The reason for not considering the second or third ionization constant of the polyprotic acid in the calculation of pH is to be explained. Concept introduction: Polyprotic acids that can lose and bases that can pick up, more than one H+ (e.g. diprotic H2A and triprotic H3A acids). For any polyprotic acid, there are more than one steps associated with the proton ionization of the acid. In one step, only one proton is dissociated from the acid. After the removal of a proton from a neutral acid, it is difficult to remove one more proton from an already proton-deficient or electron-rich species. Thus, the successive dissociations become less.

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Answer 1

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Chapter 16, Problem 80QP

Interpretation Introduction

Interpretation:

The reason for not considering the second or third ionization constant of the polyprotic acid in the calculation of pH is to be explained.

Concept introduction:

Polyprotic acids that can lose and bases that can pick up, more than one H+ (e.g. diprotic H2A and triprotic H3A acids).

For any polyprotic acid, there are more than one steps associated with the proton ionization of the acid. In one step, only one proton is dissociated from the acid.

After the removal of a proton from a neutral acid, it is difficult to remove one more proton from an already proton-deficient or electron-rich species. Thus, the successive dissociations become less.

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Answer 2

Question

Chapter 16, Problem 80QP

Interpretation Introduction

Interpretation:

The reason for not considering the second or third ionization constant of the polyprotic acid in the calculation of pH is to be explained.

Concept introduction:

Polyprotic acids that can lose and bases that can pick up, more than one H+ (e.g. diprotic H2A and triprotic H3A acids).

For any polyprotic acid, there are more than one steps associated with the proton ionization of the acid. In one step, only one proton is dissociated from the acid.

After the removal of a proton from a neutral acid, it is difficult to remove one more proton from an already proton-deficient or electron-rich species. Thus, the successive dissociations become less.

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Answer 3

Question

Chapter 16, Problem 80QP

Interpretation Introduction

Interpretation:

The reason for not considering the second or third ionization constant of the polyprotic acid in the calculation of pH is to be explained.

Concept introduction:

Polyprotic acids that can lose and bases that can pick up, more than one H+ (e.g. diprotic H2A and triprotic H3A acids).

For any polyprotic acid, there are more than one steps associated with the proton ionization of the acid. In one step, only one proton is dissociated from the acid.

After the removal of a proton from a neutral acid, it is difficult to remove one more proton from an already proton-deficient or electron-rich species. Thus, the successive dissociations become less.

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Answer 4

Question

Chapter 16, Problem 80QP

Interpretation Introduction

Interpretation:

The reason for not considering the second or third ionization constant of the polyprotic acid in the calculation of pH is to be explained.

Concept introduction:

Polyprotic acids that can lose and bases that can pick up, more than one H+ (e.g. diprotic H2A and triprotic H3A acids).

For any polyprotic acid, there are more than one steps associated with the proton ionization of the acid. In one step, only one proton is dissociated from the acid.

After the removal of a proton from a neutral acid, it is difficult to remove one more proton from an already proton-deficient or electron-rich species. Thus, the successive dissociations become less.

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Answer 5

Chapter 16, Problem 80QP

Interpretation Introduction

Interpretation:

The reason for not considering the second or third ionization constant of the polyprotic acid in the calculation of pH is to be explained.

Concept introduction:

Polyprotic acids that can lose and bases that can pick up, more than one H+ (e.g. diprotic H2A and triprotic H3A acids).

For any polyprotic acid, there are more than one steps associated with the proton ionization of the acid. In one step, only one proton is dissociated from the acid.

After the removal of a proton from a neutral acid, it is difficult to remove one more proton from an already proton-deficient or electron-rich species. Thus, the successive dissociations become less.

Answer 6

Chapter 16, Problem 80QP

Interpretation Introduction

Interpretation:

The reason for not considering the second or third ionization constant of the polyprotic acid in the calculation of pH is to be explained.

Concept introduction:

Polyprotic acids that can lose and bases that can pick up, more than one H+ (e.g. diprotic H2A and triprotic H3A acids).

For any polyprotic acid, there are more than one steps associated with the proton ionization of the acid. In one step, only one proton is dissociated from the acid.

After the removal of a proton from a neutral acid, it is difficult to remove one more proton from an already proton-deficient or electron-rich species. Thus, the successive dissociations become less.

Answer 7

Chapter 16, Problem 80QP

Interpretation Introduction

Interpretation:

The reason for not considering the second or third ionization constant of the polyprotic acid in the calculation of pH is to be explained.

Concept introduction:

Polyprotic acids that can lose and bases that can pick up, more than one H+ (e.g. diprotic H2A and triprotic H3A acids).

For any polyprotic acid, there are more than one steps associated with the proton ionization of the acid. In one step, only one proton is dissociated from the acid.

After the removal of a proton from a neutral acid, it is difficult to remove one more proton from an already proton-deficient or electron-rich species. Thus, the successive dissociations become less.

Answer 8

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Answer 10

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Answer 11

Ch. 16.1 - Practice Problem ATTEMPT What is (a) the conjugate...Ch. 16.1 - Practice Problem BUILD HSO 3 − is the conjugate...Ch. 16.1 - Practice Problem CONCEPTUALIZE Which of the models...Ch. 16.1 - 16.1.1 Which of the following pairs of species are...Ch. 16.1 - Which of the following species does not have a...Ch. 16.2 - Practice Problem ATTEMPT Identify and label the...Ch. 16.2 - Practice Problem BUILD (a) Write an equation in...Ch. 16.2 - Practice Problem CONCEPTUALIZE Write the formula...Ch. 16.2 - Calculate [ OH − ] in a solution in which [ H 3 O...Ch. 16.2 - Calculate [ H 3 O − ] in a solution in which [ OH...

Solution Summary: The author explains the reason for not considering the second or third ionization constant of the polyprotic acid in the calculation of pH.

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