From the given the p K a values of the two groups, − + N H 3 and − C O O H of valine, the more acidic one has to be determined. Concept Introduction: Greater the p K a value of a substance lower will be its acidic property. According to equation given in chapter 16 in the text book (equation 16.4), p H and p K a are related as, p H = p K a + log [ c o n j u g a t e b a s e ] [ a c i d ]

Question

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Answer 1

Question

Chapter 22, Problem 22.47SP

(a)

Interpretation Introduction

Interpretation:

From the given the $pKa$ values of the two groups, $−+NH3$ and $−COOH$ of valine, the more acidic one has to be determined.

Concept Introduction:

Greater the $pKa$ value of a substance lower will be its acidic property.

According to equation given in chapter 16 in the text book (equation 16.4),

$pH$ and $pKa$ are related as,

$pH = pKa + log [conjugate base][acid]$

(a)

Expert Solution

Answer to Problem 22.47SP

$−COOH$ group is more acidic.

Explanation of Solution

Given that $pKa$ of $−+NH3$ and $−COOH$ groups of valine are $9.62$ and $2.32$ respectively.

Since $−COOH$ has lower $pKa$ value of $2.32$ , it is more acidic.

(b)

Interpretation Introduction

Interpretation:

The predominant form of valine at $pH$ $1.0,7.0$ and $12$ has to be predicted.

Concept Introduction:

Greater the $pKa$ value of a substance lower will be its acidic property.

According to equation given in chapter 16 in the text book (equation 16.4),

$pH$ and $pKa$ are related as,

$pH = pKa + log [conjugate base][acid]$

(b)

Expert Solution

Explanation of Solution

Given data:

$pKa of carboxyl group, COOH = 2.3pKa of ammonium group, NH3+ = 9.6$

At $pH = 1,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]1 = 2.32+ log [−COO−][−COOH][−COOH][−COO−] = 21$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]1 = 9.62+ log [−NH2][−NH3+][−NH3+][−NH2] = 4.2×108$

Concentration of $−NH3+$ is greater than $−COOH$ . Hence, the predominant species is $CH(CH3)2−CH(+NH3)−COOH.$

At $pH = 7,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]7 = 2.32+ log [−COO−][−COOH][−COO−][−COOH] = 4.8×104$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]7 = 9.62+ log [−NH2][−NH3+][−NH3+][−NH2] = 4.2×102$

Concentration of $−COO−$ is greater than $−NH3+$ . Hence, the predominant species is $CH(CH3)2−CH(+NH3)−COO−.$

At $pH = 12,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]12 = 2.32+ log [−COO−][−COOH][−COO−][−COOH] = 4.8×109$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]12 = 9.62+ log [−NH2][−NH3+][−NH2][−NH3+] = 2.4×102$

Concentration of $−COO−$ is greater than $−NH3+$ . Hence, the predominant species is $CH(CH3)2−CH(NH2)−COO−.$

(c)

Interpretation Introduction

Interpretation:

Isoelectric point of valine has to be calculated.

Concept Introduction:

Isoelectric point $(pI)$ of an amino acid refers to the $pH$ at which the amino acid remains neutral and does not migrate in an electric field

It is calculated as,

$pI = pKa1+pKa22$

Where ,

$pKa1and pKa2$ are two $pKa$ values of amino acid.

(c)

Expert Solution

Answer to Problem 22.47SP

Isoelectric point of valine is calculated as $5.97$

Explanation of Solution

Given data:

$pKa1 of carboxyl group, COOH, = 2.32pKa2 of ammonium group, NH3+ = 9.62$

Therefore isoelectronic point of valine is,

$pI = pKa1+pKa22 = 2.32 + 9.622 = 5.97$

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Answer 2

Question

Chapter 22, Problem 22.47SP

(a)

Interpretation Introduction

Interpretation:

From the given the $pKa$ values of the two groups, $−+NH3$ and $−COOH$ of valine, the more acidic one has to be determined.

Concept Introduction:

Greater the $pKa$ value of a substance lower will be its acidic property.

