Upload
august-curtis
View
212
Download
0
Embed Size (px)
Citation preview
CHE116Prof. T. L. Heise
1CHE 116: General Chemistry
CHAPTER SIXTEEN
Copyright © Tyna L. Heise 2001-2002
All Rights Reserved
CHE116Prof. T. L. Heise
2Acids and Bases: Review
Properties of Acidssour tastechange with litmus
Properties of Bases
bitter taste
change with litmus
CHE116Prof. T. L. Heise
3Acids and Bases: Review
1830 - scientists have recognized that all acids contain hydrogen, but not all hydrogen bearing compounds are acids
Svante Arrhenius - linked acid behavior with the presence of an H+ and base behavior with the presence of an OH-
CHE116Prof. T. L. Heise
4Bronsted-Lowry Acids and Bases
Arrhenius’ definition is useful, but restricts acid base reactions to aqueous conditions
Johannes Bronsted and Thomas Lowry proposed a more general definition which involves the transfer of an H+ ion from one molecule to another
CHE116Prof. T. L. Heise
5Bronsted-Lowry Acids and Bases
H+ ion is simply a proton with no surrounding valence electron.Small particle interacts strongly with the nonbonding pairs of
water molecules to form hydrated hydrogen ionschemists use H+ and H3O+ interchangeably
CHE116Prof. T. L. Heise
6Bronsted-Lowry Acids and Bases
Fig 16.1
Demonstrates the interconnections possible between hydrogenated water
CHE116Prof. T. L. Heise
7Bronsted-Lowry Acids and Bases
Definitions:
Acid - any compound which transfers an H+ to another molecule
Base - any compound which accepts a transfer of an H+ from another molecule
* an acid and base always work together
CHE116Prof. T. L. Heise
8Bronsted-Lowry Acids and Bases
Definitions:
Amphoteric - some substances can be an acid or a base depending on reaction
Conjugate Acid Base pairs - two compounds that differ only in the presence of an H+. The molecule with the extra H is the acid.
CHE116Prof. T. L. Heise
10Bronsted-Lowry Acids and Bases
Sample Exercise: Write the formula for the conjugate acid of each of the following:
HSO3-
F-
PO43-
CO
CHE116Prof. T. L. Heise
11Bronsted-Lowry Acids and Bases
Sample Exercise: Write the formula for the conjugate acid of each of the following:
HSO3-
F-
PO43-
CO
Given a base, Given a base, bases accept bases accept HH++, so add an , so add an HH++ to each to each molecule.molecule.
CHE116Prof. T. L. Heise
12Bronsted-Lowry Acids and Bases
Sample Exercise: Write the formula for the conjugate acid of each of the following:
H2SO3 HSO3-
F-
PO43-
CO
Given a base, Given a base, bases accept bases accept HH++, so add an , so add an HH++ to each to each molecule.molecule.
CHE116Prof. T. L. Heise
13Bronsted-Lowry Acids and Bases
Sample Exercise: Write the formula for the conjugate acid of each of the following:
H2SO3 HSO3-
HF F-
PO43-
CO
Given a base, Given a base, bases accept bases accept HH++, so add an , so add an HH++ to each to each molecule.molecule.
CHE116Prof. T. L. Heise
14Bronsted-Lowry Acids and Bases
Sample Exercise: Write the formula for the conjugate acid of each of the following:
H2SO3 HSO3-
HF F-
HPO42- PO4
3-
CO
Given a base, Given a base, bases accept bases accept HH++, so add an , so add an HH++ to each to each molecule.molecule.
CHE116Prof. T. L. Heise
15Bronsted-Lowry Acids and Bases
Sample Exercise: Write the formula for the conjugate acid of each of the following:
H2SO3 HSO3-
HF F-
HPO42- PO4
3-
HCO+ CO
Given a base, Given a base, bases accept bases accept HH++, so add an , so add an HH++ to each to each molecule.molecule.
CHE116Prof. T. L. Heise
16Bronsted-Lowry Acids and Bases
Sample Exercise: When lithium oxide, Li2O, is dissolved in water, the solution turns basic from the reaction of the oxide ion, O2-, with water. Write the reaction that occurs, and identify the conjugate acid base pairs.
