Chemistry
Chemistry
9th Edition
ISBN: 9781133611097
Author: Steven S. Zumdahl
Publisher: Cengage Learning
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Chapter 15, Problem 59E

Consider the titration of 100.0 mL of 0.200 M acetic acid (Ka = 1.8 × 10−5) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added.

a. 0.0 mL

b. 50.0 mL

c. 100.0 mL

d. 150.0 mL

e. 200.0 mL

f. 250.0 mL

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The titration of Acetic acid with different volumes of KOH is given. The pH of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of pH of solution when 0.0mL KOH has been added to it.

Answer to Problem 59E

The value of pH of solution when 0.0mL KOH has been added is 2.72_ .

Explanation of Solution

Explanation

The concentration of H+ .is 0.00189M_ .

Given:

The concentration of Acetic acid is 0.200M .

The concentration of KOH is 0.100M .

The volume of Acetic acid is 100.0mL .

The volume of KOH is 0.0mL .

The value of Ka of Acetic acid is 1.8×105

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 100mL into L is done as,

100mL=100×0.001L=0.100L

The concentration of any species is given as,

Concentration=NumberofmolesVolumeofsolutioninlitres (1)

Rearrange the above equation to obtain the value of number of moles.

Numberofmoles=Concentration×Volumeofsolutioninlitres (2)

Substitute the concentration and volume of CH3COOH in the above equation as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.200M×0.1L=0.02moles

Make the ICE table for the reaction between CH3COOH and KOH .

CH3COOH+KOHCH3COOK++H2OInitial moles:0.0200Change:00+0.02Finalmoles:0.0200.02

The above equation shows the presence of equilibrium condition in the solution.

Make the ICE table for the dissociation reaction of CH3COOH .

CH3COOHCH3COOH+Initial(M):0.200Change(M):xxxEquilibrium(M):0.2xxx

The equilibrium ratio for the given reaction is,

Ka=[CH3COO][H+][CH3COOH]

Substitute the calculated concentration values in the above expression.

Ka=[CH3COO][H+][CH3COOH]1.8×105=(x)(x)(0.2x)M

Since, value of Ka is very small, hence, (0.2x) is taken as (0.2) .

Simplify the above equation,

1.8×105=(x)(x)(0.2x)M1.8×105=(x)(x)0.2Mx=0.00189M_

It is the concentration of H+ .

Explanation

The value of pH of solution when 0.0mL KOH has been added is 2.72_ .

The pH of the solution is calculated by the formula,

pH=log[H+]

Where,

  • [H+] is the concentration of Hydrogen ions.

Substitute the value of [H+] in the above equation.

pH=log[H+]=log(0.00189)=2.72_

The value of pH of solution when 0.0mL KOH has been added is. 2.72_ .

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The titration of Acetic acid with different volumes of KOH is given. The pH of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of pH of solution when 50.0mL KOH has been added to it.

Answer to Problem 59E

The value of pH of solution when 50.0mL KOH has been added is. 4.26_ .

Explanation of Solution

Explanation

The concentration of [H+] is. 5.5×10-5M_ .

Given

The volume of KOH is 50.0mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 50mL into L is done as,

50mL=50×0.001L=0.05L

Substitute the value of concentration and volume of KOH in equation (2) as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100M×0.05L=0.005moles

Make the ICE table for the reaction between CH3COOH and KOH .

CH3COOH+KOHCH3COOK++H2OInitial moles:0.020.0050Change:0.0050.005+0.005Finalmoles:0.01500.005

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution =VolumeofCH3COOH+VolumeofKOH=0.1L+0.05L=0.15L

Substitute the value of number of moles of CH3COOH and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.015moles0.15L=0.1M

Substitute the value of number of moles of CH3COO and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.005moles0.15L=0.033M

Make the ICE table for the dissociation reaction of CH3COOH .

CH3COOHCH3COOH+Initial(M):0.10.0330Chang(M):x+xxEquilibrium(M):0.1x0.033+xx

The equilibrium ratio for the given reaction is,

Ka=[CH3COO][H+][CH3COOH]

Substitute the calculated concentration values in the above expression.

Ka=[CH3COO][H+][CH3COOH]1.8×105=(0.033+x)(x)(0.1x)M

Since, value of Ka is very small, hence, (0.1x) is taken as (0.1) and (0.033+x) is taken as 0.033 .

