Elements Of Physical Chemistry
Elements Of Physical Chemistry
7th Edition
ISBN: 9780198796701
Author: ATKINS, P. W. (peter William), De Paula, Julio
Publisher: Oxford University Press
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Chapter 5, Problem 5.42P

(a)

Interpretation Introduction

Interpretation:

The pH of 0.10MCH3COOH(aq) has to be determined. 

Concept Introduction:

For the titration of a weak acid with a strong base, the  pH before the equivalence point depends on the excess concentration of acid and the  pH after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so pH is 7.0.  The equivalence point occurs when total volume of base has been added. 

pH=log[H3O+]

pH=log[OH]

pH+pOH=14

Calculation of Moles:

1. No. of moles (n)=WM.W

Where,

W    = Weight of a given compoundM.W = Molecular weight of the same compound

2. No. of moles (n) = (Molarity)( Volume)

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

Molarity =moles of solutevolume of solution in L

(a)

Expert Solution
Check Mark

Answer to Problem 5.42P

The pH of 0.10MCH3COOH(aq) has been determined to be 2.8723.

Explanation of Solution

Given Data:

KaforCH3COOH=1.8×105[CH3COOH(aq)]=0.10M

The equilibrium can be written as

ConcentrationCH3COOH(aq)+H2O(l)CH3COO(aq)+H3O+(aq)_Initial0.10M00Changex+x+x_Final(0.10x)xx

The equilibrium constant can be represented for the above equation,

Ka=[CH3COO][H3O+][CH3COOH]=(x)(x)(0.10x)

The value of x is very small, so can be neglected in the denominator.  Then

Ka=x20.10x=(Ka)(0.10)=1.34164×103.[H3O+]=1.34164×103

pH=log[H3O+]=log(1.34164×103)=2.8723.

Therefore, the pH of 0.10MCH3COOH(aq) has been calculated as 2.8723.

(b)

Interpretation Introduction

Interpretation:

The pH of the solution after the addition of 10.0cm3 of 0.10MNaOH(aq) has to be  determined. 

Concept Introduction:

For the titration of a weak acid with a strong base, the  pH before the equivalence point depends on the excess concentration of acid and the  pH after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so pH is 7.0.  The equivalence point occurs when total volume of base has been added. 

pH=log[H3O+]

pH=log[OH]

pH+pOH=14

Calculation of Moles:

1. No. of moles (n)=WM.W

Where,

W    = Weight of a given compoundM.W = Molecular weight of the same compound

2. No. of moles (n) = (Molarity)( Volume)

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

Molarity =moles of solutevolume of solution in L

(b)

Expert Solution
Check Mark

Answer to Problem 5.42P

The pH of the solution after the addition of 10.0cm3 of 0.10MNaOH(aq) has been  determined to be 4.56.

Explanation of Solution

Given Data:

Volume of CH3COOH            = 25.00 ml=25.00×103LConcentration of  CH3COOH    = 0.10MConcentration ofNaOH= 0.10MVolumeofNaOHadded=10.00ml=10.00×103LThetotalvolumeofthesolutionatthispoint=25.00×103L+10.00×103L=35.00×103L

The initial number of moles ofCH3COOH=Molarity×Volume=(0.10M)(25.00×103L)=2.5×103.

MolesofaddedNaOH=Molarity×Volume=(0.10M)(10.00×103L)=1×103.

The reaction occurring in the titration is the neutralization of OH (from NaOH) by H3O+ (from CH3COOH):

NaOH(aq)+CH3COOH(aq)H2O(l)+CH3COONa(aq)H3O+(aq)+OH(aq)2H2O(l)

No.ofmolesCH3COOH(aq)+NaOH(aq)H2O(l)+CH3COONa(aq)_Initial2.5×1031.0×1030Change1.0×1031.0×103+1.0×103_Final1.5×10301.0×103

Ka=[CH3COO][H3O+][CH3COOH]=(1.0×103moles35×103L)(x)(1.5×103moles35×103L)x=1.8×105×1.5=2.7×105.[H3O+]=2.7×105

pH=log[H3O+]=log(2.7×105)=4.56.

