The pH values after the addition of each proportion of the base to the acid is to be determined. The titration curve needs to be drawn. Concept introduction: A strong acid or a strong base can completely dissociate into its ions when they are in an aqueous solution. Sodium hydroxide is a strong base while nitric acid is a strong acid. When these two chemical species are reacted with each other, they form their ionic forms and a neutral solution.

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Answer 1

Question

Chapter 8, Problem 144AE

Interpretation Introduction

Interpretation:

The pH values after the addition of each proportion of the base to the acid is to be determined. The titration curve needs to be drawn.

Concept introduction:

A strong acid or a strong base can completely dissociate into its ions when they are in an aqueous solution. Sodium hydroxide is a strong base while nitric acid is a strong acid. When these two chemical species are reacted with each other, they form their ionic forms and a neutral solution.

Expert Solution & Answer

Explanation of Solution

Initial pH of the analyte solution;

Nitric acid is a strong acid that completely dissociates in water.

The amount of hydrogen ion will be equal to the concentration of nitric acid thus,

[H+]=0.1 M

The pH of the solution will be:

pH=−log[H+]=−log[0.1]=1

Addition of 4.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×4.0×10−3L=4×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−0.4)×10−3=2.1×10−3mol

The concentration of hydrogen ions =2.1×10−3mol(25.0+4.0)×10−3L=0.072mol/L

The pH of the solution =−log[0.072]=1.14

Addition of 8.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×8.0×10−3L=8×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−0.8)×10−3=1.7×10−3mol

The concentration of hydrogen ions =1.7×10−3mol(25.0+8.0)×10−3L=0.051mol/L

The pH of the solution =−log[0.051]=1.28

Addition of 12.5mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×12.5×10−3L=12.5×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−1.25)×10−3=1.25×10−3mol

The concentration of hydrogen ions =1.25×10−3mol(25.0+12.5)×10−3L=0.033mol/L

The pH of the solution =−log[0.033]=1.47

Addition of 20.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×20.0×10−3L=20×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.0)×10−3=0.5×10−3mol

The concentration of hydrogen ions =0.5×10−3mol(25.0+20.0)×10−3L=0.011mol/L

The pH of the solution =−log[0.011]=1.95

Addition of 24.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.0×10−3L=24×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.4)×10−3=0.1×10−3mol

The concentration of hydrogen ions =0.1×10−3mol(25.0+24.0)×10−3L=0.002mol/L

The pH of the solution =−log[0.002]=2.69

Addition of 24.5mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.5×10−3L=24.5×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.45)×10−3=0.05×10−3mol

The concentration of hydrogen ions =0.05×10−3mol(25.0+24.5)×10−3L=0.001mol/L

The pH of the solution =−log[0.001]=2.99

Addition of 24.9mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.9×10−3L=24.9×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.49)×10−3=0.01×10−3mol

The concentration of hydrogen ions =0.01×10−3mol(25.0+24.9)×10−3L=0.0002mol/L

The pH of the solution =−log[0.0002]=3.69

Addition of 25.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×25.0×10−3L=25.0×10−4mol

Therefore, the amount of acid remaining in the solution will be:

=(2.5−2.5)×10−3=0 .

Since, both acid and base completely react to form salt and water, the pH of the solution =7.00

Addition of 28.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×28.0×10−3L=28×10−4mol

Therefore, the amount of excess base in the solution =(2.8−2.5)×10−3=0.3×10−3mol

The concentration of hydroxide ions =0.3×10−3mol(25.0+28.0)×10−3L=0.0056mol/L

Thenthe pH of the solution can be calculated as follows:

pOH=−log[0.0056]=2.24pH=14.00−2.24=11.75

Addition of 30.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×30.0×10−3L=30×10−4mol

Therefore, the amount of excess base in the solution =(3.0−2.5)×10−3=0.5×10−3mol

The concentration of hydroxide ions =0.5×10−3mol(25.0+30.0)×10−3L=0.009mol/L

Then the pH of the solution can be calculated as follows:

pOH=−log[0.009]=2.04pH=14.00−2.24=11.95

The data obtained from the calculations is as follows:

Volume of base added (mL)	pH
0.0	1
4.0	1.14
8.0	1.28
12.5	1.47
20.0	1.95
24.0	2.69
24.5	2.99
24.9	3.69
25.0	7.0
28.0	11.75
30.0	11.95

The titration curve can be drawn as follows:

EBK CHEMICAL PRINCIPLES, Chapter 8, Problem 144AE

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Answer 2

Question

Chapter 8, Problem 144AE

Interpretation Introduction

Interpretation:

The pH values after the addition of each proportion of the base to the acid is to be determined. The titration curve needs to be drawn.

