The acids H M , H Q a n d H Z have to be arranged in order of decreasing strength. The K a values of these acids have to be determined where possible. Concept Introduction: The conjugate base of a strong acid is weak. The stronger base has the larger p H .

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 14, Problem 129QRT

Interpretation Introduction

Interpretation:

The acids HM, HQ and HZ have to be arranged in order of decreasing strength. The Ka values of these acids have to be determined where possible.

Concept Introduction:

The conjugate base of a strong acid is weak. The stronger base has the larger pH.

Expert Solution & Answer

Answer to Problem 129QRT

The decreasing order of strength of acids HM, HQ and HZ is HM>HQ>HZ. The Ka values of these acids HM, HQ and HZ are 1×10-1, 1×10-3 and 1×10-5 respectively.

Explanation of Solution

Given that, the pH values of sodium salts NaM, NaQ and NaZ are 7.0, 8.0 and 9.0 respectively. The decreasing order of pH values of these salts is given below.

NaM> NaQ> NaZ

Thus, the decreasing order of the acidic strength of the conjugate acids HM, HQ and HZ is given below.

HM>HQ>HZ

The general equation for the equilibrium can be written as given below.

A−(aq)+H2O(l)⇌HA(aq)+OH−(aq)

Where, A=M, Q or Z

The basic ionization constant for the above equation can be written as given below.

Kb=[HA][OH−][A−]

A table can be set up as given below.

A−(aq)HA(aq)OH−(aq)Initial conc.(M)0.101.0×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.1−xx10−7+x

The concentration of OH− comes from its autoionization that is 1.0×10−7.

By substituting all the values in the above equation, the value of x can be calculated.

Kb=(x)(10−7+x)(0.1−x)

1×10−7+x=[OH−]=10−pH=10−(14.00−pH)=10(pH−14.00)x=10(pH−14.00)−1×10−7

For NaZ, pH=9.0

x=10(9.0−14.00)−1×10−7=1×10−5−1×10−7=1×10−5Kb=(1×10−5)(1×10−7+1×10−5)(0.1−1×10−5)=1×10−9

For NaQ, pH=8.0

x=10(8.0−14.00)−1×10−7=1×10−6−1×10−7=1×10−6Kb=(1×10−6)(1×10−7+1×10−6)(0.1−1×10−6)=1×10−11

For NaM, pH=7.0

x=10(7.0−14.00)−1×10−7=1×10−7−1×10−7=0

This suggests that the reaction with water does not go toward products at all. That means Kb of M− may be so small that the pH of the solution is accounted for exclusively by the ionization of water. If indeed the base does provide all the OH− ions to make the solution’s pH, then [HM]=[OH−]=10−7.0=1×10−7:

Kb=(1×10−7)(1×10−7)(0.1)=1×10−13

For HZ, the value of Ka can be calculated as shown below.

Ka=KwKb=1×10−141×10−9=1×10−5

Similarly, for HQ

Ka=KwKb=1×10−141×10−11=1×10−3

For HM

Ka=KwKb=1×10−141×10−13=1×10−1

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Bacteria in the human mouth release enzymes to break up sucrose, and thenpolymerize glucose to form a bacterial plaque. When this glucose is digested, it formslactic acid, which causes tooth decay. Tooth enamel is made of calcium phosphate.Lactic acid has a Ka 1.38x10 -4 and ionizes into lactate and H3O+ ions. Assuming aninitial lactic acid concentration of 0.10 mol/L on your tooth, determine the pH(assume a 1:1:1 mole ratio) and the percent ionization. Should people be concernedwith plaque buildup? Justify your answer.

Calculate the pH at 25 of a 0.590 M aqueous solution of a weak base with a Kb value of 1.27 x 10-5.

Equimolar solutions of HF, HCOOH and HCN at 298 K have the values of Ka as 6.8×10^-4, 1.8×10^-4 and 4.8×10^-9 respectively. What will be the order of their acidic strength?

Answer 2

Question

Chapter 14, Problem 129QRT

Interpretation Introduction

Interpretation:

The acids HM, HQ and HZ have to be arranged in order of decreasing strength. The Ka values of these acids have to be determined where possible.

Concept Introduction:

The conjugate base of a strong acid is weak. The stronger base has the larger pH.

Expert Solution & Answer

Answer to Problem 129QRT

The decreasing order of strength of acids HM, HQ and HZ is HM>HQ>HZ. The Ka values of these acids HM, HQ and HZ are 1×10-1, 1×10-3 and 1×10-5 respectively.

