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Chapter 16, Problem 16.64EP

(a)

Interpretation Introduction

Interpretation:

Chemical equation for the conversion of given carboxylic acid salt to its parent carboxylic acid using HCl has to be written.

Concept Introduction:

The name of the carboxylic acid itself implies that it is acidic.  Addition of carboxylic acid to water results in ionization.  Hydrogen ion transfer occurs from carboxylic acid to water and hydronium ion is formed.  Carboxylate ion is also formed due to the loss of hydrogen ion from carboxylic acid.

Carboxylate ion is the negative ion which is formed when one or more acidic protons are lost from carboxylic acid.  Similar to carboxylic acid it reacts with strong base to form carboxylic acid salt and water.

Carboxylic acid salts when treated with a strong acid produces carboxylic acid as the product.

Carboxylic acid forms carboxylic acid salt by reacting with a strong base.  The general reaction scheme for the formation of carboxylic acid salt is given as shown below,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 16, Problem 16.64EP , additional homework tip  1

The reverse of the above reaction is conversion of carboxylic acid salt to carboxylic acid.  This is accomplished by using strong acid.  The scheme is shown below,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 16, Problem 16.64EP , additional homework tip  2

From the above chemical equation it is found that the carboxylic acid salt reacts with strong acid to form carboxylic acid.

(a)

Expert Solution
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Explanation of Solution

Given carboxylic acid salt is calcium propanoate.  The structure of calcium propanoate can be given as shown below,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 16, Problem 16.64EP , additional homework tip  3

Carboxylic acid sat is converted to carboxylic acid by reaction with strong acid.  In the problem statement it is given that the strong acid is HCl.  Calcium propanoate on reaction with hydrochloric acid gives propanoic acid as product.  The chemical equation for the reaction can be given as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 16, Problem 16.64EP , additional homework tip  4

Conclusion

Chemical equation for the conversion of given carboxylic acid salt to carboxylic acid using hydrochloric acid is written.

(b)

Interpretation Introduction

Interpretation:

Chemical equation for the conversion of given carboxylic acid salt to its parent carboxylic acid using HCl has to be written.

Concept Introduction:

The name of the carboxylic acid itself implies that it is acidic.  Addition of carboxylic acid to water results in ionization.  Hydrogen ion transfer occurs from carboxylic acid to water and hydronium ion is formed.  Carboxylate ion is also formed due to the loss of hydrogen ion from carboxylic acid.

Carboxylate ion is the negative ion which is formed when one or more acidic protons are lost from carboxylic acid.  Similar to carboxylic acid it reacts with strong base to form carboxylic acid salt and water.

Carboxylic acid salts when treated with a strong acid produces carboxylic acid as the product.

Carboxylic acid forms carboxylic acid salt by reacting with a strong base.  The general reaction scheme for the formation of carboxylic acid salt is given as shown below,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 16, Problem 16.64EP , additional homework tip  5

The reverse of the above reaction is conversion of carboxylic acid salt to carboxylic acid.  This is accomplished by using strong acid.  The scheme is shown below,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 16, Problem 16.64EP , additional homework tip  6

From the above chemical equation it is found that the carboxylic acid salt reacts with strong acid to form carboxylic acid.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given carboxylic acid salt is sodium lactate.  The structure of sodium lactate can be given as shown below,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 16, Problem 16.64EP , additional homework tip  7

Carboxylic acid sat is converted to carboxylic acid by reaction with strong acid.  In the problem statement it is given that the strong acid is HCl.  Sodium lactate on reaction with hydrochloric acid gives lactic acid as product.  The chemical equation for the reaction can be given as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 16, Problem 16.64EP , additional homework tip  8

Conclusion

Chemical equation for the conversion of given carboxylic acid salt to carboxylic acid using hydrochloric acid is written.

(c)

Interpretation Introduction

Interpretation:

Chemical equation for the conversion of given carboxylic acid salt to its parent carboxylic acid using HCl has to be written.

Concept Introduction:

The name of the carboxylic acid itself implies that it is acidic.  Addition of carboxylic acid to water results in ionization.  Hydrogen ion transfer occurs from carboxylic acid to water and hydronium ion is formed.  Carboxylate ion is also formed due to the loss of hydrogen ion from carboxylic acid.