Concept introduction:

A strong acid or a strong base can completely dissociate into its ions when they are in an aqueous solution. Sodium hydroxide is a strong base while nitric acid is a strong acid. When these two chemical species are reacted with each other, they form their ionic forms and a neutral solution.

Expert Solution & Answer

Explanation of Solution

Initial pH of the analyte solution;

Nitric acid is a strong acid that completely dissociates in water.

The amount of hydrogen ion will be equal to the concentration of nitric acid thus,

[H+]=0.1 M

The pH of the solution will be:

pH=−log[H+]=−log[0.1]=1

Addition of 4.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×4.0×10−3L=4×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−0.4)×10−3=2.1×10−3mol

The concentration of hydrogen ions =2.1×10−3mol(25.0+4.0)×10−3L=0.072mol/L

The pH of the solution =−log[0.072]=1.14

Addition of 8.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×8.0×10−3L=8×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−0.8)×10−3=1.7×10−3mol

The concentration of hydrogen ions =1.7×10−3mol(25.0+8.0)×10−3L=0.051mol/L

The pH of the solution =−log[0.051]=1.28

Addition of 12.5mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×12.5×10−3L=12.5×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−1.25)×10−3=1.25×10−3mol

The concentration of hydrogen ions =1.25×10−3mol(25.0+12.5)×10−3L=0.033mol/L

The pH of the solution =−log[0.033]=1.47

Addition of 20.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×20.0×10−3L=20×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.0)×10−3=0.5×10−3mol

The concentration of hydrogen ions =0.5×10−3mol(25.0+20.0)×10−3L=0.011mol/L

The pH of the solution =−log[0.011]=1.95

Addition of 24.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.0×10−3L=24×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.4)×10−3=0.1×10−3mol

The concentration of hydrogen ions =0.1×10−3mol(25.0+24.0)×10−3L=0.002mol/L

The pH of the solution =−log[0.002]=2.69

Addition of 24.5mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.5×10−3L=24.5×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.45)×10−3=0.05×10−3mol

The concentration of hydrogen ions =0.05×10−3mol(25.0+24.5)×10−3L=0.001mol/L

The pH of the solution =−log[0.001]=2.99

Addition of 24.9mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.9×10−3L=24.9×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.49)×10−3=0.01×10−3mol

The concentration of hydrogen ions =0.01×10−3mol(25.0+24.9)×10−3L=0.0002mol/L

The pH of the solution =−log[0.0002]=3.69

Addition of 25.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×25.0×10−3L=25.0×10−4mol

Therefore, the amount of acid remaining in the solution will be:

=(2.5−2.5)×10−3=0 .

Since, both acid and base completely react to form salt and water, the pH of the solution =7.00

Addition of 28.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×28.0×10−3L=28×10−4mol

Therefore, the amount of excess base in the solution =(2.8−2.5)×10−3=0.3×10−3mol

The concentration of hydroxide ions =0.3×10−3mol(25.0+28.0)×10−3L=0.0056mol/L

Thenthe pH of the solution can be calculated as follows:

pOH=−log[0.0056]=2.24pH=14.00−2.24=11.75

Addition of 30.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×30.0×10−3L=30×10−4mol

Therefore, the amount of excess base in the solution =(3.0−2.5)×10−3=0.5×10−3mol

The concentration of hydroxide ions =0.5×10−3mol(25.0+30.0)×10−3L=0.009mol/L

Then the pH of the solution can be calculated as follows:

pOH=−log[0.009]=2.04pH=14.00−2.24=11.95

The data obtained from the calculations is as follows:

Volume of base added (mL)	pH
0.0	1
4.0	1.14
8.0	1.28
12.5	1.47
20.0	1.95
24.0	2.69
24.5	2.99
24.9	3.69
25.0	7.0
28.0	11.75
30.0	11.95

The titration curve can be drawn as follows:

EBK CHEMICAL PRINCIPLES, Chapter 8, Problem 144AE

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Answer 3

Question

Chapter 8, Problem 144AE

Interpretation Introduction

Interpretation:

The pH values after the addition of each proportion of the base to the acid is to be determined. The titration curve needs to be drawn.