Simplify the above equation,

1.8×105=(0.033)(x)(0.1)M1.8×106=(0.033)(x)x=5.5×10-5M_

It is the concentration of H+ .

Explanation

The value of pH of solution when 50.0mL KOH has been added is. 4.26_ .

The pH of the solution is calculated by the formula,

pH=log[H+]

Substitute the value of [H+] in the above equation.

pH=log[H+]=log(5.5×105)=4.26_

The value of pH of solution when 50.0mL KOH has been added is. 4.26_ .

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The titration of Acetic acid with different volumes of KOH is given. The pH of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of pH of solution when 100.0mL KOH has been added to it.

Answer to Problem 59E

The value of pH of solution when 100.0mL KOH has been added is. 4.74_ .

Explanation of Solution

Explanation

The [H+] is. 1.8×10-5M_ .

Given

The volume of KOH is 100.0mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 100mL into L is done as,

100mL=100×0.001L=0.1L

Substitute the value of concentration and volume of KOH in equation (2) as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100M×0.1L=0.01moles

Make the ICE table for the reaction between CH3COOH and KOH .

CH3COOH+KOHCH3COOK++H2OInitial moles:0.020.010Change:0.010.01+0.01Finalmoles:0.0100.01

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution =VolumeofCH3COOH+VolumeofKOH=0.1L+0.1L=0.2L

Substitute the value of number of moles of CH3COOH and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.01moles0.2L=0.05M

Substitute the value of number of moles of CH3COO and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.01moles0.2L=0.05M

Make the ICE table for the dissociation reaction of CH3COOH .

CH3COOHCH3COOH+Initial(M):0.050.050Change(M):x+xxEquilibrium(M):0.05x0.05+xx

The equilibrium ratio for the given reaction is,

Ka=[CH3COO][H+][CH3COOH]

Substitute the calculated concentration values in the above expression.

Ka=[CH3COO][H+][CH3COOH]1.8×105=(0.05+x)(x)(0.05x)M

Since, value of Ka is very small, hence, (0.05x) is taken as (0.05) and (0.05+x) is taken as 0.05 .

Simplify the above equation,

1.8×105=(0.05)(x)(0.05)Mx=1.8×10-5M_

It is the concentration of H+ .

Explanation

The value of pH of solution when 100.0mL KOH has been added is. 4.74_ .

The pH of the solution is calculated by the formula,

pH=log[H+]

Substitute the value of [H+] in the above equation.

pH=log[H+]=log(1.8×105)=4.74_

The value of pH of solution when 100.0mL KOH has been added is. 4.74_ .

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The titration of Acetic acid with different volumes of KOH is given. The pH of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of pH of solution when 150.0mL KOH has been added to it.

Answer to Problem 59E

The value of pH of solution when 150.0mL KOH has been added is 5.22_ .

Explanation of Solution

Explanation

The [H+] is 0.6×10-5M_ .

Given

The volume of KOH is 150.0mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 150mL into L is done as,

150mL=150×0.001L=0.15L

Substitute the value of concentration and volume of KOH in equation (2) as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100M×0.15L=0.015moles

Make the ICE table for the reaction between CH3COOH and KOH .

CH3COOH+KOHCH3COOK++H2OInitial moles:0.020.0150Change:0.0150.015+0.015Finalmoles:0.00500.015

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution =VolumeofCH3COOH+VolumeofKOH=0.1L+0.15L=0.25L

Substitute the value of number of moles of CH3COOH and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.005moles0.25L=0.02M

Substitute the value of number of moles of CH3COO and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.015moles0.25L=0.06M

Make the ICE table for the dissociation reaction of CH3COOH .

CH3COOHCH3COOH+Initial(M):0.020.060Change(M):x+xxEquilibrium(M):0.02x0.06+xx

The equilibrium ratio for the given reaction is,

Ka=[CH3COO][H+][CH3COOH]

Substitute the calculated concentration values in the above expression.

Ka=[CH3COO][H+][CH3COOH]1.8×105=(0.06+x)(x)(0.02x)M

Since, value of Ka is very small, hence, (0.02x) is taken as (0.02) and (0.06+x) is taken as 0.06 .