Explanation of Solution

Given that, the pH values of sodium salts NaM, NaQ and NaZ are 7.0, 8.0 and 9.0 respectively. The decreasing order of pH values of these salts is given below.

NaM> NaQ> NaZ

Thus, the decreasing order of the acidic strength of the conjugate acids HM, HQ and HZ is given below.

HM>HQ>HZ

The general equation for the equilibrium can be written as given below.

A−(aq)+H2O(l)⇌HA(aq)+OH−(aq)

Where, A=M, Q or Z

The basic ionization constant for the above equation can be written as given below.

Kb=[HA][OH−][A−]

A table can be set up as given below.

A−(aq)HA(aq)OH−(aq)Initial conc.(M)0.101.0×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.1−xx10−7+x

The concentration of OH− comes from its autoionization that is 1.0×10−7.

By substituting all the values in the above equation, the value of x can be calculated.

Kb=(x)(10−7+x)(0.1−x)

1×10−7+x=[OH−]=10−pH=10−(14.00−pH)=10(pH−14.00)x=10(pH−14.00)−1×10−7

For NaZ, pH=9.0

x=10(9.0−14.00)−1×10−7=1×10−5−1×10−7=1×10−5Kb=(1×10−5)(1×10−7+1×10−5)(0.1−1×10−5)=1×10−9

For NaQ, pH=8.0

x=10(8.0−14.00)−1×10−7=1×10−6−1×10−7=1×10−6Kb=(1×10−6)(1×10−7+1×10−6)(0.1−1×10−6)=1×10−11

For NaM, pH=7.0

x=10(7.0−14.00)−1×10−7=1×10−7−1×10−7=0

This suggests that the reaction with water does not go toward products at all. That means Kb of M− may be so small that the pH of the solution is accounted for exclusively by the ionization of water. If indeed the base does provide all the OH− ions to make the solution’s pH, then [HM]=[OH−]=10−7.0=1×10−7:

Kb=(1×10−7)(1×10−7)(0.1)=1×10−13

For HZ, the value of Ka can be calculated as shown below.

Ka=KwKb=1×10−141×10−9=1×10−5

Similarly, for HQ

Ka=KwKb=1×10−141×10−11=1×10−3

For HM

Ka=KwKb=1×10−141×10−13=1×10−1

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 14, Problem 129QRT

Interpretation Introduction

Interpretation:

The acids HM, HQ and HZ have to be arranged in order of decreasing strength. The Ka values of these acids have to be determined where possible.

Concept Introduction:

The conjugate base of a strong acid is weak. The stronger base has the larger pH.

Expert Solution & Answer

Answer to Problem 129QRT

The decreasing order of strength of acids HM, HQ and HZ is HM>HQ>HZ. The Ka values of these acids HM, HQ and HZ are 1×10-1, 1×10-3 and 1×10-5 respectively.

Explanation of Solution

Given that, the pH values of sodium salts NaM, NaQ and NaZ are 7.0, 8.0 and 9.0 respectively. The decreasing order of pH values of these salts is given below.

NaM> NaQ> NaZ

Thus, the decreasing order of the acidic strength of the conjugate acids HM, HQ and HZ is given below.

HM>HQ>HZ

The general equation for the equilibrium can be written as given below.

A−(aq)+H2O(l)⇌HA(aq)+OH−(aq)

Where, A=M, Q or Z

The basic ionization constant for the above equation can be written as given below.

Kb=[HA][OH−][A−]

A table can be set up as given below.

A−(aq)HA(aq)OH−(aq)Initial conc.(M)0.101.0×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.1−xx10−7+x

The concentration of OH− comes from its autoionization that is 1.0×10−7.

By substituting all the values in the above equation, the value of x can be calculated.

Kb=(x)(10−7+x)(0.1−x)

1×10−7+x=[OH−]=10−pH=10−(14.00−pH)=10(pH−14.00)x=10(pH−14.00)−1×10−7

For NaZ, pH=9.0

x=10(9.0−14.00)−1×10−7=1×10−5−1×10−7=1×10−5Kb=(1×10−5)(1×10−7+1×10−5)(0.1−1×10−5)=1×10−9

For NaQ, pH=8.0

x=10(8.0−14.00)−1×10−7=1×10−6−1×10−7=1×10−6Kb=(1×10−6)(1×10−7+1×10−6)(0.1−1×10−6)=1×10−11

For NaM, pH=7.0

x=10(7.0−14.00)−1×10−7=1×10−7−1×10−7=0

This suggests that the reaction with water does not go toward products at all. That means Kb of M− may be so small that the pH of the solution is accounted for exclusively by the ionization of water. If indeed the base does provide all the OH− ions to make the solution’s pH, then [HM]=[OH−]=10−7.0=1×10−7:

Kb=(1×10−7)(1×10−7)(0.1)=1×10−13

For HZ, the value of Ka can be calculated as shown below.

