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Philip Dutton University of Windsor, Canada N9B 3P4 Prentice-Hall © 2002 General Chemistry Principles and Modern Applications Petrucci • Harwood • Herring 8 th Edition Chapter 17: Acids and Bases
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Page 1: Ch17

Philip DuttonUniversity of Windsor, Canada

N9B 3P4

Prentice-Hall © 2002

General ChemistryPrinciples and Modern Applications

Petrucci • Harwood • Herring

8th Edition

Chapter 17: Acids and Bases

Page 2: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 2 of 47

Contents

17-1 The Arrhenius Theory: A Brief Review

17-2 Brønsted-Lowry Theory of Acids and Bases

17-3 The Self-Ionization of Water and the pH Scale

17-4 Strong Acids and Strong Bases

17-5 Weak Acids and Weak Bases

17-6 Polyprotic Acids

17-7 Ions as Acids and Bases

17-8 Molecular Structure and Acid-Base Behavior

17-8 Lewis Acids and Bases

Focus On Acid Rain.

Page 3: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 3 of 47

17-1 The Arrhenius Theory: A Brief Review

HCl(g) → H+(aq) + Cl-(aq)

NaOH(s) → Na+(aq) + OH-(aq)H2O

H2O

Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → H2O(l) + Na+(aq) + Cl-(aq)

H+(aq) + OH-(aq) → H2O(l)

Arrhenius theory did not handle non OH- bases such as ammonia very well.

Page 4: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 4 of 47

17-2 Brønsted-Lowry Theory of Acids and Bases

• An acid is a proton donor.• A base is a proton acceptor.

NH3 + H2O NH4+ + OH-

NH4+ + OH- NH3 + H2O

base acid

baseacid

conjugate acid

conjugate base

Page 5: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 5 of 47

Base Ionization Constant

NH3 + H2O NH4+ + OH-

Kc= [NH3][H2O]

[NH4+][OH-]

Kb= Kc[H2O] = [NH3]

[NH4+][OH-]

= 1.810-5

base acidconjugate

acid

conjugate

base

Page 6: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 6 of 47

Acid Ionization Constant

CH3CO2H + H2O CH3CO2- + H3O+

Kc= [CH3CO2H][H2O]

[CH3CO2-][H3O+]

Ka= Kc[H2O] = = 1.810-5

[CH3CO2H]

[CH3CO2-][H3O+]

baseacidconjugate

acid

conjugate

base

Page 7: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 7 of 47

Table 17.1 Relative Strengths of Some Brønsted-Lowry Acids and Bases

HClO4 + H2O ClO4- + H3O+

NH4+ + CO3

2- NH3 + HCO3-HCl + OH- Cl- + H2O

H2O + I- OH- + HIH2O + I- OH- + HI

Page 8: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 8 of 47

17-3 The Self-Ionization of Water and the pH Scale

Page 9: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 9 of 47

Ion Product of Water

Kc= [H2O][H2O]

[H3O+][OH-]

H2O + H2O H3O+ + OH-

base acidconjugate

acid

conjugate

base

KW= Kc[H2O][H2O] = = 1.010-14[H3O+][OH-]

Page 10: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 10 of 47

pH and pOH

• The potential of the hydrogen ion was defined in 1909 as the negative of the logarithm of [H+].

pH = -log[H3O+] pOH = -log[OH-]

-logKW = -log[H3O+]-log[OH-]= -log(1.010-14)

KW = [H3O+][OH-]= 1.010-14

pKW = pH + pOH= -(-14)

pKW = pH + pOH = 14

Page 11: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 11 of 47

pH and pOH Scales

Page 12: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 12 of 47

17-4 Strong Acids and Bases

HCl CH3CO2H

Thymol Blue Indicator

pH < 1.2 < pH < 2.8 < pH

Page 13: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 13 of 47

17-5 Weak Acids and Bases

Acetic Acid HC2H3O2 or CH3CO2H

Page 14: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 14 of 47

Weak Acids

Ka= = 1.810-5

[CH3CO2H]

[CH3CO2-][H3O+]

pKa= -log(1.810-5) = 4.74

glycine H2NCH2CO2H

lactic acid CH3CH(OH) CO2H

C

OH

O

R

Page 15: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 15 of 47

Table 17.3 Ionization Constants of Weak Acids and Bases

Page 16: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 16 of 47

Example 17-5Determining a Value of KA from the pH of a Solution of a Weak Acid.

