The p H during the titration of 30.00 m L o f 0.150 M benzoic acid with 0.150 M N a O H after the addition of 0 , 10.00 , 30.00 , a n d 40.00 m L of the base. The titration curve has to be drawn and four regions of importance have to be labeled.

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Answer 1

Question

Chapter 16, Problem 16.52QE

Interpretation Introduction

Interpretation:

The pH during the titration of 30.00 mL of 0.150 M benzoic acid with 0.150 M NaOH after the addition of 0, 10.00, 30.00,and 40.00 mL of the base. The titration curve has to be drawn and four regions of importance have to be labeled.

Expert Solution & Answer

Explanation of Solution

Addition of 0 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Benzoic acid is a weak acid. The initial point in the titration when the base is not added to the system, the system is a weak acid system.

The iCe table can be set up to calculate the pH of the solution.

HA+H2O⇌H3O++A−i(M)0.15000C(M)−y+y+ye(M)0.150−yyy

Where, HA represents benzoic acid and A− represents the conjugate base of benzoic acid.

The Ka value for benzoic acid is 6.3×10−5. The expression for Ka can be written as given below,

Ka=[H3O+][A−][HA]=6.3×10−5y2(0.150−y)=6.3×10−5

This can be solved by approximation. If y≪0.150, then

y2≈6.3×10−5×0.150=0.945×10−5y=0.945×10−5=3.07×10−3=[H3O+]

The pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(3.07×10−3)=2.51.

Addition of 10 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

The neutralization reaction can be written as given,

HA(aq)+OH−(aq)→A−(aq)+H2O(l)

Calculation of milimole of acid and base:

Amount of HA=30.00 mL×(0.150 milimol H3O+mL)=4.5 milimolAmount of OH−=10.00 mL×(0.150 milimol OH−mL)=1.5 milimol

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is OH− ion.

HA+OH−→A−+H2Os(mmol)4.51.50ExcessR(mmol)−1.5−1.5+1.5+1.5f(mmol)3.001.5Excess

All the strong base is consumed. There are weak acid (HA) and its conjugate base (A−) in the solution. So this solution is an acidic buffer. By using Henderson-Hasselbalch equation, it’s pH can be calculated.

pH=pKa+lognbna=pKa+log1.53=−log(Ka)−0.3010=−log(6.3×10−5)−0.3010=4.20−0.3010=3.899.

Addition of 30.00 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Calculation of milimole of acid and base:

Amount of HA=30.00 mL×(0.150 milimol H3O+mL)=4.5 milimolAmount of OH−=30.00 mL×(0.150 milimol OH−mL)=4.5 milimolTotal volume=30.00+30.00=60.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

HA+OH−→A−+H2Os(mmol)4.54.50ExcessR(mmol)−4.5−4.5+4.5+4.5f(mmol)004.5Excessc(M)000.075

Both H3O+ and OH− ions have been completely consumed, the titration is at the equivalence point. There is only the conjugate base and water in the solution.

The iCe table can be set up to calculate the pH of the solution.

A−+H2O⇌HA+OH−i(M)0.07500C(M)−y+y+ye(M)0.075−yyy

The expression for Kb can be written as given below,

Kb=[HA][OH−][A−]=y2(0.075−y)

The value of Kb can be calculated as given below.

Kb=KwKa=1.0×10−146.3×10−5=0.16×10−9

This can be solved by approximation. If y≪0.075, then

y2≈0.16×10−9×0.075=0.012×10−9y=0.012×10−9=3.46×10−6=[OH−]

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(3.46×10−6)=5.46pH+pOH=14pH=14−pOH=14−5.46=8.54.

Addition of 40.00 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Calculation of milimole of acid and base and total volume:

Amount of HA=30.00 mL×(0.150milimol H3O+mL)=4.5 milimolAmount of OH−=40.00 mL×(0.150 milimol OH−mL)=6.00 milimolTotal volume=30.00+40.00=70.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

HA+OH−→A−+H2Os(mmol)4.56.000ExcessR(mmol)−4.5−4.5+4.5+4.5f(mmol)01.54.5Excessc(M)00.02140.0643

Now, the pH of the system can be calculated as given below.

pOH=−log[OH−]=−log(0.0214)=1.67pH+pOH=14pH=14−pOH=14−1.67=12.33.

