   Chapter 15, Problem 25E

Chapter
Section
Textbook Problem

# 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

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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