Therefore, the  pH of the solution has been calculated to be 4.56

(c)

Interpretation Introduction

Interpretation:

The volume of 0.10MNaOH(aq) is required to reach halfway to the stoichiometric point has to be determined. 

Concept Introduction:

For the titration of a weak acid with a strong base, the  pH before the equivalence point depends on the excess concentration of acid and the  pH after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so pH is 7.0.  The equivalence point occurs when total volume of base has been added. 

(c)

Expert Solution
Check Mark

Answer to Problem 5.42P

The volume of 0.10MNaOH(aq) is required to reach halfway to the stoichiometric point has been determined to be 12.5mL.

Explanation of Solution

Given Data:

VolumeofCH3COOH=25.0cm3=25.0×103L[CH3COOH(aq)]=0.10M[NaOH(aq)]=0.10M

The reaction occurring in the titration is the neutralization of OH (from NaOH) by H3O+ (from CH3COOH):

NaOH(aq)+CH3COOH(aq)H2O(l)+CH3COONa(aq)H3O+(aq)+OH(aq)2H2O(l)

Shows that 1 equivalent of acid react with 1 equivalent of base.  For achieving equivalence point, we need the equivalent quantities of acid and base.  So according to the question 25.0mL of 0.10MNaOH(aq) is required to neutralize completely 25.0mL of 0.10MCH3COOH(aq).  The volume of 0.10MNaOH(aq) is required to reach halfway to the stoichiometric point is 12.5mL.

(d)

Interpretation Introduction

Interpretation:

The pH at the halfway point has to be determined. 

Concept Introduction:

For the titration of a weak acid with a strong base, the  pH before the equivalence point depends on the excess concentration of acid and the  pH after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so pH is 7.0.  The equivalence point occurs when total volume of base has been added. 

pH=log[H3O+]

(d)

Expert Solution
Check Mark

Answer to Problem 5.42P

The pH  at the halfway point has been determined to be 3.5.

Explanation of Solution

At the equivalence point, there is not an excess of either acid or base so pH is 7.0.  The pH  at the halfway point will be 3.5.

(e)

Interpretation Introduction

Interpretation:

The volume of 0.10MNaOH(aq) required to reach the stoichiometric point has to be determined. 

Concept Introduction:

For the titration of a weak acid with a strong base, the  pH before the equivalence point depends on the excess concentration of acid and the  pH after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so pH is 7.0.  The equivalence point occurs when total volume of base has been added. 

(e)

Expert Solution
Check Mark

Answer to Problem 5.42P

The volume of 0.10MNaOH(aq) required to reach the stoichiometric point has been determined to be 25.0mL.

Explanation of Solution

Given Data:

VolumeofCH3COOH=25.0cm3=25.0×103L[CH3COOH(aq)]=0.10M[NaOH(aq)]=0.10M

The reaction occurring in the titration is the neutralization of OH (from NaOH) by H3O+ (from CH3COOH):

NaOH(aq)+CH3COOH(aq)H2O(l)+CH3COONa(aq)H3O+(aq)+OH(aq)2H2O(l)

Shows that 1 equivalent of acid react with 1 equivalent of base.  For achieving equivalence point, we need the equivalent quantities of acid and base.  So according to the question 25.0mL of 0.10MNaOH(aq) is required to neutralize completely 25.0mL of 0.10MCH3COOH(aq).

(e)

Interpretation Introduction

Interpretation:

The pH at the stoichiometric point has to be calculated. 

Concept Introduction:

For the titration of a weak acid with a strong base, the  pH before the equivalence point depends on the excess concentration of acid and the  pH after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so pH is 7.0.  The equivalence point occurs when total volume of base has been added. 

pH=log[H3O+]

(e)

Expert Solution
Check Mark

Answer to Problem 5.42P

The pH at the stoichiometric point has  been calculated to be 7.0.