Butyric acid, HC4H7O2 (or CH3CH2CH2CO2H) is used to make compounds employed in artificial flavorings and syrups. A 0.250 M aqueous solution of HC4H7O2 is found to have a pH of 2.72. Determine KA for butyric acid.

HC4H7O2 + H2O C4H77O2 + H3O+ Ka = ?

Solution:

For HC4H7O2 KA is likely to be much larger than KW. Therefore assume self-ionization of water is unimportant.

Page 17: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 17 of 47

Example 17-5

HC4H7O2 + H2O C4H7O2 + H3O+

Initial conc. 0.250 M 0 0

Changes -x M +x M +x M

Eqlbrm conc. (0.250-x) M x M x M

Page 18: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 18 of 47

Example 17-5

Log[H3O+] = -pH = -2.72

HC4H7O2 + H2O C4H77O2 + H3O+

[H3O+] = 10-2.72 = 1.910-3 = x

[H3O+] [C4H7O2-]

[HC4H7O2] Ka=

1.910-3 · 1.910-3

(0.250 – 1.910-3)=

Ka= 1.510-5 Check assumption: Ka >> KW.

Page 19: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 19 of 47

Percent Ionization

HA + H2O H3O+ + A-

Degree of ionization =[H3O+] from HA

[HA] originally

Percent ionization =[H3O+] from HA

[HA] originally 100%

Page 20: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 20 of 47

Percent Ionization

Ka =[H3O+][A-]

[HA]

Ka =n

H3O+ A-n

HAn

1

V

Page 21: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 21 of 47

17-6 Polyprotic Acids

H3PO4 + H2O H3O+ + H2PO4-

H2PO4- + H2O H3O+ + HPO4

2-

HPO42- + H2O H3O+ + PO4

3-

Phosphoric acid:

A triprotic acid.

Ka = 7.110-3

Ka = 6.310-8

Ka = 4.210-13

Page 22: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 22 of 47

Phosphoric Acid

• Ka1 >> Ka2

• All H3O+ is formed in the first ionization step.

• H2PO4- essentially does not ionize further.

• Assume [H2PO4-] = [H3O+].

• [HPO42-] Ka2

regardless of solution molarity.

Page 23: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 23 of 47

Table 17.4 Ionization Constants of Some Polyprotic Acids

Page 24: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 24 of 47

Example 17-9Calculating Ion Concentrations in a Polyprotic Acid Solution.

For a 3.0 M H3PO4 solution, calculate:

(a) [H3O+]; (b) [H2PO4-]; (c) [HPO4

2-] (d) [PO43-]

H3PO4 + H2O H2PO4- + H3O+

Initial conc. 3.0 M 0 0

Changes -x M +x M +x M

Eqlbrm conc. (3.0-x) M x M x M

Page 25: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 25 of 47

Example 17-9

H3PO4 + H2O H2PO4- + H3O+

[H3O+] [H2PO4-]

[H3PO4] Ka=

x · x

(3.0 – x)=

Assume that x << 3.0

= 7.110-3

x2 = (3.0)(7.110-3) x = 0.14 M

[H2PO4-] = [H3O+] = 0.14 M

Page 26: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 26 of 47

Example 17-9

H2PO4- + H2O HPO4

2- + H3O+

[H3O+] [HPO42-]

[H2PO4-]

Ka=y · (0.14 + y)

(0.14 - y)= = 6.310-8

Initial conc. 0.14 M 0 0.14 M

Changes -y M +y M +y M

Eqlbrm conc. (0.14 - y) M y M (0.14 +y) M

y << 0.14 M y = [HPO42-] = 6.310-8

Page 27: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 27 of 47

Example 17-9

HPO4- + H2O PO4

3- + H3O+

[H3O+] [HPO42-]

[H2PO4-]

Ka=(0.14)[PO4

3-]

6.310-8 = = 4.210-13 M

[PO43-] = 1.910-19 M

Page 28: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 28 of 47

Sulfuric Acid

Sulfuric acid:

A diprotic acid.

H2SO4 + H2O H3O+ + HSO4-

HSO4- + H2O H3O+ + SO4

2-

Ka = very large

Ka = 1.96

Page 29: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 29 of 47

General Approach to Solution Equilibrium Calculations

• Identify species present in any significant amounts in solution (excluding H2O).

• Write equations that include these species.– Number of equations = number of unknowns.

• Equilibrium constant expressions.