Concept introduction:

A strong acid or a strong base can completely dissociate into its ions when they are in an aqueous solution. Sodium hydroxide is a strong base while nitric acid is a strong acid. When these two chemical species are reacted with each other, they form their ionic forms and a neutral solution.

Expert Solution & Answer

Explanation of Solution

Initial pH of the analyte solution;

Nitric acid is a strong acid that completely dissociates in water.

The amount of hydrogen ion will be equal to the concentration of nitric acid thus,

[H+]=0.1 M

The pH of the solution will be:

pH=−log[H+]=−log[0.1]=1

Addition of 4.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×4.0×10−3L=4×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−0.4)×10−3=2.1×10−3mol

The concentration of hydrogen ions =2.1×10−3mol(25.0+4.0)×10−3L=0.072mol/L

The pH of the solution =−log[0.072]=1.14

Addition of 8.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×8.0×10−3L=8×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−0.8)×10−3=1.7×10−3mol

The concentration of hydrogen ions =1.7×10−3mol(25.0+8.0)×10−3L=0.051mol/L

The pH of the solution =−log[0.051]=1.28

Addition of 12.5mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×12.5×10−3L=12.5×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−1.25)×10−3=1.25×10−3mol

The concentration of hydrogen ions =1.25×10−3mol(25.0+12.5)×10−3L=0.033mol/L

The pH of the solution =−log[0.033]=1.47

Addition of 20.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×20.0×10−3L=20×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.0)×10−3=0.5×10−3mol

The concentration of hydrogen ions =0.5×10−3mol(25.0+20.0)×10−3L=0.011mol/L

The pH of the solution =−log[0.011]=1.95

Addition of 24.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.0×10−3L=24×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.4)×10−3=0.1×10−3mol

The concentration of hydrogen ions =0.1×10−3mol(25.0+24.0)×10−3L=0.002mol/L

The pH of the solution =−log[0.002]=2.69

Addition of 24.5mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.5×10−3L=24.5×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.45)×10−3=0.05×10−3mol

The concentration of hydrogen ions =0.05×10−3mol(25.0+24.5)×10−3L=0.001mol/L

The pH of the solution =−log[0.001]=2.99

Addition of 24.9mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.9×10−3L=24.9×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.49)×10−3=0.01×10−3mol

The concentration of hydrogen ions =0.01×10−3mol(25.0+24.9)×10−3L=0.0002mol/L

The pH of the solution =−log[0.0002]=3.69

Addition of 25.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×25.0×10−3L=25.0×10−4mol

Therefore, the amount of acid remaining in the solution will be:

=(2.5−2.5)×10−3=0 .

Since, both acid and base completely react to form salt and water, the pH of the solution =7.00

Addition of 28.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×28.0×10−3L=28×10−4mol

Therefore, the amount of excess base in the solution =(2.8−2.5)×10−3=0.3×10−3mol

The concentration of hydroxide ions =0.3×10−3mol(25.0+28.0)×10−3L=0.0056mol/L

Thenthe pH of the solution can be calculated as follows:

pOH=−log[0.0056]=2.24pH=14.00−2.24=11.75

Addition of 30.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×30.0×10−3L=30×10−4mol

Therefore, the amount of excess base in the solution =(3.0−2.5)×10−3=0.5×10−3mol

The concentration of hydroxide ions =0.5×10−3mol(25.0+30.0)×10−3L=0.009mol/L

Then the pH of the solution can be calculated as follows:

pOH=−log[0.009]=2.04pH=14.00−2.24=11.95

The data obtained from the calculations is as follows:

Volume of base added (mL)	pH
0.0	1
4.0	1.14
8.0	1.28
12.5	1.47
20.0	1.95
24.0	2.69
24.5	2.99
24.9	3.69
25.0	7.0
28.0	11.75
30.0	11.95

The titration curve can be drawn as follows:

EBK CHEMICAL PRINCIPLES, Chapter 8, Problem 144AE

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Answer 4

Question

Chapter 8, Problem 144AE

Interpretation Introduction

Interpretation:

The pH values after the addition of each proportion of the base to the acid is to be determined. The titration curve needs to be drawn.

