Foundations of College Chemistry, Binder Ready Version
Foundations of College Chemistry, Binder Ready Version
15th Edition
ISBN: 9781119083900
Author: Morris Hein, Susan Arena, Cary Willard
Publisher: WILEY
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Chapter 16, Problem 23PE

(a)

Interpretation Introduction

Interpretation:

Percent ionization and pH of solution of 1.0 M phenol (HC6H5O) has to be calculated.

Concept Introduction:

Consider a weak acid HA that is in equilibrium with its ions and is as follows:

  HA(aq)H+(aq)+A(aq)

The equilibrium constant for ionization of weak acid is called ionization constant Ka and is expressed as follows:

  Ka=[H+][A][HA]

Percent ionization for a weak acid is defined as the ratio of concentration of H+ or A to initial concentration of HA that is multiplied with 100 and is expressed as follows:

  Percent ionization=([H+]or[A][HA])100

pH is defined as the concentration of hydrogen ion. It also explains about the acidity of solution.

(a)

Expert Solution
Check Mark

Explanation of Solution

The ionization of phenol, HC6H5O is as follows:

  HC6H5O(aq)H+(aq)+C6H5O(aq)

The expression for Ka for given reaction is as follows:

  Ka=[H+][C6H5O][HC6H5O]        (1)

Through chemical equation it is evident that one C6H5O is formed for every H+. Thus concentration of H+ is equal to concentration of C6H5O. Thus consider x for H+ and C6H5O and value of x is small so concentration of HC6H5O can be considered as 1.0M.

Rearrange equation (1) for [H+].

  [H+]=Ka[HC6H5O][C6H5O]        (2)

Substitute x for [C6H5O], x for [H+], 1.3×1010 for Ka and 1.0 for [HC6H5O] in equation (2).

  x=(1.3×1010)(1.0)xx2=1.3×1010

Solve this equation for x.

  x=1.14×105

Hence, [H+] in of 1.0 M phenol, HC6H5O is 1.14×105 M.

pH is defined as negative logarithm of concentration of H+. It is calculated as follows:

  pH=log[H+]        (3)

Substitute 1.14×105 for H+ in equation (3).

  pH=log[1.14×105]=log1.14+5log10=0.056+5=4.94

Hence, pH in 1.0 M phenol, HC6H5O is 4.94.

The percent ionization of phenol, HC6H5O  is as follows:

  Percent ionization=([H+][HC6H5O])100        (4)

Substitute 1.14×105 M for H+ and 1.0 M phenol, HC6H5O in equation (4).

  Percent ionization=(1.14×105 M1.0 M)100=0.0011 %

Hence, percent ionization of phenol, HC6H5O is 0.0011 %.

(b)

Interpretation Introduction

Interpretation:

Percent ionization and pH of solution of 0.10 M phenol (HC6H5O) has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

The ionization of phenol (HC6H5O) is as follows:

  HC6H5O(aq)H+(aq)+C6H5O(aq)

The expression for Ka for given reaction is as follows:

  Ka=[H+][C6H5O][HC6H5O]        (1)

Through chemical equation it is evident that one C6H5O is formed for every H+. Thus concentration of H+ is equal to concentration of C6H5O. Thus consider x for H+ and C6H5O and value of x is small so concentration of HC6H5O can be considered as 0.10 M.

Rearrange equation (1) for [H+].

  [H+]=Ka[HC6H5O][C6H5O]        (2)

Substitute x for [C6H5O], x for [H+], 1.3×1010 for Ka and 0.10 for [HC6H5O] in equation (2).

  x=(1.3×1010)(0.10)xx2=1.3×1011

Solve this equation for x.

  x=3.6×106

Hence, [H+] in of 0.10 M phenol (HC6H5O) is 3.6×106 M.

pH is defined as negative logarithm of concentration of H+. It is calculated as follows:

  pH=log[H+]        (3)

Substitute 3.6×106 for H+ in equation (3).

  pH=log[3.6×106]=log3.6+6log10=0.556+6=5.44

Hence, pH in 0.10 M phenol, HC6H5O is 5.44.

The percent ionization of phenol, HC6H5O  is as follows:

  Percent ionization=([H+][HC7H5O2])100        (4)

Substitute 3.6×106 M for H+ and 0.10 M for [HC6H5O] in equation (4).

Hence, percent ionization of phenol, HC6H5O is 0.0036 %.

(c)

Interpretation Introduction

Interpretation:

Percent ionization and pH of solution of 0.010 M phenol (HC6H5O) has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Explanation of Solution

The ionization of phenol, HC6H5O is as follows:

  HC6H5O(aq)H+(aq)+C6H5O(aq)

The expression for Ka for given reaction is as follows:

  Ka=[H+][C6H5O][HC6H5O]        (1)

Through chemical equation it is evident that one C6H5O is formed for every H+. Thus concentration of H+ is equal to concentration of C6H5O. Thus consider x for H+ and C6H5O and value of x is small so concentration of HC6H5O can be considered as 0.010 M.

Rearrange equation (1) for [H+].