Carboxylate ion is the negative ion which is formed when one or more acidic protons are lost from carboxylic acid.  Similar to carboxylic acid it reacts with strong base to form carboxylic acid salt and water.

Carboxylic acid salts when treated with a strong acid produces carboxylic acid as the product.

Carboxylic acid forms carboxylic acid salt by reacting with a strong base.  The general reaction scheme for the formation of carboxylic acid salt is given as shown below,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 16, Problem 16.64EP , additional homework tip  9

The reverse of the above reaction is conversion of carboxylic acid salt to carboxylic acid.  This is accomplished by using strong acid.  The scheme is shown below,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 16, Problem 16.64EP , additional homework tip  10

From the above chemical equation it is found that the carboxylic acid salt reacts with strong acid to form carboxylic acid.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given carboxylic acid salt is magnesium succinate.  The structure of magnesium succinate can be given as shown below,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 16, Problem 16.64EP , additional homework tip  11

Carboxylic acid sat is converted to carboxylic acid by reaction with strong acid.  In the problem statement it is given that the strong acid is HCl.  Magnesium succinate on reaction with hydrochloric acid gives succinic acid as product.  The chemical equation for the reaction can be given as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 16, Problem 16.64EP , additional homework tip  12

Conclusion

Chemical equation for the conversion of given carboxylic acid salt to carboxylic acid using hydrochloric acid is written.

(d)

Interpretation Introduction

Interpretation:

Chemical equation for the conversion of given carboxylic acid salt to its parent carboxylic acid using HCl has to be written.

Concept Introduction:

The name of the carboxylic acid itself implies that it is acidic.  Addition of carboxylic acid to water results in ionization.  Hydrogen ion transfer occurs from carboxylic acid to water and hydronium ion is formed.  Carboxylate ion is also formed due to the loss of hydrogen ion from carboxylic acid.

Carboxylate ion is the negative ion which is formed when one or more acidic protons are lost from carboxylic acid.  Similar to carboxylic acid it reacts with strong base to form carboxylic acid salt and water.

Carboxylic acid salts when treated with a strong acid produces carboxylic acid as the product.

Carboxylic acid forms carboxylic acid salt by reacting with a strong base.  The general reaction scheme for the formation of carboxylic acid salt is given as shown below,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 16, Problem 16.64EP , additional homework tip  13

The reverse of the above reaction is conversion of carboxylic acid salt to carboxylic acid.  This is accomplished by using strong acid.  The scheme is shown below,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 16, Problem 16.64EP , additional homework tip  14

From the above chemical equation it is found that the carboxylic acid salt reacts with strong acid to form carboxylic acid.

(d)

Expert Solution
Check Mark

Explanation of Solution

Given carboxylic acid salt is potassium benzoate.  The structure of potassium benzoate can be given as shown below,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 16, Problem 16.64EP , additional homework tip  15

Carboxylic acid sat is converted to carboxylic acid by reaction with strong acid.  In the problem statement it is given that the strong acid is HCl.  Potassium benzoate on reaction with hydrochloric acid gives benzoic acid as product.  The chemical equation for the reaction can be given as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 16, Problem 16.64EP , additional homework tip  16

Conclusion

Chemical equation for the conversion of given carboxylic acid salt to carboxylic acid using hydrochloric acid is written.