Concept introduction:

A strong acid or a strong base can completely dissociate into its ions when they are in an aqueous solution. Sodium hydroxide is a strong base while nitric acid is a strong acid. When these two chemical species are reacted with each other, they form their ionic forms and a neutral solution.

Expert Solution & Answer

Explanation of Solution

Initial pH of the analyte solution;

Nitric acid is a strong acid that completely dissociates in water.

The amount of hydrogen ion will be equal to the concentration of nitric acid thus,

[H+]=0.1 M

The pH of the solution will be:

pH=−log[H+]=−log[0.1]=1

Addition of 4.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×4.0×10−3L=4×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−0.4)×10−3=2.1×10−3mol

The concentration of hydrogen ions =2.1×10−3mol(25.0+4.0)×10−3L=0.072mol/L

The pH of the solution =−log[0.072]=1.14

Addition of 8.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×8.0×10−3L=8×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−0.8)×10−3=1.7×10−3mol

The concentration of hydrogen ions =1.7×10−3mol(25.0+8.0)×10−3L=0.051mol/L

The pH of the solution =−log[0.051]=1.28

Addition of 12.5mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×12.5×10−3L=12.5×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−1.25)×10−3=1.25×10−3mol

The concentration of hydrogen ions =1.25×10−3mol(25.0+12.5)×10−3L=0.033mol/L

The pH of the solution =−log[0.033]=1.47

Addition of 20.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×20.0×10−3L=20×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.0)×10−3=0.5×10−3mol

The concentration of hydrogen ions =0.5×10−3mol(25.0+20.0)×10−3L=0.011mol/L

The pH of the solution =−log[0.011]=1.95

Addition of 24.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.0×10−3L=24×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.4)×10−3=0.1×10−3mol

The concentration of hydrogen ions =0.1×10−3mol(25.0+24.0)×10−3L=0.002mol/L

The pH of the solution =−log[0.002]=2.69

Addition of 24.5mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.5×10−3L=24.5×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.45)×10−3=0.05×10−3mol

The concentration of hydrogen ions =0.05×10−3mol(25.0+24.5)×10−3L=0.001mol/L

The pH of the solution =−log[0.001]=2.99

Addition of 24.9mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.9×10−3L=24.9×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.49)×10−3=0.01×10−3mol

The concentration of hydrogen ions =0.01×10−3mol(25.0+24.9)×10−3L=0.0002mol/L

The pH of the solution =−log[0.0002]=3.69

Addition of 25.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×25.0×10−3L=25.0×10−4mol

Therefore, the amount of acid remaining in the solution will be:

=(2.5−2.5)×10−3=0 .

Since, both acid and base completely react to form salt and water, the pH of the solution =7.00

Addition of 28.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×28.0×10−3L=28×10−4mol

Therefore, the amount of excess base in the solution =(2.8−2.5)×10−3=0.3×10−3mol

The concentration of hydroxide ions =0.3×10−3mol(25.0+28.0)×10−3L=0.0056mol/L

Thenthe pH of the solution can be calculated as follows:

pOH=−log[0.0056]=2.24pH=14.00−2.24=11.75

Addition of 30.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×30.0×10−3L=30×10−4mol

Therefore, the amount of excess base in the solution =(3.0−2.5)×10−3=0.5×10−3mol

The concentration of hydroxide ions =0.5×10−3mol(25.0+30.0)×10−3L=0.009mol/L

Then the pH of the solution can be calculated as follows:

pOH=−log[0.009]=2.04pH=14.00−2.24=11.95

The data obtained from the calculations is as follows:

Volume of base added (mL)	pH
0.0	1
4.0	1.14
8.0	1.28
12.5	1.47
20.0	1.95
24.0	2.69
24.5	2.99
24.9	3.69
25.0	7.0
28.0	11.75
30.0	11.95

The titration curve can be drawn as follows:

EBK CHEMICAL PRINCIPLES, Chapter 8, Problem 144AE

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Answer 5

Chapter 8, Problem 144AE

Interpretation Introduction

Interpretation:

The pH values after the addition of each proportion of the base to the acid is to be determined. The titration curve needs to be drawn.

