Student Solutions Manual for Bettelheim/Brown/Campbell/Farrell/Torres' Introduction to General, Organic and Biochemistry, 11th
Student Solutions Manual for Bettelheim/Brown/Campbell/Farrell/Torres' Introduction to General, Organic and Biochemistry, 11th
11th Edition
ISBN: 9781305081055
Author: Bettelheim, Frederick A.
Publisher: Cengage Learning
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Chapter 18, Problem 18.33P
Interpretation Introduction

(a)

Interpretation:

The ratio of A- to [HA] with the help of pH and pKa values should be calculated.

Concept Introduction:

The pH of a solution indicates the number of H3 O+ ions in the solution. The mathematical relation between pH and H3 O+ is as given below;.

pH =  log [H3O+][H3O+] = 10pH

Acid dissociation constant Ka is related to pKa as;.

pKa =  log KaKa = 10pKa.

Expert Solution
Check Mark

Answer to Problem 18.33P

103.

Explanation of Solution

Given Information:

pKa = 5.0.

pH = 2.0.

Calculation:

From pH of solution, concentration of hydroxide ion can be calculated as follows:

[H3O+] = 10pH= 102  = 0.01 

Also,

 Ka = 10pKa= 105 

Since,

[A][HA]=Ka[H3O+]

Thus, by putting the values, the ratio [A][HA] can be calculated as follows:

[A][HA]=105102=103.

Interpretation Introduction

(b)

Interpretation:

The ratio of A- to [HA] with the help of pH and pKa values should be calculated.

Concept Introduction:

The pH of a solution indicates the number of H3 O+ ions in the solution. The mathematical relation between pH and H3 O+ is as given below;.

pH =  log [H3O+][H3O+] = 10pH

Acid dissociation constant Ka is related to pKa as;.

pKa =  log KaKa = 10pKa.

Expert Solution
Check Mark

Answer to Problem 18.33P

Given:

pKa = 5.0.

pH = 5.0.

Hence [H3 O+ ] = 10-pH = 10-5.

And Ka = 10-pKa = 10-5.

Since Student Solutions Manual for Bettelheim/Brown/Campbell/Farrell/Torres' Introduction to General, Organic and Biochemistry, 11th, Chapter 18, Problem 18.33P

Plug the values of Ka and H3 O+ ions to calculate the ratio of [A- ] to [HA].

[A- ] / [HA] = 10-5 / 10-5 = 1.

Explanation of Solution

Given Information:

pKa = 5.0.

pH = 5.0.

Calculation:

From pH of solution, concentration of hydroxide ion can be calculated as follows:

[H3O+] = 10pH= 105  = 1.0×105  

Also,

 Ka = 10pKa= 105 

Since,

[A][HA]=Ka[H3O+]

Thus, by putting the values, the ratio [A][HA] can be calculated as follows:

[A][HA]=105105=1.

Interpretation Introduction

(c)

Interpretation:

The ratio of A- to [HA] with the help of pH and pKa values should be calculated.

Concept Introduction:

The pH of a solution indicates the number of H3 O+ ions in the solution. The mathematical relation between pH and H3 O+ is as given below;.

pH =  log [H3O+][H3O+] = 10pH

Acid dissociation constant Ka is related to pKa as;.

pKa =  log KaKa = 10pKa.

Expert Solution
Check Mark

Answer to Problem 18.33P

102.

Explanation of Solution

Given Information:

pKa = 5.0.

pH = 7.0.

Calculation:

From pH of solution, concentration of hydroxide ion can be calculated as follows:

[H3O+] = 10pH= 107  = 1.0× 107  

Also,

 Ka = 10pKa= 105 

Since,

[A][HA]=Ka[H3O+]

Thus, by putting the values, the ratio [A][HA] can be calculated as follows:

[A][HA]=1051.0× 107  =102.

Interpretation Introduction

(d)

Interpretation:

The ratio of A- to [HA] with the help of pH and pKa values should be calculated.

Concept Introduction:

The pH of a solution indicates the number of H3 O+ ions in the solution. The mathematical relation between pH and H3 O+ is as given below;.

pH =  log [H3O+][H3O+] = 10pH

Acid dissociation constant Ka is related to pKa as;.

pKa =  log KaKa = 10pKa.