According to equation given in chapter 16 in the text book (equation 16.4),

$pH$ and $pKa$ are related as,

$pH = pKa + log [conjugate base][acid]$

(a)

Expert Solution

Answer to Problem 22.47SP

$−COOH$ group is more acidic.

Explanation of Solution

Given that $pKa$ of $−+NH3$ and $−COOH$ groups of valine are $9.62$ and $2.32$ respectively.

Since $−COOH$ has lower $pKa$ value of $2.32$ , it is more acidic.

(b)

Interpretation Introduction

Interpretation:

The predominant form of valine at $pH$ $1.0,7.0$ and $12$ has to be predicted.

Concept Introduction:

Greater the $pKa$ value of a substance lower will be its acidic property.

According to equation given in chapter 16 in the text book (equation 16.4),

$pH$ and $pKa$ are related as,

$pH = pKa + log [conjugate base][acid]$

(b)

Expert Solution

Explanation of Solution

Given data:

$pKa of carboxyl group, COOH = 2.3pKa of ammonium group, NH3+ = 9.6$

At $pH = 1,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]1 = 2.32+ log [−COO−][−COOH][−COOH][−COO−] = 21$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]1 = 9.62+ log [−NH2][−NH3+][−NH3+][−NH2] = 4.2×108$

Concentration of $−NH3+$ is greater than $−COOH$ . Hence, the predominant species is $CH(CH3)2−CH(+NH3)−COOH.$

At $pH = 7,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]7 = 2.32+ log [−COO−][−COOH][−COO−][−COOH] = 4.8×104$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]7 = 9.62+ log [−NH2][−NH3+][−NH3+][−NH2] = 4.2×102$

Concentration of $−COO−$ is greater than $−NH3+$ . Hence, the predominant species is $CH(CH3)2−CH(+NH3)−COO−.$

At $pH = 12,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]12 = 2.32+ log [−COO−][−COOH][−COO−][−COOH] = 4.8×109$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]12 = 9.62+ log [−NH2][−NH3+][−NH2][−NH3+] = 2.4×102$

Concentration of $−COO−$ is greater than $−NH3+$ . Hence, the predominant species is $CH(CH3)2−CH(NH2)−COO−.$

(c)

Interpretation Introduction

Interpretation:

Isoelectric point of valine has to be calculated.

Concept Introduction:

Isoelectric point $(pI)$ of an amino acid refers to the $pH$ at which the amino acid remains neutral and does not migrate in an electric field

It is calculated as,

$pI = pKa1+pKa22$

Where ,

$pKa1and pKa2$ are two $pKa$ values of amino acid.

(c)

Expert Solution

Answer to Problem 22.47SP

Isoelectric point of valine is calculated as $5.97$

Explanation of Solution

Given data:

$pKa1 of carboxyl group, COOH, = 2.32pKa2 of ammonium group, NH3+ = 9.62$

Therefore isoelectronic point of valine is,

$pI = pKa1+pKa22 = 2.32 + 9.622 = 5.97$

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Answer 3

Question

Chapter 22, Problem 22.47SP

(a)

Interpretation Introduction

Interpretation:

From the given the $pKa$ values of the two groups, $−+NH3$ and $−COOH$ of valine, the more acidic one has to be determined.

Concept Introduction:

Greater the $pKa$ value of a substance lower will be its acidic property.

According to equation given in chapter 16 in the text book (equation 16.4),

$pH$ and $pKa$ are related as,

$pH = pKa + log [conjugate base][acid]$

(a)

Expert Solution

Answer to Problem 22.47SP

$−COOH$ group is more acidic.

Explanation of Solution

Given that $pKa$ of $−+NH3$ and $−COOH$ groups of valine are $9.62$ and $2.32$ respectively.

Since $−COOH$ has lower $pKa$ value of $2.32$ , it is more acidic.

(b)

Interpretation Introduction

Interpretation:

The predominant form of valine at $pH$ $1.0,7.0$ and $12$ has to be predicted.

Concept Introduction:

Greater the $pKa$ value of a substance lower will be its acidic property.