Ka=KwKb=1×10−141×10−9=1×10−5

Similarly, for HQ

Ka=KwKb=1×10−141×10−11=1×10−3

For HM

Ka=KwKb=1×10−141×10−13=1×10−1

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 14, Problem 129QRT

Interpretation Introduction

Interpretation:

The acids HM, HQ and HZ have to be arranged in order of decreasing strength. The Ka values of these acids have to be determined where possible.

Concept Introduction:

The conjugate base of a strong acid is weak. The stronger base has the larger pH.

Expert Solution & Answer

Answer to Problem 129QRT

The decreasing order of strength of acids HM, HQ and HZ is HM>HQ>HZ. The Ka values of these acids HM, HQ and HZ are 1×10-1, 1×10-3 and 1×10-5 respectively.

Explanation of Solution

Given that, the pH values of sodium salts NaM, NaQ and NaZ are 7.0, 8.0 and 9.0 respectively. The decreasing order of pH values of these salts is given below.

NaM> NaQ> NaZ

Thus, the decreasing order of the acidic strength of the conjugate acids HM, HQ and HZ is given below.

HM>HQ>HZ

The general equation for the equilibrium can be written as given below.

A−(aq)+H2O(l)⇌HA(aq)+OH−(aq)

Where, A=M, Q or Z

The basic ionization constant for the above equation can be written as given below.

Kb=[HA][OH−][A−]

A table can be set up as given below.

A−(aq)HA(aq)OH−(aq)Initial conc.(M)0.101.0×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.1−xx10−7+x

The concentration of OH− comes from its autoionization that is 1.0×10−7.

By substituting all the values in the above equation, the value of x can be calculated.

Kb=(x)(10−7+x)(0.1−x)

1×10−7+x=[OH−]=10−pH=10−(14.00−pH)=10(pH−14.00)x=10(pH−14.00)−1×10−7

For NaZ, pH=9.0

x=10(9.0−14.00)−1×10−7=1×10−5−1×10−7=1×10−5Kb=(1×10−5)(1×10−7+1×10−5)(0.1−1×10−5)=1×10−9

For NaQ, pH=8.0

x=10(8.0−14.00)−1×10−7=1×10−6−1×10−7=1×10−6Kb=(1×10−6)(1×10−7+1×10−6)(0.1−1×10−6)=1×10−11

For NaM, pH=7.0

x=10(7.0−14.00)−1×10−7=1×10−7−1×10−7=0

This suggests that the reaction with water does not go toward products at all. That means Kb of M− may be so small that the pH of the solution is accounted for exclusively by the ionization of water. If indeed the base does provide all the OH− ions to make the solution’s pH, then [HM]=[OH−]=10−7.0=1×10−7:

Kb=(1×10−7)(1×10−7)(0.1)=1×10−13

For HZ, the value of Ka can be calculated as shown below.

Ka=KwKb=1×10−141×10−9=1×10−5

Similarly, for HQ

Ka=KwKb=1×10−141×10−11=1×10−3

For HM

Ka=KwKb=1×10−141×10−13=1×10−1

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 14, Problem 129QRT

Interpretation Introduction

Interpretation:

The acids HM, HQ and HZ have to be arranged in order of decreasing strength. The Ka values of these acids have to be determined where possible.

Concept Introduction:

The conjugate base of a strong acid is weak. The stronger base has the larger pH.

Expert Solution & Answer

Answer to Problem 129QRT

The decreasing order of strength of acids HM, HQ and HZ is HM>HQ>HZ. The Ka values of these acids HM, HQ and HZ are 1×10-1, 1×10-3 and 1×10-5 respectively.

Explanation of Solution

Given that, the pH values of sodium salts NaM, NaQ and NaZ are 7.0, 8.0 and 9.0 respectively. The decreasing order of pH values of these salts is given below.

NaM> NaQ> NaZ

Thus, the decreasing order of the acidic strength of the conjugate acids HM, HQ and HZ is given below.

HM>HQ>HZ

The general equation for the equilibrium can be written as given below.

A−(aq)+H2O(l)⇌HA(aq)+OH−(aq)

Where, A=M, Q or Z

The basic ionization constant for the above equation can be written as given below.

Kb=[HA][OH−][A−]

A table can be set up as given below.

A−(aq)HA(aq)OH−(aq)Initial conc.(M)0.101.0×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.1−xx10−7+x

The concentration of OH− comes from its autoionization that is 1.0×10−7.