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Chapter 16 Solutions

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card

Ch. 16.3 - Prob. 1QQCh. 16.3 - Prob. 2QQCh. 16.3 - Prob. 3QQCh. 16.4 - Prob. 1QQCh. 16.4 - Prob. 2QQCh. 16.4 - Prob. 3QQCh. 16.5 - Prob. 1QQCh. 16.5 - Prob. 2QQCh. 16.5 - Prob. 3QQCh. 16.6 - Prob. 1QQCh. 16.6 - Prob. 2QQCh. 16.7 - Prob. 1QQCh. 16.7 - Prob. 2QQCh. 16.7 - Prob. 3QQCh. 16.8 - Prob. 1QQCh. 16.8 - Prob. 2QQCh. 16.8 - Prob. 3QQCh. 16.8 - Prob. 4QQCh. 16.9 - Prob. 1QQCh. 16.9 - Prob. 2QQCh. 16.10 - Prob. 1QQCh. 16.10 - Prob. 2QQCh. 16.11 - Prob. 1QQCh. 16.11 - Prob. 2QQCh. 16.11 - Prob. 3QQCh. 16.12 - Prob. 1QQCh. 16.12 - Prob. 2QQCh. 16.12 - Prob. 3QQCh. 16.12 - Prob. 4QQCh. 16.13 - Prob. 1QQCh. 16.13 - Prob. 2QQCh. 16.14 - Prob. 1QQCh. 16.14 - Prob. 2QQCh. 16.14 - Prob. 3QQCh. 16.15 - Prob. 1QQCh. 16.15 - Prob. 2QQCh. 16.15 - Prob. 3QQCh. 16.16 - Prob. 1QQCh. 16.16 - Prob. 2QQCh. 16.16 - Prob. 3QQCh. 16.17 - Prob. 1QQCh. 16.17 - Prob. 2QQCh. 16.18 - Prob. 1QQCh. 16.18 - Prob. 2QQCh. 16.18 - Prob. 3QQCh. 16.19 - Prob. 1QQCh. 16.19 - Prob. 2QQCh. 16.19 - Prob. 3QQCh. 16.19 - Prob. 4QQCh. 16.20 - Prob. 1QQCh. 16.20 - Prob. 2QQCh. 16.20 - Prob. 3QQCh. 16.20 - Prob. 4QQCh. 16 - Prob. 16.1EPCh. 16 - Prob. 16.2EPCh. 16 - Prob. 16.3EPCh. 16 - Prob. 16.4EPCh. 16 - Prob. 16.5EPCh. 16 - Prob. 16.6EPCh. 16 - Prob. 16.7EPCh. 16 - Prob. 16.8EPCh. 16 - Prob. 16.9EPCh. 16 - Prob. 16.10EPCh. 16 - Prob. 16.11EPCh. 16 - Prob. 16.12EPCh. 16 - Prob. 16.13EPCh. 16 - Prob. 16.14EPCh. 16 - Prob. 16.15EPCh. 16 - Prob. 16.16EPCh. 16 - Prob. 16.17EPCh. 16 - Prob. 16.18EPCh. 16 - Prob. 16.19EPCh. 16 - Prob. 16.20EPCh. 16 - Prob. 16.21EPCh. 16 - Prob. 16.22EPCh. 16 - Prob. 16.23EPCh. 16 - Prob. 16.24EPCh. 16 - Prob. 16.25EPCh. 16 - Prob. 16.26EPCh. 16 - Prob. 16.27EPCh. 16 - Prob. 16.28EPCh. 16 - Prob. 16.29EPCh. 16 - Prob. 16.30EPCh. 16 - Prob. 16.31EPCh. 16 - Prob. 16.32EPCh. 16 - Prob. 16.33EPCh. 16 - Prob. 16.34EPCh. 16 - Prob. 16.35EPCh. 16 - Prob. 16.36EPCh. 16 - Prob. 16.37EPCh. 16 - Prob. 16.38EPCh. 16 - Prob. 16.39EPCh. 16 - Prob. 16.40EPCh. 16 - Determine the maximum number of hydrogen bonds...Ch. 16 - Prob. 16.42EPCh. 16 - Prob. 16.43EPCh. 16 - Prob. 16.44EPCh. 16 - Prob. 16.45EPCh. 16 - Prob. 16.46EPCh. 16 - Prob. 16.47EPCh. 16 - Prob. 16.48EPCh. 16 - Prob. 16.49EPCh. 16 - Prob. 16.50EPCh. 16 - Prob. 16.51EPCh. 16 - Prob. 16.52EPCh. 16 - Prob. 16.53EPCh. 16 - Prob. 16.54EPCh. 16 - Prob. 16.55EPCh. 16 - Prob. 16.56EPCh. 16 - Give the IUPAC name for each of the following...Ch. 16 - Give the IUPAC name for each of the following...Ch. 16 - Prob. 