According to equation given in chapter 16 in the text book (equation 16.4),

$pH$ and $pKa$ are related as,

$pH = pKa + log [conjugate base][acid]$

(b)

Expert Solution

Explanation of Solution

Given data:

$pKa of carboxyl group, COOH = 2.3pKa of ammonium group, NH3+ = 9.6$

At $pH = 1,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]1 = 2.32+ log [−COO−][−COOH][−COOH][−COO−] = 21$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]1 = 9.62+ log [−NH2][−NH3+][−NH3+][−NH2] = 4.2×108$

Concentration of $−NH3+$ is greater than $−COOH$ . Hence, the predominant species is $CH(CH3)2−CH(+NH3)−COOH.$

At $pH = 7,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]7 = 2.32+ log [−COO−][−COOH][−COO−][−COOH] = 4.8×104$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]7 = 9.62+ log [−NH2][−NH3+][−NH3+][−NH2] = 4.2×102$

Concentration of $−COO−$ is greater than $−NH3+$ . Hence, the predominant species is $CH(CH3)2−CH(+NH3)−COO−.$

At $pH = 12,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]12 = 2.32+ log [−COO−][−COOH][−COO−][−COOH] = 4.8×109$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]12 = 9.62+ log [−NH2][−NH3+][−NH2][−NH3+] = 2.4×102$

Concentration of $−COO−$ is greater than $−NH3+$ . Hence, the predominant species is $CH(CH3)2−CH(NH2)−COO−.$

(c)

Interpretation Introduction

Interpretation:

Isoelectric point of valine has to be calculated.

Concept Introduction:

Isoelectric point $(pI)$ of an amino acid refers to the $pH$ at which the amino acid remains neutral and does not migrate in an electric field

It is calculated as,

$pI = pKa1+pKa22$

Where ,

$pKa1and pKa2$ are two $pKa$ values of amino acid.

(c)

Expert Solution

Answer to Problem 22.47SP

Isoelectric point of valine is calculated as $5.97$

Explanation of Solution

Given data:

$pKa1 of carboxyl group, COOH, = 2.32pKa2 of ammonium group, NH3+ = 9.62$

Therefore isoelectronic point of valine is,

$pI = pKa1+pKa22 = 2.32 + 9.622 = 5.97$

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Answer 4

Question

Chapter 22, Problem 22.47SP

(a)

Interpretation Introduction

Interpretation:

From the given the $pKa$ values of the two groups, $−+NH3$ and $−COOH$ of valine, the more acidic one has to be determined.

Concept Introduction:

Greater the $pKa$ value of a substance lower will be its acidic property.

According to equation given in chapter 16 in the text book (equation 16.4),

$pH$ and $pKa$ are related as,

$pH = pKa + log [conjugate base][acid]$

(a)

Expert Solution

Answer to Problem 22.47SP

$−COOH$ group is more acidic.

Explanation of Solution

Given that $pKa$ of $−+NH3$ and $−COOH$ groups of valine are $9.62$ and $2.32$ respectively.

Since $−COOH$ has lower $pKa$ value of $2.32$ , it is more acidic.

(b)

Interpretation Introduction

Interpretation:

The predominant form of valine at $pH$ $1.0,7.0$ and $12$ has to be predicted.

Concept Introduction:

Greater the $pKa$ value of a substance lower will be its acidic property.

According to equation given in chapter 16 in the text book (equation 16.4),

$pH$ and $pKa$ are related as,

$pH = pKa + log [conjugate base][acid]$

(b)

Expert Solution

Explanation of Solution

Given data:

$pKa of carboxyl group, COOH = 2.3pKa of ammonium group, NH3+ = 9.6$

At $pH = 1,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]1 = 2.32+ log [−COO−][−COOH][−COOH][−COO−] = 21$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]1 = 9.62+ log [−NH2][−NH3+][−NH3+][−NH2] = 4.2×108$

Concentration of $−NH3+$ is greater than $−COOH$ . Hence, the predominant species is $CH(CH3)2−CH(+NH3)−COOH.$

At $pH = 7,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]7 = 2.32+ log [−COO−][−COOH][−COO−][−COOH] = 4.8×104$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]7 = 9.62+ log [−NH2][−NH3+][−NH3+][−NH2] = 4.2×102$