• Material balance equations.

• Electroneutrality condition.

• Solve the system of equations for the unknowns.

Page 30: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 30 of 47

17-7 Ions as Acids and Bases

NH4+ + H2O NH3 + H3O+

baseacid

CH3CO2- + H2O CH3CO2H + OH-

base acid

[NH3] [H3O+] [OH-] Ka= [NH4

+] [OH-]

[NH3] [H3O+] Ka= [NH4

+] = ?

=KW

Kb

= 1.010-14

1.810-5= 5.610-10

Ka Kb = Kw

Page 31: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 31 of 47

Hydrolysis

• Water (hydro) causing cleavage (lysis) of a bond.

Na+ + H2O → Na+ + H2O

NH4+ + H2O → NH3 + H3O+

Cl- + H2O → Cl- + H2O

No reaction

No reaction

Hydrolysis

Page 32: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 32 of 47

17-8 Molecular Structure and Acid-Base Behavior

• Why is HCl a strong acid, but HF is a weak one?

• Why is CH3CO2H a stronger acid than CH3CH2OH?

• There is a relationship between molecular structure and acid strength.

• Bond dissociation energies are measured in the gas phase and not in solution.

Page 33: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 33 of 47

Strengths of Binary Acids

HI HBr HCl HF

160.9 > 141.4 > 127.4 > 91.7 pm

297 < 368 < 431 < 569 kJ/mol

Bond length

Bond energy

109 > 108 > 1.3106 >> 6.610-4 Acid strength

HF + H2O → [F-·····H3O+] F- + H3O+

ion pairH-bonding

free ions

Page 34: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 34 of 47

Strengths of Oxoacids

• Factors promoting electron withdrawal from the OH bond to the oxygen atom:– High electronegativity (EN) of the central atom.

– A large number of terminal O atoms in the molecule.

H-O-Cl H-O-Br

ENCl = 3.0 ENBr= 2.8

Ka = 2.910-8 Ka = 2.110-9

Page 35: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 35 of 47

S OO

O

O

H H····

····

-

2+

··

··

·· ···· ··

-

S OO

O

H H····

····

-

+

··

··

·· ··

S OO

O

O

H H····

····

·· ···· ··

S OO

O

H H····

···· ··

·· ··

Ka 103 Ka =1.310-2

Page 36: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 36 of 47

Strengths of Organic Acids

C OC

O

H H····

·· ··H

H

OCH H····

H

H

C

H

H

Ka = 1.810-5 Ka =1.310-16

acetic acid ethanol

Page 37: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 37 of 47

Focus on the Anions Formed

OCH····

H

H

C

H

H

C

O

C

O

H

-··

····

H

H

····

C

O

C

O

H

-······

H

H

····

··

-

Page 38: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 38 of 47

Structural Effects

C

H

H

C

O

C

O

H

-··

····

H

H

····

CH

H

H

C

O

O

-··

····

····

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

Ka = 1.810-5

Ka = 1.310-5

Page 39: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 39 of 47

Structural Effects

C

O

C

O

H

-··

····

H

H

····

Ka = 1.810-5

Ka = 1.410-3

C

O

C

O

H

-··

····

H

Cl

····

Page 40: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 40 of 47

Strengths of Amines as Bases

NH

H

H

·· NBr

H

H

··

pKb = 4.74 pKa = 7.61

ammonia bromamine

Page 41: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 41 of 47

Strengths of Amines as Bases

CH

H

H

C

H

H

CH

H

H

C

H

H

CH

H

H

C

H

H

pKb = 4.74 pKa = 3.38 pKb = 3.37

methylamine ethylamine propylamine

NH2 NH2NH2

Page 42: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 42 of 47

Resonance Effects

Page 43: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 43 of 47

Inductive Effects

Page 44: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 44 of 47

17-9 Lewis Acids and Bases

• Lewis Acid– A species (atom, ion or molecule) that is an electron

pair acceptor.

• Lewis Base– A species that is an electron pair donor.

base acid adduct

Page 45: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 45 of 47

Showing Electron Movement

Page 46: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 46 of 47

Focus On Acid Rain

CO2 + H2O H2CO3

H2CO3 + H2O HCO3- + H3O+

3 NO2 + H2O 2 HNO3 + NO

Page 47: Ch17

Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 47 of 47

Chapter 17 Questions

Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding.

Choose a variety of problems from the text as examples.

Practice good techniques and get coaching from people who have been here before.


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