Titration curve:

The titration curve is plotted between volume of base added and the corresponding pH values.

Volume of NaOH(mL)pH02.5110.003.9030.008.5440.0012.33

The titration curve with four important regions is given below.

Chemistry Principles And Practice, Chapter 16, Problem 16.52QE

Figure

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1. Consider the following molecular-level diagrams of a titration. O-HA molecule -Aion °° о ° (a) о (b) (c) (d) a. Which diagram best illustrates the microscopic representation for the EQUIVALENCE POINT in a titration of a weak acid (HA) with sodium. hydroxide? (e)

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Answer 2

Question

Chapter 16, Problem 16.52QE

Interpretation Introduction

Interpretation:

The pH during the titration of 30.00 mL of 0.150 M benzoic acid with 0.150 M NaOH after the addition of 0, 10.00, 30.00,and 40.00 mL of the base. The titration curve has to be drawn and four regions of importance have to be labeled.

Expert Solution & Answer

Explanation of Solution

Addition of 0 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Benzoic acid is a weak acid. The initial point in the titration when the base is not added to the system, the system is a weak acid system.

The iCe table can be set up to calculate the pH of the solution.

HA+H2O⇌H3O++A−i(M)0.15000C(M)−y+y+ye(M)0.150−yyy

Where, HA represents benzoic acid and A− represents the conjugate base of benzoic acid.

The Ka value for benzoic acid is 6.3×10−5. The expression for Ka can be written as given below,

Ka=[H3O+][A−][HA]=6.3×10−5y2(0.150−y)=6.3×10−5

This can be solved by approximation. If y≪0.150, then

y2≈6.3×10−5×0.150=0.945×10−5y=0.945×10−5=3.07×10−3=[H3O+]

The pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(3.07×10−3)=2.51.

Addition of 10 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

The neutralization reaction can be written as given,

HA(aq)+OH−(aq)→A−(aq)+H2O(l)

Calculation of milimole of acid and base:

Amount of HA=30.00 mL×(0.150 milimol H3O+mL)=4.5 milimolAmount of OH−=10.00 mL×(0.150 milimol OH−mL)=1.5 milimol

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is OH− ion.

HA+OH−→A−+H2Os(mmol)4.51.50ExcessR(mmol)−1.5−1.5+1.5+1.5f(mmol)3.001.5Excess

All the strong base is consumed. There are weak acid (HA) and its conjugate base (A−) in the solution. So this solution is an acidic buffer. By using Henderson-Hasselbalch equation, it’s pH can be calculated.

pH=pKa+lognbna=pKa+log1.53=−log(Ka)−0.3010=−log(6.3×10−5)−0.3010=4.20−0.3010=3.899.

Addition of 30.00 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Calculation of milimole of acid and base:

Amount of HA=30.00 mL×(0.150 milimol H3O+mL)=4.5 milimolAmount of OH−=30.00 mL×(0.150 milimol OH−mL)=4.5 milimolTotal volume=30.00+30.00=60.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

HA+OH−→A−+H2Os(mmol)4.54.50ExcessR(mmol)−4.5−4.5+4.5+4.5f(mmol)004.5Excessc(M)000.075

Both H3O+ and OH− ions have been completely consumed, the titration is at the equivalence point. There is only the conjugate base and water in the solution.

The iCe table can be set up to calculate the pH of the solution.

A−+H2O⇌HA+OH−i(M)0.07500C(M)−y+y+ye(M)0.075−yyy

The expression for Kb can be written as given below,

Kb=[HA][OH−][A−]=y2(0.075−y)

The value of Kb can be calculated as given below.

Kb=KwKa=1.0×10−146.3×10−5=0.16×10−9

This can be solved by approximation. If y≪0.075, then

y2≈0.16×10−9×0.075=0.012×10−9y=0.012×10−9=3.46×10−6=[OH−]

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(3.46×10−6)=5.46pH+pOH=14pH=14−pOH=14−5.46=8.54.