Concentration of $−COO−$ is greater than $−NH3+$ . Hence, the predominant species is $CH(CH3)2−CH(+NH3)−COO−.$

At $pH = 12,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]12 = 2.32+ log [−COO−][−COOH][−COO−][−COOH] = 4.8×109$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]12 = 9.62+ log [−NH2][−NH3+][−NH2][−NH3+] = 2.4×102$

Concentration of $−COO−$ is greater than $−NH3+$ . Hence, the predominant species is $CH(CH3)2−CH(NH2)−COO−.$

(c)

Interpretation Introduction

Interpretation:

Isoelectric point of valine has to be calculated.

Concept Introduction:

Isoelectric point $(pI)$ of an amino acid refers to the $pH$ at which the amino acid remains neutral and does not migrate in an electric field

It is calculated as,

$pI = pKa1+pKa22$

Where ,

$pKa1and pKa2$ are two $pKa$ values of amino acid.

(c)

Expert Solution

Answer to Problem 22.47SP

Isoelectric point of valine is calculated as $5.97$

Explanation of Solution

Given data:

$pKa1 of carboxyl group, COOH, = 2.32pKa2 of ammonium group, NH3+ = 9.62$

Therefore isoelectronic point of valine is,

$pI = pKa1+pKa22 = 2.32 + 9.622 = 5.97$

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Answer 5

Chapter 22, Problem 22.47SP

(a)

Interpretation Introduction

Interpretation:

From the given the $pKa$ values of the two groups, $−+NH3$ and $−COOH$ of valine, the more acidic one has to be determined.

Concept Introduction:

Greater the $pKa$ value of a substance lower will be its acidic property.

According to equation given in chapter 16 in the text book (equation 16.4),

$pH$ and $pKa$ are related as,

$pH = pKa + log [conjugate base][acid]$

(a)

Expert Solution

Answer to Problem 22.47SP

$−COOH$ group is more acidic.

Explanation of Solution

Given that $pKa$ of $−+NH3$ and $−COOH$ groups of valine are $9.62$ and $2.32$ respectively.

Since $−COOH$ has lower $pKa$ value of $2.32$ , it is more acidic.

(b)

Interpretation Introduction

Interpretation:

The predominant form of valine at $pH$ $1.0,7.0$ and $12$ has to be predicted.

Concept Introduction:

Greater the $pKa$ value of a substance lower will be its acidic property.

According to equation given in chapter 16 in the text book (equation 16.4),

$pH$ and $pKa$ are related as,

$pH = pKa + log [conjugate base][acid]$

(b)

Expert Solution

Explanation of Solution

Given data:

$pKa of carboxyl group, COOH = 2.3pKa of ammonium group, NH3+ = 9.6$

At $pH = 1,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]1 = 2.32+ log [−COO−][−COOH][−COOH][−COO−] = 21$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]1 = 9.62+ log [−NH2][−NH3+][−NH3+][−NH2] = 4.2×108$

Concentration of $−NH3+$ is greater than $−COOH$ . Hence, the predominant species is $CH(CH3)2−CH(+NH3)−COOH.$

At $pH = 7,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]7 = 2.32+ log [−COO−][−COOH][−COO−][−COOH] = 4.8×104$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]7 = 9.62+ log [−NH2][−NH3+][−NH3+][−NH2] = 4.2×102$

Concentration of $−COO−$ is greater than $−NH3+$ . Hence, the predominant species is $CH(CH3)2−CH(+NH3)−COO−.$

At $pH = 12,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]12 = 2.32+ log [−COO−][−COOH][−COO−][−COOH] = 4.8×109$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]12 = 9.62+ log [−NH2][−NH3+][−NH2][−NH3+] = 2.4×102$

Concentration of $−COO−$ is greater than $−NH3+$ . Hence, the predominant species is $CH(CH3)2−CH(NH2)−COO−.$

(c)

Interpretation Introduction

Interpretation:

Isoelectric point of valine has to be calculated.