Concept introduction:

A strong acid or a strong base can completely dissociate into its ions when they are in an aqueous solution. Sodium hydroxide is a strong base while nitric acid is a strong acid. When these two chemical species are reacted with each other, they form their ionic forms and a neutral solution.

Expert Solution & Answer

Explanation of Solution

Initial pH of the analyte solution;

Nitric acid is a strong acid that completely dissociates in water.

The amount of hydrogen ion will be equal to the concentration of nitric acid thus,

[H+]=0.1 M

The pH of the solution will be:

pH=−log[H+]=−log[0.1]=1

Addition of 4.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×4.0×10−3L=4×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−0.4)×10−3=2.1×10−3mol

The concentration of hydrogen ions =2.1×10−3mol(25.0+4.0)×10−3L=0.072mol/L

The pH of the solution =−log[0.072]=1.14

Addition of 8.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×8.0×10−3L=8×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−0.8)×10−3=1.7×10−3mol

The concentration of hydrogen ions =1.7×10−3mol(25.0+8.0)×10−3L=0.051mol/L

The pH of the solution =−log[0.051]=1.28

Addition of 12.5mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×12.5×10−3L=12.5×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−1.25)×10−3=1.25×10−3mol

The concentration of hydrogen ions =1.25×10−3mol(25.0+12.5)×10−3L=0.033mol/L

The pH of the solution =−log[0.033]=1.47

Addition of 20.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×20.0×10−3L=20×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.0)×10−3=0.5×10−3mol

The concentration of hydrogen ions =0.5×10−3mol(25.0+20.0)×10−3L=0.011mol/L

The pH of the solution =−log[0.011]=1.95

Addition of 24.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.0×10−3L=24×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.4)×10−3=0.1×10−3mol

The concentration of hydrogen ions =0.1×10−3mol(25.0+24.0)×10−3L=0.002mol/L

The pH of the solution =−log[0.002]=2.69

Addition of 24.5mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.5×10−3L=24.5×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.45)×10−3=0.05×10−3mol

The concentration of hydrogen ions =0.05×10−3mol(25.0+24.5)×10−3L=0.001mol/L

The pH of the solution =−log[0.001]=2.99

Addition of 24.9mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.9×10−3L=24.9×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.49)×10−3=0.01×10−3mol

The concentration of hydrogen ions =0.01×10−3mol(25.0+24.9)×10−3L=0.0002mol/L

The pH of the solution =−log[0.0002]=3.69

Addition of 25.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×25.0×10−3L=25.0×10−4mol

Therefore, the amount of acid remaining in the solution will be:

=(2.5−2.5)×10−3=0 .

Since, both acid and base completely react to form salt and water, the pH of the solution =7.00

Addition of 28.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×28.0×10−3L=28×10−4mol

Therefore, the amount of excess base in the solution =(2.8−2.5)×10−3=0.3×10−3mol

The concentration of hydroxide ions =0.3×10−3mol(25.0+28.0)×10−3L=0.0056mol/L

Thenthe pH of the solution can be calculated as follows:

pOH=−log[0.0056]=2.24pH=14.00−2.24=11.75

Addition of 30.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×30.0×10−3L=30×10−4mol

Therefore, the amount of excess base in the solution =(3.0−2.5)×10−3=0.5×10−3mol

The concentration of hydroxide ions =0.5×10−3mol(25.0+30.0)×10−3L=0.009mol/L

Then the pH of the solution can be calculated as follows:

pOH=−log[0.009]=2.04pH=14.00−2.24=11.95

The data obtained from the calculations is as follows:

Volume of base added (mL)	pH
0.0	1
4.0	1.14
8.0	1.28
12.5	1.47
20.0	1.95
24.0	2.69
24.5	2.99
24.9	3.69
25.0	7.0
28.0	11.75
30.0	11.95

The titration curve can be drawn as follows:

EBK CHEMICAL PRINCIPLES, Chapter 8, Problem 144AE

Answer 6

Chapter 8, Problem 144AE

Interpretation Introduction

Interpretation:

The pH values after the addition of each proportion of the base to the acid is to be determined. The titration curve needs to be drawn.