  [H+]=Ka[HC6H5O][C6H5O]        (2)

Substitute x for [C6H5O], x for [H+], 1.3×1010 for Ka and 0.010 for [HC6H5O] in equation (2).

  x=(1.3×1010)(0.010)xx2=3×1013

Solve this equation for x.

  x=5.48×107

Hence, [H+] in of 0.010 M phenol, HC6H5O is 5.48×107 M.

pH is defined as negative logarithm of concentration of H+. It is calculated as follows:

  pH=log[H+]        (3)

Substitute 5.48×107 for H+ in equation (3).

  pH=log[5.48×107]=log5.48+7log10=0.739+7=6.261

Hence, pH in 0.010 M phenol, HC6H5O is 6.261.

The percent ionization of phenol, HC6H5O  is as follows:

  Percent ionization=([H+][HC7H5O2])100        (4)

Substitute 5.48×107 M for H+ and 0.010 M for [HC6H5O]  in equation (4).

  Percent ionization=(5.48×107 M0.010 M)100=0.0055 %

Hence, percent ionization of phenol (HC6H5O) is. 0.0055 %.

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Chapter 16 Solutions

Foundations of College Chemistry, Binder Ready Version

Ch. 16.6 - Prob. 16.11PCh. 16.6 - Prob. 16.12PCh. 16.7 - Prob. 16.13PCh. 16.7 - Prob. 16.14PCh. 16.7 - Prob. 16.15PCh. 16.8 - Prob. 16.16PCh. 16 - Prob. 1RQCh. 16 - Prob. 2RQCh. 16 - Prob. 3RQCh. 16 - Prob. 4RQCh. 16 - Prob. 5RQCh. 16 - Prob. 6RQCh. 16 - Prob. 7RQCh. 16 - Prob. 8RQCh. 16 - Prob. 9RQCh. 16 - Prob. 10RQCh. 16 - Prob. 11RQCh. 16 - Prob. 12RQCh. 16 - Prob. 13RQCh. 16 - Prob. 14RQCh. 16 - Prob. 15RQCh. 16 - Prob. 16RQCh. 16 - Prob. 17RQCh. 16 - Prob. 18RQCh. 16 - Prob. 19RQCh. 16 - Prob. 20RQCh. 16 - Prob. 21RQCh. 16 - Prob. 22RQCh. 16 - Prob. 23RQCh. 16 - Prob. 24RQCh. 16 - Prob. 25RQCh. 16 - Prob. 26RQCh. 16 - Prob. 27RQCh. 16 - Prob. 1PECh. 16 - Prob. 2PECh. 16 - Prob. 3PECh. 16 - Prob. 4PECh. 16 - Prob. 5PECh. 16 - Prob. 6PECh. 16 - Prob. 7PECh. 16 - Prob. 8PECh. 16 - Prob. 9PECh. 16 - Prob. 10PECh. 16 - Prob. 11PECh. 16 - Prob. 12PECh. 16 - Prob. 13PECh. 16 - Prob. 14PECh. 16 - Prob. 15PECh. 16 - Prob. 16PECh. 16 - Prob. 17PECh. 16 - Prob. 18PECh. 16 - Prob. 19PECh. 16 - Prob. 20PECh. 16 - Prob. 21PECh. 16 - Prob. 22PECh. 16 - Prob. 23PECh. 16 - Prob. 24PECh. 16 - Prob. 25PECh. 16 - Prob. 26PECh. 16 - Prob. 27PECh. 16 - Prob. 28PECh. 16 - Prob. 29PECh. 16 - Prob. 30PECh. 16 - Prob. 31PECh. 16 - Prob. 32PECh. 16 - Prob. 33PECh. 16 - Prob. 34PECh. 16 - Prob. 35PECh. 16 - Prob. 36PECh. 16 - Prob. 37PECh. 16 - Prob. 38PECh. 16 - Prob. 39PECh. 16 - Prob. 40PECh. 16 - Prob. 41PECh. 16 - Prob. 42PECh. 16 - Prob. 43PECh. 16 - Prob. 44PECh. 16 - Prob. 45PECh. 16 - Prob. 46PECh. 16 - Prob. 47PECh. 16 - Prob. 48PECh. 16 - Prob. 49AECh. 16 - Prob. 50AECh. 16 - Prob. 51AECh. 16 - Prob. 52AECh. 16 - Prob. 53AECh. 16 - Prob. 54AECh. 16 - Prob. 55AECh. 16 - Prob. 56AECh. 16 - Prob. 57AECh. 16 - Prob. 58AECh. 16 - Prob. 59AECh. 16 - Prob. 60AECh. 16 - Prob. 61AECh. 16 - Prob. 62AECh. 16 - Prob. 63AECh. 16 - Prob. 64AECh. 16 - Prob. 65AECh. 16 - Prob. 66AECh. 16 - Prob. 67AECh. 16 - Prob. 68AECh. 16 - Prob. 69AECh. 16 - Prob. 70AECh. 16 - Prob. 71AECh. 16 - Prob. 72AECh. 16 - Prob. 73AECh. 16 - Prob. 74AECh. 16 - Prob. 75AECh. 16 - Prob. 76AECh. 16 - Prob. 77AECh. 16 - Prob. 78AECh. 16 - Prob. 79AECh. 16 - Prob. 80AECh. 16 - Prob. 81AECh. 16 - Prob. 83AECh. 16 - Prob. 84AECh. 16 - Prob. 85AECh. 16 - Prob. 86CECh. 16 - Prob. 87CECh. 16 - Prob. 88CECh. 16 - Prob. 89CE
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