Addition of 40.00 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Calculation of milimole of acid and base and total volume:

Amount of HA=30.00 mL×(0.150milimol H3O+mL)=4.5 milimolAmount of OH−=40.00 mL×(0.150 milimol OH−mL)=6.00 milimolTotal volume=30.00+40.00=70.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

HA+OH−→A−+H2Os(mmol)4.56.000ExcessR(mmol)−4.5−4.5+4.5+4.5f(mmol)01.54.5Excessc(M)00.02140.0643

Now, the pH of the system can be calculated as given below.

pOH=−log[OH−]=−log(0.0214)=1.67pH+pOH=14pH=14−pOH=14−1.67=12.33.

Titration curve:

The titration curve is plotted between volume of base added and the corresponding pH values.

Volume of NaOH(mL)pH02.5110.003.9030.008.5440.0012.33

The titration curve with four important regions is given below.

Chemistry Principles And Practice, Chapter 16, Problem 16.52QE

Figure

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Answer 3

Question

Chapter 16, Problem 16.52QE

Interpretation Introduction

Interpretation:

The pH during the titration of 30.00 mL of 0.150 M benzoic acid with 0.150 M NaOH after the addition of 0, 10.00, 30.00,and 40.00 mL of the base. The titration curve has to be drawn and four regions of importance have to be labeled.

Expert Solution & Answer

Explanation of Solution

Addition of 0 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Benzoic acid is a weak acid. The initial point in the titration when the base is not added to the system, the system is a weak acid system.

The iCe table can be set up to calculate the pH of the solution.

HA+H2O⇌H3O++A−i(M)0.15000C(M)−y+y+ye(M)0.150−yyy

Where, HA represents benzoic acid and A− represents the conjugate base of benzoic acid.

The Ka value for benzoic acid is 6.3×10−5. The expression for Ka can be written as given below,

Ka=[H3O+][A−][HA]=6.3×10−5y2(0.150−y)=6.3×10−5

This can be solved by approximation. If y≪0.150, then

y2≈6.3×10−5×0.150=0.945×10−5y=0.945×10−5=3.07×10−3=[H3O+]

The pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(3.07×10−3)=2.51.

Addition of 10 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

The neutralization reaction can be written as given,

HA(aq)+OH−(aq)→A−(aq)+H2O(l)

Calculation of milimole of acid and base:

Amount of HA=30.00 mL×(0.150 milimol H3O+mL)=4.5 milimolAmount of OH−=10.00 mL×(0.150 milimol OH−mL)=1.5 milimol

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is OH− ion.

HA+OH−→A−+H2Os(mmol)4.51.50ExcessR(mmol)−1.5−1.5+1.5+1.5f(mmol)3.001.5Excess

All the strong base is consumed. There are weak acid (HA) and its conjugate base (A−) in the solution. So this solution is an acidic buffer. By using Henderson-Hasselbalch equation, it’s pH can be calculated.

pH=pKa+lognbna=pKa+log1.53=−log(Ka)−0.3010=−log(6.3×10−5)−0.3010=4.20−0.3010=3.899.

Addition of 30.00 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Calculation of milimole of acid and base:

Amount of HA=30.00 mL×(0.150 milimol H3O+mL)=4.5 milimolAmount of OH−=30.00 mL×(0.150 milimol OH−mL)=4.5 milimolTotal volume=30.00+30.00=60.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

HA+OH−→A−+H2Os(mmol)4.54.50ExcessR(mmol)−4.5−4.5+4.5+4.5f(mmol)004.5Excessc(M)000.075

Both H3O+ and OH− ions have been completely consumed, the titration is at the equivalence point. There is only the conjugate base and water in the solution.

The iCe table can be set up to calculate the pH of the solution.

A−+H2O⇌HA+OH−i(M)0.07500C(M)−y+y+ye(M)0.075−yyy

The expression for Kb can be written as given below,

Kb=[HA][OH−][A−]=y2(0.075−y)

The value of Kb can be calculated as given below.

Kb=KwKa=1.0×10−146.3×10−5=0.16×10−9

This can be solved by approximation. If y≪0.075, then

y2≈0.16×10−9×0.075=0.012×10−9y=0.012×10−9=3.46×10−6=[OH−]

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(3.46×10−6)=5.46pH+pOH=14pH=14−pOH=14−5.46=8.54.

