Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN: 9781305079373
Author: William L. Masterton, Cecile N. Hurley
Publisher: Cengage Learning
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Chapter 23, Problem 29QAP

For alanine, K a1 = 5.1 × 10 5 , K a2 = 1.8 × 10 10 . Calculate the ratios [ Z ] / [ C + ] and [ Z ] / [ A + ] at pH
(a) 2.00.
(b) 6.00.
(c) 10.50.
What is the principal species at each pH?

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

The value of [Z][C+], [Z][A] and principal species at pH 2.00 needs to be determined.

Concept introduction:

General formula of alanine is shown as follows:

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  1

The pH is calculated as negative logarithm of molar hydrogen ion concentration.

pH=log[H+]

Answer to Problem 29QAP

[Z][C]==5.1×101

[Z][A]=5.5×107

The principal species at pH2.00 is cationic form.

Explanation of Solution

Under equilibrium condition, the Zwitter ionic form of alanine is shown as follows:

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  2

Under acidic condition, oxygen atom accepts the proton Zwitter ion exists in cationic form.

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  3

Under basic condition, base abstracts proton form nitrogen atom and it forms the anionic form.

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  4

Dissociation of cation is represented as follows:

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  5

Ka1=[Z][H+][C]

Dissociation of anion is represented as follows:

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  6

Ka2=[A][H+][Z]

Given value of Ka1andKa2 as follows:

Ka1=5.1×103,Ka2=1.8×1010

At pH=2.00

pH=log[H+]

So, [H+]=10pH

Take antilog both sides,

[H+]=102.00=0.01M=1×102

[Z][C] can be calculated as follows:

Here, Z is Zwitter ion and C is cationic form of Zwitter ion.

5.1×103=[Z]×1.0×102[C][Z][C]=5.1×1031.0×102=5.1×101

[Z][A] can be calculated as follows:

Here, Z is Zwitter ion and A is anionic form of Zwitter ion.

1.8×1010=[A]×1.0×102[Z]

Therefore,

[Z][A]=1.0×10-21.8×10-10=5.5×107

The principal species at pH2.00 is cationic form.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

The value of [Z][C+], [Z][A] and principal species at pH 6.00 needs to be determined.

Concept introduction:

General formula of alanine is shown as follows:

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  7

The pH is calculated as negative logarithm of molar hydrogen ion concentration.

pH=log[H+]

Answer to Problem 29QAP

[Z][C]=5.1×103

[Z][A]=5.5×103

The principal species at pH6.00 is cationic and anionic form.

Explanation of Solution

Under equilibrium condition zwitter ionic form of alanine is:

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  8

Under acidic condition oxygen atom accepts the proton zwitter ion exists in cationic form.

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  9

Under basic condition base abstracts proton form nitrogen atom and it forms the anionic form.

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  10

Dissociation of cation is-

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  11

Ka1=[Z][H+][C]

Dissociation of anion is-

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  12

Ka2=[A][H+][Z]

Given value of Ka1andKa2 as follows:

Ka1=5.1×103,Ka2=1.8×1010

At pH=6.00

pH=log[H+]

So, [H+]=10pH

Take antilog both sides,

[H+]=106.00=1.0×106M

[Z][C] can be calculated as:

Here, Z is zwitter ion and C is cationic form of zwitter ion.

5.1×103=[Z]×1.0×106[C]

[Z][C]=5.1×1031.0×106=5.1×103

[Z][A] can be calculated as follows:

Here, Z is zwitter ion and A is anionic form of zwitter ion.

1.8×1010=[A]×1.0×106[Z]

[Z][A]=1.0×1061.8×1010=5.5×103

The principal species at pH6.00 is cationic and anionic form.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

The value of [Z][C+], [Z][A] and principal species at pH 10.50 needs to be determined.

Concept introduction:

General formula of alanine is shown as follows:

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  13

The pH is calculated as negative logarithm of molar hydrogen ion concentration.

pH=log[H+]

Answer to Problem 29QAP

[Z][C]=1.6×108

[Z][A]=1.8×101

The principal species at pH=10.50 is anionic form.

Explanation of Solution

Under equilibrium condition zwitter ionic form of alanine is:

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  14

Under acidic condition oxygen atom accepts the proton zwitter ion exists in cationic form.

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  15

Under basic condition base abstracts proton form nitrogen atom and it forms the anionic form.

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  16

Dissociation of cation is-

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  17

Ka1=[Z][H+][C]

Dissociation of anion is-

Chemistry: Principles and Reactions, Chapter 23, Problem 29QAP , additional homework tip  18

Ka2=[A][H+][Z]

Given value of Ka1andKa2 as follows:

Ka1=5.1×103,Ka2=1.8×1010

At pH=10.50

pH=log[H+]

So, [H+]=10pH

Take antilog both sides,

[H+]=1010.50=3.2×1011M

[Z][C] can be calculated as:

Here, Z is zwitter ion and C is cationic form of zwitter ion.

5.1×103=[Z]×3.2×1011[C]

So,[Z][C]=5.1×1033.2×1011=1.6×108

[Z][A] can be calculated as follows:

Here, Z is zwitter ion and A is anionic form of zwitter ion.

1.8×1010=[A]×3.2×1011[Z]

[Z][A]=3.2×10111.8×1010=1.8×101

The principal species at pH=10.50 is anionic form.

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Chapter 23 Solutions

Chemistry: Principles and Reactions

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