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 14, Problem 9QAP

Calculate [H+] and pH in a solution in which lactic acid, HC3H5O3, is 0.250 M and the lactate ion, C3H5O3-, is

(a) 0.250 M (b) 0.125 M

(c) 0.0800 M (d) 0.0500 M

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The [H+] and pH of the solution with molarity of lactic acid 0.50 M and lactate ion 0.250 M needs to be determined.

Concept Introduction :

For an acid dissociation reaction, the [H+] can be calculated from acid dissociation constant as follows:

  HAH++A

Here,

  Ka=[H+][A][HA]

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

  pH=log[H+]

Answer to Problem 9QAP

  1.4×104 M and 3.85

Explanation of Solution

The equilibrium reaction is as follows:

  HC3H5O3(aq)H+(aq) + C3H5O3(aq)

The expression of Ka for this reaction is as follows:

  Ka[H+][C3H5O3-][HC3H5O3] 

The above expression can be written as follows:

  [H+] = Ka[C3H5O3][ HC3H5O3] 

The Ka value of lactic acid HC3H5O3 is 1.4 x 10-4, the concentrated of lactic acid (HC3H5O3) is 0.250 M, and for lactate ion C3H5O3- is 0.250 M.

Substitute these values in the above expression as follows:

  [H+] = Ka[C3H5O3][ HC3H5O3] 

  = (1.4x 10 4) (0.250 M)(0.250 M)

  = 1.4×104

Now, the pH of the solution can be calculated by using the [H+] as follows:

  pH = log 10[ H +]    =  log 10[1.4× 10 4]    = 3.85

So, the pH of the solution is 3.85 and [H+] is 1.4×104 M.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The [H+] and pH of the solution with molarity of lactic acid 0.50 M and lactate ion 0.125 M needs to be determined.

Concept Introduction :

For an acid dissociation reaction, the [H+] can be calculated from acid dissociation constant as follows:

  HAH++A

Here,

  Ka=[H+][A][HA]

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

  pH=log[H+]

Answer to Problem 9QAP

  2.8×104 M and 3.55.

Explanation of Solution

The equilibrium reaction is as follows:

  HC3H5O3(aq)H+(aq) + C3H5O3(aq)

The expression of Ka for this reaction is as follows:

  Ka[H+][C3H5O3-][HC3H5O3] 

The above expression can be written as follows:

  [H+] = Ka[C3H5O3][ HC3H5O3] 

The Ka value of lactic acid HC3H5O3 is 1.4 x 10-4, the concentrated of lactic acid (HC3H5O3) is 0.250 M, and for lactate ion C3H5O3- is 0.125 M.

Substitute these values in the above expression as follows:

  [H+] = Ka[C3H5O3][ HC3H5O3] 

  = (1.4x 10 4) (0.250 M)(0.125 M)

  = 2.8×104

Now, the pH of the solution can be calculated by using the [H+] as follows:

pH = -log10 [H+]

= -log10 [2.8 x 10-4]

= 3.55

So, the pH of the solution is 3.55 and [H+] is 2.8×104 M.

(C)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The [H+] and pH of the solution with molarity of lactic acid 0.50 M and lactate ion 0.08 M needs to be determined.

Concept Introduction :

For an acid dissociation reaction, the [H+] can be calculated from acid dissociation constant as follows:

  HAH++A

Here,

  Ka=[H+][A][HA]

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

  pH=log[H+]

Answer to Problem 9QAP

  4.4×104 M and 3.36

Explanation of Solution

The equilibrium reaction is as follows:

  HC3H5O3(aq)H+(aq) + C3H5O3(aq)

The expression of Ka for this reaction is as follows:

  Ka[H+][C3H5O3-][HC3H5O3] 

The above expression can be written as follows:

  [H+] = Ka[C3H5O3][ HC3H5O3] 

The Ka value of lactic acid HC3H5O3 is 1.4 x 10-4, the concentrated of lactic acid (HC3H5O3) is 0.250 M, and for lactate ion C3H5O3- is 0.0800 M.

Substitute these values in the above expression as follows:

  [H+] = Ka[C3H5O3][ HC3H5O3] 

  = (1.4x 10 4) (0.125 M)(0.0800 M)

  = 4.4×104

Now, the pH of the solution can be calculated by using the [H+] as follows:

pH = -log10 [H+]

= -log104.4×104

= 3.36

So, the pH of the solution is 3.36 and [H+] is 4.4×104 M.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The [H+] and pH of the solution with molarity of lactic acid 0.50 M and lactate ion 0.0500 M needs to be determined.

Concept Introduction :

For an acid dissociation reaction, the [H+] can be calculated from acid dissociation constant as follows:

  HAH++A

Here,

  Ka=[H+][A][HA]

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

  pH=log[H+]

Answer to Problem 9QAP

  7×104 M and 3.15.

Explanation of Solution

The equilibrium reaction is as follows:

  HC3H5O3(aq)H+(aq) + C3H5O3(aq)

The expression of Ka for this reaction is as follows:

  Ka[H+][C3H5O3-][HC3H5O3] 

The above expression can be written as follows:

  [H+] = Ka[C3H5O3][ HC3H5O3] 

The Ka value of lactic acid HC3H5O3 is 1.4 x 10-4, the concentrated of lactic acid (HC3H5O3) is 0.250 M, and for lactate ion C3H5O3- is 0.0500 M.

Substitute these values in the above expression as follows:

  [H+] = Ka[C3H5O3][ HC3H5O3] 

  = (1.4x 10 4) (0.125 M)(0.0500 M)

  = 7×104

Now, the pH of the solution can be calculated by using the [H+] as follows:

pH = -log10 [H+]

= -log107×104

= 3.15

So, the pH of the solution is 3.15 and [OH-] is 7×104 M.

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Chapter 14 Solutions

Chemistry: Principles and Reactions

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