Addition of 40.00 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Calculation of milimole of acid and base and total volume:

Amount of HA=30.00 mL×(0.150milimol H3O+mL)=4.5 milimolAmount of OH−=40.00 mL×(0.150 milimol OH−mL)=6.00 milimolTotal volume=30.00+40.00=70.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

HA+OH−→A−+H2Os(mmol)4.56.000ExcessR(mmol)−4.5−4.5+4.5+4.5f(mmol)01.54.5Excessc(M)00.02140.0643

Now, the pH of the system can be calculated as given below.

pOH=−log[OH−]=−log(0.0214)=1.67pH+pOH=14pH=14−pOH=14−1.67=12.33.

Titration curve:

The titration curve is plotted between volume of base added and the corresponding pH values.

Volume of NaOH(mL)pH02.5110.003.9030.008.5440.0012.33

The titration curve with four important regions is given below.

Chemistry Principles And Practice, Chapter 16, Problem 16.52QE

Figure

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Answer 4

Question

Chapter 16, Problem 16.52QE

Interpretation Introduction

Interpretation:

The pH during the titration of 30.00 mL of 0.150 M benzoic acid with 0.150 M NaOH after the addition of 0, 10.00, 30.00,and 40.00 mL of the base. The titration curve has to be drawn and four regions of importance have to be labeled.

Expert Solution & Answer

Explanation of Solution

Addition of 0 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Benzoic acid is a weak acid. The initial point in the titration when the base is not added to the system, the system is a weak acid system.

The iCe table can be set up to calculate the pH of the solution.

HA+H2O⇌H3O++A−i(M)0.15000C(M)−y+y+ye(M)0.150−yyy

Where, HA represents benzoic acid and A− represents the conjugate base of benzoic acid.

The Ka value for benzoic acid is 6.3×10−5. The expression for Ka can be written as given below,

Ka=[H3O+][A−][HA]=6.3×10−5y2(0.150−y)=6.3×10−5

This can be solved by approximation. If y≪0.150, then

y2≈6.3×10−5×0.150=0.945×10−5y=0.945×10−5=3.07×10−3=[H3O+]

The pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(3.07×10−3)=2.51.

Addition of 10 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

The neutralization reaction can be written as given,

HA(aq)+OH−(aq)→A−(aq)+H2O(l)

Calculation of milimole of acid and base:

Amount of HA=30.00 mL×(0.150 milimol H3O+mL)=4.5 milimolAmount of OH−=10.00 mL×(0.150 milimol OH−mL)=1.5 milimol

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is OH− ion.

HA+OH−→A−+H2Os(mmol)4.51.50ExcessR(mmol)−1.5−1.5+1.5+1.5f(mmol)3.001.5Excess

All the strong base is consumed. There are weak acid (HA) and its conjugate base (A−) in the solution. So this solution is an acidic buffer. By using Henderson-Hasselbalch equation, it’s pH can be calculated.

pH=pKa+lognbna=pKa+log1.53=−log(Ka)−0.3010=−log(6.3×10−5)−0.3010=4.20−0.3010=3.899.

Addition of 30.00 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Calculation of milimole of acid and base:

Amount of HA=30.00 mL×(0.150 milimol H3O+mL)=4.5 milimolAmount of OH−=30.00 mL×(0.150 milimol OH−mL)=4.5 milimolTotal volume=30.00+30.00=60.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

HA+OH−→A−+H2Os(mmol)4.54.50ExcessR(mmol)−4.5−4.5+4.5+4.5f(mmol)004.5Excessc(M)000.075

Both H3O+ and OH− ions have been completely consumed, the titration is at the equivalence point. There is only the conjugate base and water in the solution.

The iCe table can be set up to calculate the pH of the solution.

A−+H2O⇌HA+OH−i(M)0.07500C(M)−y+y+ye(M)0.075−yyy

The expression for Kb can be written as given below,

Kb=[HA][OH−][A−]=y2(0.075−y)

The value of Kb can be calculated as given below.

Kb=KwKa=1.0×10−146.3×10−5=0.16×10−9

This can be solved by approximation. If y≪0.075, then

y2≈0.16×10−9×0.075=0.012×10−9y=0.012×10−9=3.46×10−6=[OH−]

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(3.46×10−6)=5.46pH+pOH=14pH=14−pOH=14−5.46=8.54.