By substituting all the values in the above equation, the value of x can be calculated.

Kb=(x)(10−7+x)(0.1−x)

1×10−7+x=[OH−]=10−pH=10−(14.00−pH)=10(pH−14.00)x=10(pH−14.00)−1×10−7

For NaZ, pH=9.0

x=10(9.0−14.00)−1×10−7=1×10−5−1×10−7=1×10−5Kb=(1×10−5)(1×10−7+1×10−5)(0.1−1×10−5)=1×10−9

For NaQ, pH=8.0

x=10(8.0−14.00)−1×10−7=1×10−6−1×10−7=1×10−6Kb=(1×10−6)(1×10−7+1×10−6)(0.1−1×10−6)=1×10−11

For NaM, pH=7.0

x=10(7.0−14.00)−1×10−7=1×10−7−1×10−7=0

This suggests that the reaction with water does not go toward products at all. That means Kb of M− may be so small that the pH of the solution is accounted for exclusively by the ionization of water. If indeed the base does provide all the OH− ions to make the solution’s pH, then [HM]=[OH−]=10−7.0=1×10−7:

Kb=(1×10−7)(1×10−7)(0.1)=1×10−13

For HZ, the value of Ka can be calculated as shown below.

Ka=KwKb=1×10−141×10−9=1×10−5

Similarly, for HQ

Ka=KwKb=1×10−141×10−11=1×10−3

For HM

Ka=KwKb=1×10−141×10−13=1×10−1

Answer 6

Chapter 14, Problem 129QRT

Interpretation Introduction

Interpretation:

The acids HM, HQ and HZ have to be arranged in order of decreasing strength. The Ka values of these acids have to be determined where possible.

Concept Introduction:

The conjugate base of a strong acid is weak. The stronger base has the larger pH.

Expert Solution & Answer

Answer to Problem 129QRT

The decreasing order of strength of acids HM, HQ and HZ is HM>HQ>HZ. The Ka values of these acids HM, HQ and HZ are 1×10-1, 1×10-3 and 1×10-5 respectively.

Explanation of Solution

Given that, the pH values of sodium salts NaM, NaQ and NaZ are 7.0, 8.0 and 9.0 respectively. The decreasing order of pH values of these salts is given below.

NaM> NaQ> NaZ

Thus, the decreasing order of the acidic strength of the conjugate acids HM, HQ and HZ is given below.

HM>HQ>HZ

The general equation for the equilibrium can be written as given below.

A−(aq)+H2O(l)⇌HA(aq)+OH−(aq)

Where, A=M, Q or Z

The basic ionization constant for the above equation can be written as given below.

Kb=[HA][OH−][A−]

A table can be set up as given below.

A−(aq)HA(aq)OH−(aq)Initial conc.(M)0.101.0×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.1−xx10−7+x

The concentration of OH− comes from its autoionization that is 1.0×10−7.

By substituting all the values in the above equation, the value of x can be calculated.

Kb=(x)(10−7+x)(0.1−x)

1×10−7+x=[OH−]=10−pH=10−(14.00−pH)=10(pH−14.00)x=10(pH−14.00)−1×10−7

For NaZ, pH=9.0

x=10(9.0−14.00)−1×10−7=1×10−5−1×10−7=1×10−5Kb=(1×10−5)(1×10−7+1×10−5)(0.1−1×10−5)=1×10−9

For NaQ, pH=8.0

x=10(8.0−14.00)−1×10−7=1×10−6−1×10−7=1×10−6Kb=(1×10−6)(1×10−7+1×10−6)(0.1−1×10−6)=1×10−11

For NaM, pH=7.0

x=10(7.0−14.00)−1×10−7=1×10−7−1×10−7=0

This suggests that the reaction with water does not go toward products at all. That means Kb of M− may be so small that the pH of the solution is accounted for exclusively by the ionization of water. If indeed the base does provide all the OH− ions to make the solution’s pH, then [HM]=[OH−]=10−7.0=1×10−7:

Kb=(1×10−7)(1×10−7)(0.1)=1×10−13

For HZ, the value of Ka can be calculated as shown below.

Ka=KwKb=1×10−141×10−9=1×10−5

Similarly, for HQ

Ka=KwKb=1×10−141×10−11=1×10−3

For HM

Ka=KwKb=1×10−141×10−13=1×10−1

Answer 7

Chapter 14, Problem 129QRT

Interpretation Introduction

Interpretation:

The acids HM, HQ and HZ have to be arranged in order of decreasing strength. The Ka values of these acids have to be determined where possible.