Explanation of Solution

Given Data:

VolumeofCH3COOH=25.0cm3=25.0×103L[CH3COOH(aq)]=0.10M[NaOH(aq)]=0.10M

The reaction occurring in the titration is the neutralization of OH (from NaOH) by H3O+ (from CH3COOH):

NaOH(aq)+CH3COOH(aq)H2O(l)+CH3COONa(aq)H3O+(aq)+OH(aq)2H2O(l)

Shows that 1 equivalent of acid react with 1 equivalent of base.  For achieving equivalence point, we need the equivalent quantities of acid and base.  So according to the question 25.0mL of 0.10MNaOH(aq) is required to neutralize completely 25.0mL of 0.10MCH3COOH(aq).  At the equivalence point, there is not an excess of either acid or base so pH is 7.0.

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Chapter 5 Solutions

Elements Of Physical Chemistry

Ch. 5 - Prob. 5D.1STCh. 5 - Prob. 5D.2STCh. 5 - Prob. 5D.3STCh. 5 - Prob. 5D.4STCh. 5 - Prob. 5D.5STCh. 5 - Prob. 5E.1STCh. 5 - Prob. 5E.2STCh. 5 - Prob. 5F.1STCh. 5 - Prob. 5F.2STCh. 5 - Prob. 5F.3STCh. 5 - Prob. 5F.4STCh. 5 - Prob. 5F.5STCh. 5 - Prob. 5G.1STCh. 5 - Prob. 5G.2STCh. 5 - Prob. 5G.3STCh. 5 - Prob. 5H.1STCh. 5 - Prob. 5H.2STCh. 5 - Prob. 5H.3STCh. 5 - Prob. 5I.1STCh. 5 - Prob. 5I.2STCh. 5 - Prob. 5I.3STCh. 5 - Prob. 5I.4STCh. 5 - Prob. 5I.5STCh. 5 - Prob. 5I.6STCh. 5 - Prob. 5J.1STCh. 5 - Prob. 5J.2STCh. 5 - Prob. 5J.3STCh. 5 - Prob. 5J.4STCh. 5 - Prob. 5J.5STCh. 5 - Prob. 5A.1ECh. 5 - Prob. 5A.2ECh. 5 - Prob. 5A.3ECh. 5 - Prob. 5A.4ECh. 5 - Prob. 5A.5ECh. 5 - Prob. 5A.6ECh. 5 - Prob. 5A.7ECh. 5 - Prob. 5A.8ECh. 5 - Prob. 5A.9ECh. 5 - Prob. 5A.10ECh. 5 - Prob. 5A.11ECh. 5 - Prob. 5A.12ECh. 5 - Prob. 5A.13ECh. 5 - Prob. 5B.1ECh. 5 - Prob. 5B.2ECh. 5 - Prob. 5B.3ECh. 5 - Prob. 5B.4ECh. 5 - Prob. 5B.5ECh. 5 - Prob. 5C.1ECh. 5 - Prob. 5C.2ECh. 5 - Prob. 5C.3ECh. 5 - Prob. 5C.4ECh. 5 - Prob. 5D.1ECh. 5 - Prob. 5D.2ECh. 5 - Prob. 5D.3ECh. 5 - Prob. 5D.4ECh. 5 - Prob. 5E.1ECh. 5 - Prob. 5E.2ECh. 5 - Prob. 5F.4ECh. 5 - Prob. 5G.1ECh. 5 - Prob. 5G.2ECh. 5 - Prob. 5G.3ECh. 5 - Prob. 5H.1ECh. 5 - Prob. 5H.2ECh. 5 - Prob. 5H.3ECh. 5 - Prob. 5H.4ECh. 5 - Prob. 