16.59EPCh. 16 - Give the common name for each of the carboxylic...Ch. 16 - Prob. 16.61EPCh. 16 - Write a chemical equation for the preparation of...Ch. 16 - Prob. 16.63EPCh. 16 - Prob. 16.64EPCh. 16 - Prob. 16.65EPCh. 16 - Prob. 16.66EPCh. 16 - Prob. 16.67EPCh. 16 - Prob. 16.68EPCh. 16 - Prob. 16.69EPCh. 16 - Prob. 16.70EPCh. 16 - Prob. 16.71EPCh. 16 - Prob. 16.72EPCh. 16 - Prob. 16.73EPCh. 16 - Prob. 16.74EPCh. 16 - Prob. 16.75EPCh. 16 - Prob. 16.76EPCh. 16 - Prob. 16.77EPCh. 16 - Prob. 16.78EPCh. 16 - Prob. 16.79EPCh. 16 - Prob. 16.80EPCh. 16 - Prob. 16.81EPCh. 16 - Prob. 16.82EPCh. 16 - Prob. 16.83EPCh. 16 - Prob. 16.84EPCh. 16 - Prob. 16.85EPCh. 16 - Prob. 16.86EPCh. 16 - Prob. 16.87EPCh. 16 - Prob. 16.88EPCh. 16 - Prob. 16.89EPCh. 16 - Prob. 16.90EPCh. 16 - Prob. 16.91EPCh. 16 - Prob. 16.92EPCh. 16 - Assign an IUPAC name to each of the following...Ch. 16 - Prob. 16.94EPCh. 16 - Prob. 16.95EPCh. 16 - Prob. 16.96EPCh. 16 - Prob. 16.97EPCh. 16 - Prob. 16.98EPCh. 16 - Prob. 16.99EPCh. 16 - Prob. 16.100EPCh. 16 - How many carbon atoms are present in a molecule of...Ch. 16 - Prob. 16.102EPCh. 16 - Prob. 16.103EPCh. 16 - Prob. 16.104EPCh. 16 - Prob. 16.105EPCh. 16 - Prob. 16.106EPCh. 16 - Prob. 16.107EPCh. 16 - Prob. 16.108EPCh. 16 - Prob. 16.109EPCh. 16 - Prob. 16.110EPCh. 16 - Prob. 16.111EPCh. 16 - Prob. 16.112EPCh. 16 - Prob. 16.113EPCh. 16 - Prob. 16.114EPCh. 16 - Prob. 16.115EPCh. 16 - Prob. 16.116EPCh. 16 - Prob. 16.117EPCh. 16 - Prob. 16.118EPCh. 16 - Prob. 16.119EPCh. 16 - Prob. 16.120EPCh. 16 - Prob. 16.121EPCh. 16 - Prob. 16.122EPCh. 16 - Prob. 16.123EPCh. 16 - Prob. 16.124EPCh. 16 - Prob. 16.125EPCh. 16 - Prob. 16.126EPCh. 16 - Prob. 16.127EPCh. 16 - Prob. 16.128EPCh. 16 - Prob. 16.129EPCh. 16 - Prob. 16.130EPCh. 16 - Prob. 16.131EPCh. 16 - Prob. 16.132EPCh. 16 - Prob. 16.133EPCh. 16 - Prob. 16.134EPCh. 16 - Prob. 16.135EPCh. 16 - Prob. 16.136EPCh. 16 - Prob. 16.137EPCh. 16 - Prob. 16.138EPCh. 16 - Prob. 16.139EPCh. 16 - Prob. 16.140EPCh. 16 - Prob. 16.141EPCh. 16 - Prob. 16.142EPCh. 16 - Prob. 16.143EPCh. 16 - Prob. 16.144EPCh. 16 - Prob. 16.145EPCh. 16 - Prob. 16.146EPCh. 16 - Prob. 16.147EPCh. 16 - Prob. 16.148EPCh. 16 - Draw a condensed structural formula for the...Ch. 16 - Prob. 16.150EPCh. 16 - Prob. 16.151EPCh. 16 - Prob. 16.152EPCh. 16 - Prob. 16.153EPCh. 16 - Prob. 16.154EPCh. 16 - Prob. 16.155EPCh. 16 - Prob. 16.156EPCh. 16 - Prob. 16.157EPCh. 16 - Prob. 16.158EPCh. 16 - Prob. 16.159EPCh. 16 - Prob. 16.160EPCh. 16 - Prob. 16.161EPCh. 16 - Prob. 16.162EPCh. 16 - Prob. 16.163EPCh. 16 - Prob. 16.164EP
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