Simplify the above equation,

1.8×105=(0.06)(x)(0.02)Mx=0.6×10-5M_

It is the concentration of H+ .

Explanation

The value of pH of solution when 150.0mL KOH has been added is 5.22_ .

The pH of the solution is calculated by the formula,

pH=log[H+]

Substitute the value of [H+] in the above equation.

pH=log[H+]=log(0.6×105)=5.22_

The value of pH of solution when 150.0mL KOH has been added is 5.22_ .

(e)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The titration of Acetic acid with different volumes of KOH is given. The pH of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of pH of solution when 200.0mL KOH has been added to it.

Answer to Problem 59E

The value of pH of solution when 200.0mL KOH has been added is 8.8_ .

Explanation of Solution

Explanation

The [H+] is 0.61×10-5M_ .

Given

The volume of KOH is 200.0mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 200mL into L is done as,

200mL=200×0.001L=0.2L

Substitute the value of concentration and volume of KOH in equation (2) as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100M×0.2L=0.02moles

Make the ICE table for the reaction between CH3COOH and KOH .

CH3COOH+KOHCH3COOK++H2OInitial moles:0.020.020Change:0.020.02+0.02Finalmoles:000.02

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution =VolumeofCH3COOH+VolumeofKOH=0.1L+0.2L=0.3L

Substitute the value of number of moles of CH3COOK+ and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.02moles0.3L=0.067M

The reaction is now written as,

CH3COOK++H2OCH3COOH+KOH

Make the ICE table for the above reaction.

CH3COOK++H2OCH3COOH+KOHInitial moles:0.067000Change:x0+xxFinalmoles:0.067x0xx

The equilibrium ratio for the given reaction is,

Kb=[CH3COOH][KOH][CH3COOK+]

The value of Kb for acetate ion is 5.6×1010 .

Substitute the calculated concentration values in the above expression.

Kb=[CH3COOH][KOH][CH3COOK+]5.6×1010=(x)(x)(0.067x)M

Since, value of Kb is very small, hence, (0.067x) is taken as (0.067) .

Simplify the above equation,

5.6×1010=(x)(x)(0.067)Mx=0.61×10-5M_

It is the concentration of H+ .

Explanation

The value of pH of solution when 200.0mL KOH has been added is 8.8_ .

The pOH of the solution is calculated by the formula,

pOH=log[OH]

Where,

  • [OH] is the concentration of Hydroxide ions.

Substitute the value of [OH] in the above equation.

pOH=log[OH]=log(0.61×105)=5.21

The relationship between pOH is given as,

pH+pOH=14

Substitute the value of pOH in the above equation as,

pH+pOH=14pH+5.21=14pH=8.8_

The value of pH of solution when 200.0mL KOH has been added is 8.8_ .

(f)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The titration of Acetic acid with different volumes of KOH is given. The pH of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of pH of solution when 250.0mL KOH has been added to it.

Answer to Problem 59E

The value of pH of solution when 250.0mL KOH has been added is 12.15_ .

Explanation of Solution

Explanation

The [OH] is 0.014M_ .

Given

The volume of KOH is 250.0mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 250mL into L is done as,

250mL=250×0.001L=0.25L

Substitute the value of concentration and volume of KOH in equation (2) as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100M×0.25L=0.025moles

Make the ICE table for the reaction between CH3COOH and KOH .

CH3COOH+KOHCH3COOK++H2OInitial moles:0.020.0250Change:0.020.02+0.02Finalmoles:00.0050.02

There is presence of excess Hydroxide ions in solution.

Total volume of solution =VolumeofCH3COOH+VolumeofKOH=0.1L+0.25L=0.35L

Substitute the value of number of moles of KOH and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.005moles0.35L=0.014M_

Explanation

The value of pH of solution when 250.0mL KOH has been added is 12.15_ .

The pOH of the solution is calculated by the formula,

pOH=log[OH]

Substitute the value of [OH] in the above equation.

pOH=log[OH]=log(0.014)=1.85

The relationship between pOH is given as,

pH+pOH=14

Substitute the value of pOH in the above equation.

pH+pOH=14pH+1.85=14pH=12.15_

The value of pH of solution when 250.0mL KOH has been added is 12.15_ .

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Chapter 15 Solutions

Chemistry

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