5I.1ECh. 5 - Prob. 5I.2ECh. 5 - Prob. 5I.3ECh. 5 - Prob. 5I.4ECh. 5 - Prob. 5J.1ECh. 5 - Prob. 5J.2ECh. 5 - Prob. 5J.3ECh. 5 - Prob. 5J.4ECh. 5 - Prob. 5.1DQCh. 5 - Prob. 5.2DQCh. 5 - Prob. 5.3DQCh. 5 - Prob. 5.4DQCh. 5 - Prob. 5.5DQCh. 5 - Prob. 5.6DQCh. 5 - Prob. 5.7DQCh. 5 - Prob. 5.8DQCh. 5 - Prob. 5.9DQCh. 5 - Prob. 5.10DQCh. 5 - Prob. 5.11DQCh. 5 - Prob. 5.12DQCh. 5 - Prob. 5.13DQCh. 5 - Prob. 5.14DQCh. 5 - Prob. 5.15DQCh. 5 - Prob. 5.16DQCh. 5 - Prob. 5.17DQCh. 5 - Prob. 5.1PCh. 5 - Prob. 5.2PCh. 5 - Prob. 5.3PCh. 5 - Prob. 5.4PCh. 5 - Prob. 5.5PCh. 5 - Prob. 5.6PCh. 5 - Prob. 5.7PCh. 5 - Prob. 5.8PCh. 5 - Prob. 5.9PCh. 5 - Prob. 5.10PCh. 5 - Prob. 5.11PCh. 5 - Prob. 5.12PCh. 5 - Prob. 5.13PCh. 5 - Prob. 5.14PCh. 5 - Prob. 5.15PCh. 5 - Prob. 5.16PCh. 5 - Prob. 5.17PCh. 5 - Prob. 5.18PCh. 5 - Prob. 5.19PCh. 5 - Prob. 5.20PCh. 5 - Prob. 5.21PCh. 5 - Prob. 5.22PCh. 5 - Prob. 5.23PCh. 5 - Prob. 5.24PCh. 5 - Prob. 5.25PCh. 5 - Prob. 5.27PCh. 5 - Prob. 5.28PCh. 5 - Prob. 5.29PCh. 5 - Prob. 5.30PCh. 5 - Prob. 5.31PCh. 5 - Prob. 5.32PCh. 5 - Prob. 5.33PCh. 5 - Prob. 5.34PCh. 5 - Prob. 5.35PCh. 5 - Prob. 5.36PCh. 5 - Prob. 5.37PCh. 5 - Prob. 5.38PCh. 5 - Prob. 5.39PCh. 5 - Prob. 5.40PCh. 5 - Prob. 5.41PCh. 5 - Prob. 5.42PCh. 5 - Prob. 5.43PCh. 5 - Prob. 5.44PCh. 5 - Prob. 5.45PCh. 5 - Prob. 5.46PCh. 5 - Prob. 5.47PCh. 5 - Prob. 5.48PCh. 5 - Prob. 5.49PCh. 5 - Prob. 5.50PCh. 5 - Prob. 5.51PCh. 5 - Prob. 5.52PCh. 5 - Prob. 5.53PCh. 5 - Prob. 5.54PCh. 5 - Prob. 5.55PCh. 5 - Prob. 5.56PCh. 5 - Prob. 5.58PCh. 5 - Prob. 5.59PCh. 5 - Prob. 5.60PCh. 5 - Prob. 5.61PCh. 5 - Prob. 5.62PCh. 5 - Prob. 5.63PCh. 5 - Prob. 5.64PCh. 5 - Prob. 5.66PCh. 5 - Prob. 5.67PCh. 5 - Prob. 5.68PCh. 5 - Prob. 5.69PCh. 5 - Prob. 5.70PCh. 5 - Prob. 5.71PCh. 5 - Prob. 5.72PCh. 5 - Prob. 5.73PCh. 5 - Prob. 5.74PCh. 5 - Prob. 5.76PCh. 5 - Prob. 5.77PCh. 5 - Prob. 5.78PCh. 5 - Prob. 5.79PCh. 5 - Prob. 5.1PRCh. 5 - Prob. 5.2PRCh. 5 - Prob. 5.3PRCh. 5 - Prob. 5.4PR
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