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.

Question

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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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There are 48 pairs of students in the following table. Each pair has quantitatively determined the mass of taurine in a 250 mL can of the popular energy drink marketed as “Munster” using High Performance Liquid Chromatography (HPLC). The class results are presented below: QUESTION: Summarise and report these results including an indication of measurement uncertainty. In both calculation samples calculate if an outlier is present, max value, number of samples, mean, standard deviation, g (suspect), g (critical) and t (critical). Mass of Taurine (mg) Mass of Taurine (mg) (Table continued) 152.01 152.23 151.87 151.45 154.11 152.64 152.98 153.24 152.88 151.45 153.49 152.48 150.68 152.33 151.52 153.63 152.48 151.68 153.17 153.40 153.77 153.67 152.34 153.16 152.57 153.02 152.86 151.50 151.23 152.57 152.72 151.54 146.47 152.38 152.44 152.54 152.53 152.54 151.32…

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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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