Concept Introduction:

The conjugate base of a strong acid is weak. The stronger base has the larger pH.

Answer 8

Expert Solution & Answer

Answer to Problem 129QRT

The decreasing order of strength of acids HM, HQ and HZ is HM>HQ>HZ. The Ka values of these acids HM, HQ and HZ are 1×10-1, 1×10-3 and 1×10-5 respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given that, the pH values of sodium salts NaM, NaQ and NaZ are 7.0, 8.0 and 9.0 respectively. The decreasing order of pH values of these salts is given below.

NaM> NaQ> NaZ

Thus, the decreasing order of the acidic strength of the conjugate acids HM, HQ and HZ is given below.

HM>HQ>HZ

The general equation for the equilibrium can be written as given below.

A−(aq)+H2O(l)⇌HA(aq)+OH−(aq)

Where, A=M, Q or Z

The basic ionization constant for the above equation can be written as given below.

Kb=[HA][OH−][A−]

A table can be set up as given below.

A−(aq)HA(aq)OH−(aq)Initial conc.(M)0.101.0×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.1−xx10−7+x

The concentration of OH− comes from its autoionization that is 1.0×10−7.

By substituting all the values in the above equation, the value of x can be calculated.

Kb=(x)(10−7+x)(0.1−x)

1×10−7+x=[OH−]=10−pH=10−(14.00−pH)=10(pH−14.00)x=10(pH−14.00)−1×10−7

For NaZ, pH=9.0

x=10(9.0−14.00)−1×10−7=1×10−5−1×10−7=1×10−5Kb=(1×10−5)(1×10−7+1×10−5)(0.1−1×10−5)=1×10−9

For NaQ, pH=8.0

x=10(8.0−14.00)−1×10−7=1×10−6−1×10−7=1×10−6Kb=(1×10−6)(1×10−7+1×10−6)(0.1−1×10−6)=1×10−11

For NaM, pH=7.0

x=10(7.0−14.00)−1×10−7=1×10−7−1×10−7=0

This suggests that the reaction with water does not go toward products at all. That means Kb of M− may be so small that the pH of the solution is accounted for exclusively by the ionization of water. If indeed the base does provide all the OH− ions to make the solution’s pH, then [HM]=[OH−]=10−7.0=1×10−7:

Kb=(1×10−7)(1×10−7)(0.1)=1×10−13

For HZ, the value of Ka can be calculated as shown below.

Ka=KwKb=1×10−141×10−9=1×10−5

Similarly, for HQ

Ka=KwKb=1×10−141×10−11=1×10−3

For HM

Ka=KwKb=1×10−141×10−13=1×10−1

Answer 12

Ch. 14.1 - Identify each molecule or ion as a Brnsted-Lowry...Ch. 14.1 - Using Le Chatelier’s Principle Use Le Chatelier’s...Ch. 14.1 - Prob. 14.3E Ch. 14.1 - Complete the table. Ch. 14.1 - Prob. 14.4E Ch. 14.1 - Prob. 14.5E Ch. 14.2 - Prob. 14.6CE Ch. 14.2 - Prob. 14.7E Ch. 14.3 - Prob. 14.2PSP Ch. 14.3 - Prob. 14.3PSP

The acids H M , H Q a n d H Z have to be arranged in order of decreasing strength. The K a values of these acids have to be determined where possible. Concept Introduction: The conjugate base of a strong acid is weak. The stronger base has the larger p H .

Concept explainers

Videos

Interpretation:

The acids HM, HQ and HZ have to be arranged in order of decreasing strength. The Ka values of these acids have to be determined where possible.

Concept Introduction:

The conjugate base of a strong acid is weak. The stronger base has the larger pH.

Answer to Problem 129QRT

Explanation of Solution

Want to see more full solutions like this?

Chapter 14 Solutions

The acids H M , H Q a n d H Z have to be arranged in order of decreasing strength. The K a values of these acids have to be determined where possible. Concept Introduction: The conjugate base of a strong acid is weak. The stronger base has the larger p H .

Concept explainers

Videos

Interpretation: The acids HM, HQ and HZ have to be arranged in order of decreasing strength. The Ka values of these acids have to be determined where possible. Concept Introduction: The conjugate base of a strong acid is weak. The stronger base has the larger pH.

Answer to Problem 129QRT

Explanation of Solution

Want to see more full solutions like this?

Chapter 14 Solutions

Interpretation:

The acids HM, HQ and HZ have to be arranged in order of decreasing strength. The Ka values of these acids have to be determined where possible.

Concept Introduction:

The conjugate base of a strong acid is weak. The stronger base has the larger pH.