Expert Solution
Check Mark

Answer to Problem 18.33P

104.

Explanation of Solution

Given Information:

pKa = 5.0.

pH = 9.0.

Calculation:

From pH of solution, concentration of hydroxide ion can be calculated as follows:

[H3O+] = 10pH= 109  = 1.0× 109 

Also,

 Ka = 10pKa= 105 

Since,

[A][HA]=Ka[H3O+]

Thus, by putting the values, the ratio [A][HA] can be calculated as follows:

[A][HA]=1051.0× 109  =104.

Interpretation Introduction

(e)

Interpretation:

The ratio of A- to [HA] with the help of pH and pKa values should be calculated.

Concept Introduction:

The pH of a solution indicates the number of H3 O+ ions in the solution. The mathematical relation between pH and H3 O+ is as given below;.

pH =  log [H3O+][H3O+] = 10pH

Acid dissociation constant Ka is related to pKa as;.

pKa =  log KaKa = 10pKa.

Expert Solution
Check Mark

Answer to Problem 18.33P

106.

Explanation of Solution

Given Information:

pKa = 5.0.

pH = 11.0.

Calculation:

From pH of solution, concentration of hydroxide ion can be calculated as follows:

[H3O+] = 10pH= 1011  = 1.0× 1011 

Also,

 Ka = 10pKa= 105 

Since,

[A][HA]=Ka[H3O+]

Thus, by putting the values, the ratio [A][HA] can be calculated as follows:

[A][HA]=1051.0× 1011  =106.

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Chapter 18 Solutions

Student Solutions Manual for Bettelheim/Brown/Campbell/Farrell/Torres' Introduction to General, Organic and Biochemistry, 11th

Ch. 18 - Prob. 18.11PCh. 18 - Prob. 18.12PCh. 18 - Prob. 18.13PCh. 18 - 18-14 Answer true or false. (a) Carboxylic acids...Ch. 18 - 18-15 Draw a structural formula for the dimer...Ch. 18 - 18-16 Propanedioic (malonic) acid forms an...Ch. 18 - 18-17 Hexanoic (caproic) acid has a solubility in...Ch. 18 - 18-18 Propanoic acid and methyl acetate are...Ch. 18 - 18-19 The following compounds have approximately...Ch. 18 - Prob. 18.20PCh. 18 - Prob. 18.21PCh. 18 - Prob. 18.22PCh. 18 - 18-23 Characterize the structural features...Ch. 18 - Prob. 18.24PCh. 18 - Prob. 18.25PCh. 18 - 18-26 Answer true or false. (a) Carboxylic acids...Ch. 18 - Prob. 18.27PCh. 18 - 18-28 Arrange these compounds in order of...Ch. 18 - 18-29 Complete the equations for these acid—base...Ch. 18 - 18-30 Complete the equations for these acid-base...Ch. 18 - 18-31 Formic acid is one of the components...Ch. 18 - Prob. 18.32PCh. 18 - Prob. 18.33PCh. 18 - Prob. 18.34PCh. 18 - Prob. 18.35PCh. 18 - Prob. 18.36PCh. 18 - Prob. 18.37PCh. 18 - 18-38 Which is the stronger base: CH3CH2NH2 or...Ch. 18 - Prob. 18.39PCh. 18 - Prob. 18.40PCh. 18 - 18-41 Complete these examples of Fischer...Ch. 18 - Prob. 18.42PCh. 18 - Prob. 18.43PCh. 18 - Prob. 18.44PCh. 18 - Prob. 18.45PCh. 18 - 18-46 Procaine (its hydrochloride salt is marketed...Ch. 18 - 18-47 Methylparaben and propylparaben are used as...Ch. 18 - 18-48 4-Aminobenzoic acid is prepared from benzoic...Ch. 18 - Prob. 18.49PCh. 18 - Prob. 18.50PCh. 18 - Prob. 18.51PCh. 18 - Prob. 18.52PCh. 18 - Prob. 18.53PCh. 18 - Prob. 18.54PCh. 18 - Prob. 18.55P
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