Concept Introduction:

Isoelectric point $(pI)$ of an amino acid refers to the $pH$ at which the amino acid remains neutral and does not migrate in an electric field

It is calculated as,

$pI = pKa1+pKa22$

Where ,

$pKa1and pKa2$ are two $pKa$ values of amino acid.

(c)

Expert Solution

Answer to Problem 22.47SP

Isoelectric point of valine is calculated as $5.97$

Explanation of Solution

Given data:

$pKa1 of carboxyl group, COOH, = 2.32pKa2 of ammonium group, NH3+ = 9.62$

Therefore isoelectronic point of valine is,

$pI = pKa1+pKa22 = 2.32 + 9.622 = 5.97$

Answer 6

Chapter 22, Problem 22.47SP

(a)

Interpretation Introduction

Interpretation:

From the given the $pKa$ values of the two groups, $−+NH3$ and $−COOH$ of valine, the more acidic one has to be determined.

Concept Introduction:

Greater the $pKa$ value of a substance lower will be its acidic property.

According to equation given in chapter 16 in the text book (equation 16.4),

$pH$ and $pKa$ are related as,

$pH = pKa + log [conjugate base][acid]$

(a)

Expert Solution

Answer to Problem 22.47SP

$−COOH$ group is more acidic.

Explanation of Solution

Given that $pKa$ of $−+NH3$ and $−COOH$ groups of valine are $9.62$ and $2.32$ respectively.

Since $−COOH$ has lower $pKa$ value of $2.32$ , it is more acidic.

Answer 7

Chapter 22, Problem 22.47SP

(a)

Interpretation Introduction

Interpretation:

From the given the $pKa$ values of the two groups, $−+NH3$ and $−COOH$ of valine, the more acidic one has to be determined.

Concept Introduction:

Greater the $pKa$ value of a substance lower will be its acidic property.

According to equation given in chapter 16 in the text book (equation 16.4),

$pH$ and $pKa$ are related as,

$pH = pKa + log [conjugate base][acid]$

Answer 8

(a)

Expert Solution

Answer to Problem 22.47SP

$−COOH$ group is more acidic.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given that $pKa$ of $−+NH3$ and $−COOH$ groups of valine are $9.62$ and $2.32$ respectively.

Since $−COOH$ has lower $pKa$ value of $2.32$ , it is more acidic.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The predominant form of valine at $pH$ $1.0,7.0$ and $12$ has to be predicted.

Concept Introduction:

Greater the $pKa$ value of a substance lower will be its acidic property.

According to equation given in chapter 16 in the text book (equation 16.4),

$pH$ and $pKa$ are related as,

$pH = pKa + log [conjugate base][acid]$

(b)

Expert Solution

Explanation of Solution

Given data:

$pKa of carboxyl group, COOH = 2.3pKa of ammonium group, NH3+ = 9.6$

At $pH = 1,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]1 = 2.32+ log [−COO−][−COOH][−COOH][−COO−] = 21$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]1 = 9.62+ log [−NH2][−NH3+][−NH3+][−NH2] = 4.2×108$

Concentration of $−NH3+$ is greater than $−COOH$ . Hence, the predominant species is $CH(CH3)2−CH(+NH3)−COOH.$

At $pH = 7,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]7 = 2.32+ log [−COO−][−COOH][−COO−][−COOH] = 4.8×104$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]7 = 9.62+ log [−NH2][−NH3+][−NH3+][−NH2] = 4.2×102$

Concentration of $−COO−$ is greater than $−NH3+$ . Hence, the predominant species is $CH(CH3)2−CH(+NH3)−COO−.$

At $pH = 12,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]12 = 2.32+ log [−COO−][−COOH][−COO−][−COOH] = 4.8×109$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]12 = 9.62+ log [−NH2][−NH3+][−NH2][−NH3+] = 2.4×102$

Concentration of $−COO−$ is greater than $−NH3+$ . Hence, the predominant species is $CH(CH3)2−CH(NH2)−COO−.$

Answer 14

(b)

Interpretation Introduction

Interpretation:

The predominant form of valine at $pH$ $1.0,7.0$ and $12$ has to be predicted.