Concept introduction:

A strong acid or a strong base can completely dissociate into its ions when they are in an aqueous solution. Sodium hydroxide is a strong base while nitric acid is a strong acid. When these two chemical species are reacted with each other, they form their ionic forms and a neutral solution.

Expert Solution & Answer

Explanation of Solution

Initial pH of the analyte solution;

Nitric acid is a strong acid that completely dissociates in water.

The amount of hydrogen ion will be equal to the concentration of nitric acid thus,

[H+]=0.1 M

The pH of the solution will be:

pH=−log[H+]=−log[0.1]=1

Addition of 4.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×4.0×10−3L=4×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−0.4)×10−3=2.1×10−3mol

The concentration of hydrogen ions =2.1×10−3mol(25.0+4.0)×10−3L=0.072mol/L

The pH of the solution =−log[0.072]=1.14

Addition of 8.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×8.0×10−3L=8×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−0.8)×10−3=1.7×10−3mol

The concentration of hydrogen ions =1.7×10−3mol(25.0+8.0)×10−3L=0.051mol/L

The pH of the solution =−log[0.051]=1.28

Addition of 12.5mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×12.5×10−3L=12.5×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−1.25)×10−3=1.25×10−3mol

The concentration of hydrogen ions =1.25×10−3mol(25.0+12.5)×10−3L=0.033mol/L

The pH of the solution =−log[0.033]=1.47

Addition of 20.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×20.0×10−3L=20×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.0)×10−3=0.5×10−3mol

The concentration of hydrogen ions =0.5×10−3mol(25.0+20.0)×10−3L=0.011mol/L

The pH of the solution =−log[0.011]=1.95

Addition of 24.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.0×10−3L=24×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.4)×10−3=0.1×10−3mol

The concentration of hydrogen ions =0.1×10−3mol(25.0+24.0)×10−3L=0.002mol/L

The pH of the solution =−log[0.002]=2.69

Addition of 24.5mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.5×10−3L=24.5×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.45)×10−3=0.05×10−3mol

The concentration of hydrogen ions =0.05×10−3mol(25.0+24.5)×10−3L=0.001mol/L

The pH of the solution =−log[0.001]=2.99

Addition of 24.9mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.9×10−3L=24.9×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.49)×10−3=0.01×10−3mol

The concentration of hydrogen ions =0.01×10−3mol(25.0+24.9)×10−3L=0.0002mol/L

The pH of the solution =−log[0.0002]=3.69

Addition of 25.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×25.0×10−3L=25.0×10−4mol

Therefore, the amount of acid remaining in the solution will be:

=(2.5−2.5)×10−3=0 .

Since, both acid and base completely react to form salt and water, the pH of the solution =7.00

Addition of 28.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×28.0×10−3L=28×10−4mol

Therefore, the amount of excess base in the solution =(2.8−2.5)×10−3=0.3×10−3mol

The concentration of hydroxide ions =0.3×10−3mol(25.0+28.0)×10−3L=0.0056mol/L

Thenthe pH of the solution can be calculated as follows:

pOH=−log[0.0056]=2.24pH=14.00−2.24=11.75

Addition of 30.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×30.0×10−3L=30×10−4mol

Therefore, the amount of excess base in the solution =(3.0−2.5)×10−3=0.5×10−3mol

The concentration of hydroxide ions =0.5×10−3mol(25.0+30.0)×10−3L=0.009mol/L

Then the pH of the solution can be calculated as follows:

pOH=−log[0.009]=2.04pH=14.00−2.24=11.95

The data obtained from the calculations is as follows:

Volume of base added (mL)	pH
0.0	1
4.0	1.14
8.0	1.28
12.5	1.47
20.0	1.95
24.0	2.69
24.5	2.99
24.9	3.69
25.0	7.0
28.0	11.75
30.0	11.95

The titration curve can be drawn as follows:

EBK CHEMICAL PRINCIPLES, Chapter 8, Problem 144AE

Answer 7

Chapter 8, Problem 144AE

Interpretation Introduction

Interpretation:

The pH values after the addition of each proportion of the base to the acid is to be determined. The titration curve needs to be drawn.