Addition of 40.00 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Calculation of milimole of acid and base and total volume:

Amount of HA=30.00 mL×(0.150milimol H3O+mL)=4.5 milimolAmount of OH−=40.00 mL×(0.150 milimol OH−mL)=6.00 milimolTotal volume=30.00+40.00=70.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

HA+OH−→A−+H2Os(mmol)4.56.000ExcessR(mmol)−4.5−4.5+4.5+4.5f(mmol)01.54.5Excessc(M)00.02140.0643

Now, the pH of the system can be calculated as given below.

pOH=−log[OH−]=−log(0.0214)=1.67pH+pOH=14pH=14−pOH=14−1.67=12.33.

Titration curve:

The titration curve is plotted between volume of base added and the corresponding pH values.

Volume of NaOH(mL)pH02.5110.003.9030.008.5440.0012.33

The titration curve with four important regions is given below.

Chemistry Principles And Practice, Chapter 16, Problem 16.52QE

Figure

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Answer 5

Chapter 16, Problem 16.52QE

Interpretation Introduction

Interpretation:

The pH during the titration of 30.00 mL of 0.150 M benzoic acid with 0.150 M NaOH after the addition of 0, 10.00, 30.00,and 40.00 mL of the base. The titration curve has to be drawn and four regions of importance have to be labeled.

Expert Solution & Answer

Explanation of Solution

Addition of 0 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Benzoic acid is a weak acid. The initial point in the titration when the base is not added to the system, the system is a weak acid system.

The iCe table can be set up to calculate the pH of the solution.

HA+H2O⇌H3O++A−i(M)0.15000C(M)−y+y+ye(M)0.150−yyy

Where, HA represents benzoic acid and A− represents the conjugate base of benzoic acid.

The Ka value for benzoic acid is 6.3×10−5. The expression for Ka can be written as given below,

Ka=[H3O+][A−][HA]=6.3×10−5y2(0.150−y)=6.3×10−5

This can be solved by approximation. If y≪0.150, then

y2≈6.3×10−5×0.150=0.945×10−5y=0.945×10−5=3.07×10−3=[H3O+]

The pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(3.07×10−3)=2.51.

Addition of 10 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

The neutralization reaction can be written as given,

HA(aq)+OH−(aq)→A−(aq)+H2O(l)

Calculation of milimole of acid and base:

Amount of HA=30.00 mL×(0.150 milimol H3O+mL)=4.5 milimolAmount of OH−=10.00 mL×(0.150 milimol OH−mL)=1.5 milimol

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is OH− ion.

HA+OH−→A−+H2Os(mmol)4.51.50ExcessR(mmol)−1.5−1.5+1.5+1.5f(mmol)3.001.5Excess

All the strong base is consumed. There are weak acid (HA) and its conjugate base (A−) in the solution. So this solution is an acidic buffer. By using Henderson-Hasselbalch equation, it’s pH can be calculated.

pH=pKa+lognbna=pKa+log1.53=−log(Ka)−0.3010=−log(6.3×10−5)−0.3010=4.20−0.3010=3.899.

Addition of 30.00 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Calculation of milimole of acid and base:

Amount of HA=30.00 mL×(0.150 milimol H3O+mL)=4.5 milimolAmount of OH−=30.00 mL×(0.150 milimol OH−mL)=4.5 milimolTotal volume=30.00+30.00=60.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

HA+OH−→A−+H2Os(mmol)4.54.50ExcessR(mmol)−4.5−4.5+4.5+4.5f(mmol)004.5Excessc(M)000.075

Both H3O+ and OH− ions have been completely consumed, the titration is at the equivalence point. There is only the conjugate base and water in the solution.

The iCe table can be set up to calculate the pH of the solution.

A−+H2O⇌HA+OH−i(M)0.07500C(M)−y+y+ye(M)0.075−yyy

The expression for Kb can be written as given below,

Kb=[HA][OH−][A−]=y2(0.075−y)

The value of Kb can be calculated as given below.

Kb=KwKa=1.0×10−146.3×10−5=0.16×10−9

This can be solved by approximation. If y≪0.075, then

y2≈0.16×10−9×0.075=0.012×10−9y=0.012×10−9=3.46×10−6=[OH−]

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(3.46×10−6)=5.46pH+pOH=14pH=14−pOH=14−5.46=8.54.

