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 pH of a solution is calculated by the formula, pH=log[H+]

The sum, pH+pOH=14

To find: The pH

(d)

Interpretation Introduction

Interpretation:

The pH value after 0.020mole of HCl is added to 1.00L of each of the given four solutions in Exercise 21 is 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 pH of a solution is calculated by the formula, pH=log[H+]

The sum, pH+pOH=14

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