Concept introduction:

A strong acid or a strong base can completely dissociate into its ions when they are in an aqueous solution. Sodium hydroxide is a strong base while nitric acid is a strong acid. When these two chemical species are reacted with each other, they form their ionic forms and a neutral solution.

Answer 8

Expert Solution & Answer

Explanation of Solution

Initial pH of the analyte solution;

Nitric acid is a strong acid that completely dissociates in water.

The amount of hydrogen ion will be equal to the concentration of nitric acid thus,

[H+]=0.1 M

The pH of the solution will be:

pH=−log[H+]=−log[0.1]=1

Addition of 4.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×4.0×10−3L=4×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−0.4)×10−3=2.1×10−3mol

The concentration of hydrogen ions =2.1×10−3mol(25.0+4.0)×10−3L=0.072mol/L

The pH of the solution =−log[0.072]=1.14

Addition of 8.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×8.0×10−3L=8×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−0.8)×10−3=1.7×10−3mol

The concentration of hydrogen ions =1.7×10−3mol(25.0+8.0)×10−3L=0.051mol/L

The pH of the solution =−log[0.051]=1.28

Addition of 12.5mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×12.5×10−3L=12.5×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−1.25)×10−3=1.25×10−3mol

The concentration of hydrogen ions =1.25×10−3mol(25.0+12.5)×10−3L=0.033mol/L

The pH of the solution =−log[0.033]=1.47

Addition of 20.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×20.0×10−3L=20×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.0)×10−3=0.5×10−3mol

The concentration of hydrogen ions =0.5×10−3mol(25.0+20.0)×10−3L=0.011mol/L

The pH of the solution =−log[0.011]=1.95

Addition of 24.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.0×10−3L=24×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.4)×10−3=0.1×10−3mol

The concentration of hydrogen ions =0.1×10−3mol(25.0+24.0)×10−3L=0.002mol/L

The pH of the solution =−log[0.002]=2.69

Addition of 24.5mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.5×10−3L=24.5×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.45)×10−3=0.05×10−3mol

The concentration of hydrogen ions =0.05×10−3mol(25.0+24.5)×10−3L=0.001mol/L

The pH of the solution =−log[0.001]=2.99

Addition of 24.9mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.9×10−3L=24.9×10−4mol

Therefore, the amount of acid remaining in the solution =(2.5−2.49)×10−3=0.01×10−3mol

The concentration of hydrogen ions =0.01×10−3mol(25.0+24.9)×10−3L=0.0002mol/L

The pH of the solution =−log[0.0002]=3.69

Addition of 25.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×25.0×10−3L=25.0×10−4mol

Therefore, the amount of acid remaining in the solution will be:

=(2.5−2.5)×10−3=0 .

Since, both acid and base completely react to form salt and water, the pH of the solution =7.00

Addition of 28.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×28.0×10−3L=28×10−4mol

Therefore, the amount of excess base in the solution =(2.8−2.5)×10−3=0.3×10−3mol

The concentration of hydroxide ions =0.3×10−3mol(25.0+28.0)×10−3L=0.0056mol/L

Thenthe pH of the solution can be calculated as follows:

pOH=−log[0.0056]=2.24pH=14.00−2.24=11.75

Addition of 30.0mLof base:

Total amount of acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×30.0×10−3L=30×10−4mol

Therefore, the amount of excess base in the solution =(3.0−2.5)×10−3=0.5×10−3mol

The concentration of hydroxide ions =0.5×10−3mol(25.0+30.0)×10−3L=0.009mol/L

Then the pH of the solution can be calculated as follows:

pOH=−log[0.009]=2.04pH=14.00−2.24=11.95

The data obtained from the calculations is as follows:

Volume of base added (mL)	pH
0.0	1
4.0	1.14
8.0	1.28
12.5	1.47
20.0	1.95
24.0	2.69
24.5	2.99
24.9	3.69
25.0	7.0
28.0	11.75
30.0	11.95

The titration curve can be drawn as follows:

EBK CHEMICAL PRINCIPLES, Chapter 8, Problem 144AE

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

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