Addition of 40.00 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Calculation of milimole of acid and base and total volume:

Amount of HA=30.00 mL×(0.150milimol H3O+mL)=4.5 milimolAmount of OH−=40.00 mL×(0.150 milimol OH−mL)=6.00 milimolTotal volume=30.00+40.00=70.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

HA+OH−→A−+H2Os(mmol)4.56.000ExcessR(mmol)−4.5−4.5+4.5+4.5f(mmol)01.54.5Excessc(M)00.02140.0643

Now, the pH of the system can be calculated as given below.

pOH=−log[OH−]=−log(0.0214)=1.67pH+pOH=14pH=14−pOH=14−1.67=12.33.

Titration curve:

The titration curve is plotted between volume of base added and the corresponding pH values.

Volume of NaOH(mL)pH02.5110.003.9030.008.5440.0012.33

The titration curve with four important regions is given below.

Chemistry Principles And Practice, Chapter 16, Problem 16.52QE

Figure

Answer 6

Chapter 16, Problem 16.52QE

Interpretation Introduction

Interpretation:

The pH during the titration of 30.00 mL of 0.150 M benzoic acid with 0.150 M NaOH after the addition of 0, 10.00, 30.00,and 40.00 mL of the base. The titration curve has to be drawn and four regions of importance have to be labeled.

Expert Solution & Answer

Explanation of Solution

Addition of 0 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Benzoic acid is a weak acid. The initial point in the titration when the base is not added to the system, the system is a weak acid system.

The iCe table can be set up to calculate the pH of the solution.

HA+H2O⇌H3O++A−i(M)0.15000C(M)−y+y+ye(M)0.150−yyy

Where, HA represents benzoic acid and A− represents the conjugate base of benzoic acid.

The Ka value for benzoic acid is 6.3×10−5. The expression for Ka can be written as given below,

Ka=[H3O+][A−][HA]=6.3×10−5y2(0.150−y)=6.3×10−5

This can be solved by approximation. If y≪0.150, then

y2≈6.3×10−5×0.150=0.945×10−5y=0.945×10−5=3.07×10−3=[H3O+]

The pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(3.07×10−3)=2.51.

Addition of 10 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

The neutralization reaction can be written as given,

HA(aq)+OH−(aq)→A−(aq)+H2O(l)

Calculation of milimole of acid and base:

Amount of HA=30.00 mL×(0.150 milimol H3O+mL)=4.5 milimolAmount of OH−=10.00 mL×(0.150 milimol OH−mL)=1.5 milimol

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is OH− ion.

HA+OH−→A−+H2Os(mmol)4.51.50ExcessR(mmol)−1.5−1.5+1.5+1.5f(mmol)3.001.5Excess

All the strong base is consumed. There are weak acid (HA) and its conjugate base (A−) in the solution. So this solution is an acidic buffer. By using Henderson-Hasselbalch equation, it’s pH can be calculated.

pH=pKa+lognbna=pKa+log1.53=−log(Ka)−0.3010=−log(6.3×10−5)−0.3010=4.20−0.3010=3.899.

Addition of 30.00 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Calculation of milimole of acid and base:

Amount of HA=30.00 mL×(0.150 milimol H3O+mL)=4.5 milimolAmount of OH−=30.00 mL×(0.150 milimol OH−mL)=4.5 milimolTotal volume=30.00+30.00=60.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

HA+OH−→A−+H2Os(mmol)4.54.50ExcessR(mmol)−4.5−4.5+4.5+4.5f(mmol)004.5Excessc(M)000.075

Both H3O+ and OH− ions have been completely consumed, the titration is at the equivalence point. There is only the conjugate base and water in the solution.

The iCe table can be set up to calculate the pH of the solution.

A−+H2O⇌HA+OH−i(M)0.07500C(M)−y+y+ye(M)0.075−yyy

The expression for Kb can be written as given below,

Kb=[HA][OH−][A−]=y2(0.075−y)

The value of Kb can be calculated as given below.