Concept Introduction:

Greater the $pKa$ value of a substance lower will be its acidic property.

According to equation given in chapter 16 in the text book (equation 16.4),

$pH$ and $pKa$ are related as,

$pH = pKa + log [conjugate base][acid]$

Answer 15

(b)

Expert Solution

Explanation of Solution

Given data:

$pKa of carboxyl group, COOH = 2.3pKa of ammonium group, NH3+ = 9.6$

At $pH = 1,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]1 = 2.32+ log [−COO−][−COOH][−COOH][−COO−] = 21$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]1 = 9.62+ log [−NH2][−NH3+][−NH3+][−NH2] = 4.2×108$

Concentration of $−NH3+$ is greater than $−COOH$ . Hence, the predominant species is $CH(CH3)2−CH(+NH3)−COOH.$

At $pH = 7,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]7 = 2.32+ log [−COO−][−COOH][−COO−][−COOH] = 4.8×104$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]7 = 9.62+ log [−NH2][−NH3+][−NH3+][−NH2] = 4.2×102$

Concentration of $−COO−$ is greater than $−NH3+$ . Hence, the predominant species is $CH(CH3)2−CH(+NH3)−COO−.$

At $pH = 12,$

Concentration of $−COOH$

$pH = pKa + log [conjugate base][acid]12 = 2.32+ log [−COO−][−COOH][−COO−][−COOH] = 4.8×109$

Concentration of $−NH3+$

$pH = pKa + log [conjugate base][acid]12 = 9.62+ log [−NH2][−NH3+][−NH2][−NH3+] = 2.4×102$

Concentration of $−COO−$ is greater than $−NH3+$ . Hence, the predominant species is $CH(CH3)2−CH(NH2)−COO−.$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

(c)

Interpretation Introduction

Interpretation:

Isoelectric point of valine has to be calculated.

Concept Introduction:

Isoelectric point $(pI)$ of an amino acid refers to the $pH$ at which the amino acid remains neutral and does not migrate in an electric field

It is calculated as,

$pI = pKa1+pKa22$

Where ,

$pKa1and pKa2$ are two $pKa$ values of amino acid.

(c)

Expert Solution

Answer to Problem 22.47SP

Isoelectric point of valine is calculated as $5.97$

Explanation of Solution

Given data:

$pKa1 of carboxyl group, COOH, = 2.32pKa2 of ammonium group, NH3+ = 9.62$

Therefore isoelectronic point of valine is,

$pI = pKa1+pKa22 = 2.32 + 9.622 = 5.97$

Answer 20

(c)

Interpretation Introduction

Interpretation:

Isoelectric point of valine has to be calculated.

Concept Introduction:

Isoelectric point $(pI)$ of an amino acid refers to the $pH$ at which the amino acid remains neutral and does not migrate in an electric field

It is calculated as,

$pI = pKa1+pKa22$

Where ,

$pKa1and pKa2$ are two $pKa$ values of amino acid.

Answer 21

(c)

Expert Solution

Answer to Problem 22.47SP

Isoelectric point of valine is calculated as $5.97$

Answer 22

(c)

Expert Solution

Answer 23

(c)

Expert Solution

Answer 24

Expert Solution

Answer 25

Explanation of Solution

Given data:

$pKa1 of carboxyl group, COOH, = 2.32pKa2 of ammonium group, NH3+ = 9.62$

Therefore isoelectronic point of valine is,

$pI = pKa1+pKa22 = 2.32 + 9.622 = 5.97$

Answer 26

Ch. 22.2 - Prob. 1RC Ch. 22.3 - Prob. 1RC Ch. 22.4 - Prob. 1RC Ch. 22 - Prob. 22.1QP Ch. 22 - Prob. 22.2QP Ch. 22 - Prob. 22.3QP Ch. 22 - Prob. 22.4QP Ch. 22 - Prob. 22.6QP Ch. 22 - Prob. 22.7QP Ch. 22 - Prob. 22.8QP

Answer to Problem 22.47SP

Explanation of Solution

Explanation of Solution

Answer to Problem 22.47SP

Explanation of Solution

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