Kb=KwKa=1.0×10−146.3×10−5=0.16×10−9

This can be solved by approximation. If y≪0.075, then

y2≈0.16×10−9×0.075=0.012×10−9y=0.012×10−9=3.46×10−6=[OH−]

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(3.46×10−6)=5.46pH+pOH=14pH=14−pOH=14−5.46=8.54.

Addition of 40.00 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Calculation of milimole of acid and base and total volume:

Amount of HA=30.00 mL×(0.150milimol H3O+mL)=4.5 milimolAmount of OH−=40.00 mL×(0.150 milimol OH−mL)=6.00 milimolTotal volume=30.00+40.00=70.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

HA+OH−→A−+H2Os(mmol)4.56.000ExcessR(mmol)−4.5−4.5+4.5+4.5f(mmol)01.54.5Excessc(M)00.02140.0643

Now, the pH of the system can be calculated as given below.

pOH=−log[OH−]=−log(0.0214)=1.67pH+pOH=14pH=14−pOH=14−1.67=12.33.

Titration curve:

The titration curve is plotted between volume of base added and the corresponding pH values.

Volume of NaOH(mL)pH02.5110.003.9030.008.5440.0012.33

The titration curve with four important regions is given below.

Chemistry Principles And Practice, Chapter 16, Problem 16.52QE

Figure

Answer 7

Chapter 16, Problem 16.52QE

Interpretation Introduction

Interpretation:

The pH during the titration of 30.00 mL of 0.150 M benzoic acid with 0.150 M NaOH after the addition of 0, 10.00, 30.00,and 40.00 mL of the base. The titration curve has to be drawn and four regions of importance have to be labeled.

Answer 8

Expert Solution & Answer

Explanation of Solution

Addition of 0 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Benzoic acid is a weak acid. The initial point in the titration when the base is not added to the system, the system is a weak acid system.

The iCe table can be set up to calculate the pH of the solution.

HA+H2O⇌H3O++A−i(M)0.15000C(M)−y+y+ye(M)0.150−yyy

Where, HA represents benzoic acid and A− represents the conjugate base of benzoic acid.

The Ka value for benzoic acid is 6.3×10−5. The expression for Ka can be written as given below,

Ka=[H3O+][A−][HA]=6.3×10−5y2(0.150−y)=6.3×10−5

This can be solved by approximation. If y≪0.150, then

y2≈6.3×10−5×0.150=0.945×10−5y=0.945×10−5=3.07×10−3=[H3O+]

The pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(3.07×10−3)=2.51.

Addition of 10 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

The neutralization reaction can be written as given,

HA(aq)+OH−(aq)→A−(aq)+H2O(l)

Calculation of milimole of acid and base:

Amount of HA=30.00 mL×(0.150 milimol H3O+mL)=4.5 milimolAmount of OH−=10.00 mL×(0.150 milimol OH−mL)=1.5 milimol

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is OH− ion.

HA+OH−→A−+H2Os(mmol)4.51.50ExcessR(mmol)−1.5−1.5+1.5+1.5f(mmol)3.001.5Excess

All the strong base is consumed. There are weak acid (HA) and its conjugate base (A−) in the solution. So this solution is an acidic buffer. By using Henderson-Hasselbalch equation, it’s pH can be calculated.

pH=pKa+lognbna=pKa+log1.53=−log(Ka)−0.3010=−log(6.3×10−5)−0.3010=4.20−0.3010=3.899.

Addition of 30.00 mL of 0.150 M NaOHto30.00 mL of 0.150 Mbenzoic acid:

Calculation of milimole of acid and base:

Amount of HA=30.00 mL×(0.150 milimol H3O+mL)=4.5 milimolAmount of OH−=30.00 mL×(0.150 milimol OH−mL)=4.5 milimolTotal volume=30.00+30.00=60.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

HA+OH−→A−+H2Os(mmol)4.54.50ExcessR(mmol)−4.5−4.5+4.5+4.5f(mmol)004.5Excessc(M)000.075

Both H3O+ and OH− ions have been completely consumed, the titration is at the equivalence point. There is only the conjugate base and water in the solution.

The iCe table can be set up to calculate the pH of the solution.

A−+H2O⇌HA+OH−i(M)0.07500C(M)−y+y+ye(M)0.075−yyy