CHE116Prof. T. L. Heise
17Bronsted-Lowry Acids and Bases
Sample Exercise: When lithium oxide, Li2O, is dissolved in water, the solution turns basic from the reaction of the oxide ion, O2-, with water. Write the reaction that occurs, and identify the conjugate acid base pairs.
O2- + H2O OH- + OH-
CHE116Prof. T. L. Heise
18Bronsted-Lowry Acids and Bases
Sample Exercise: When lithium oxide, Li2O, is dissolved in water, the solution turns basic from the reaction of the oxide ion, O2-, with water. Write the reaction that occurs, and identify the conjugate acid base pairs.
O2- + H2O OH- + OH-
CHE116Prof. T. L. Heise
19Bronsted-Lowry Acids and Bases
Sample Exercise: When lithium oxide, Li2O, is dissolved in water, the solution turns basic from the reaction of the oxide ion, O2-, with water. Write the reaction that occurs, and identify the conjugate acid base pairs.
O2- + H2O OH- + OH-
acidacid base base
basebase acidacid
CHE116Prof. T. L. Heise
20Bronsted-Lowry Acids and Bases
Relative Strengths of Acids and Bases:the more readily a substance donates an H+, the less readily it’s
conjugate base will accept one the more readily a substance accepts an H+, the less readily it’s
conjugate acid will donate onethe stronger one of the substances is, the weaker it’s conjugate
CHE116Prof. T. L. Heise
21Bronsted-Lowry Acids and Bases
Relative Strengths of Acids and Bases:strong acids completely transfer their protons to water, leaving no
undissociated molecules weak acids are those that only partly dissociate in aqueous
solution and therefore exist in the solution as a mixture of acid molecules and component ions
CHE116Prof. T. L. Heise
22Bronsted-Lowry Acids and Bases
Relative Strengths of Acids and Bases:negligible acids are those that have hydrogen but do not donate them
at all, their conjugate bases would be extremely strong, reacting with water to complete their octet and leaving OH- behind.
In every acid base reaction, the position of the equilibrium favors transfer of H+ from stronger side to weaker side
CHE116Prof. T. L. Heise
24Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right:
a) PO43-(aq)+H2O(l)HPO4
2-(aq)+OH-(aq)
b) NH4+(aq)+OH-(aq) NH3(aq)+H2O(l)
CHE116Prof. T. L. Heise
25Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right:
a) PO43-(aq)+H2O(l)HPO4
2-(aq)+OH-(aq)
2 acids are: H2O and HPO42-
2 bases are: PO43- and OH-
CHE116Prof. T. L. Heise
26Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right:
a) PO43-(aq)+H2O(l)HPO4
2-(aq)+OH-(aq)
2 acids are: H2O and HPO42-
2 bases are: PO43- and OH-
red indicates strength
CHE116Prof. T. L. Heise
27Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right:
a) PO43-(aq)+H2O(l)HPO4
2-(aq)+OH-(aq)
2 acids are: H2O and HPO42-
2 bases are: PO43- and OH-
reverse reaction favored
CHE116Prof. T. L. Heise
28Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right:
a) PO43-(aq)+H2O(l)HPO4
2-(aq)+OH-(aq)
2 acids are: H2O and HPO42-
2 bases are: PO43- and OH-
shifts left
CHE116Prof. T. L. Heise
29Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right:
b) NH4+(aq)+OH-(aq) NH3(aq)+H2O(l)
CHE116Prof. T. L. Heise
30Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right:
b) NH4+(aq)+OH-(aq) NH3(aq)+H2O(l)
2 acids are: NH4+ and H2O
2 bases are: NH3 and OH-
CHE116Prof. T. L. Heise
31Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right:
b) NH4+(aq)+OH-(aq) NH3(aq)+H2O(l)
2 acids are: NH4+ and H2O
2 bases are: NH3 and OH-
red indicates strength
CHE116Prof. T. L. Heise
32Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right:
b) NH4+(aq)+OH-(aq) NH3(aq)+H2O(l)
2 acids are: NH4+ and H2O
2 bases are: NH3 and OH-
favors forward reaction
CHE116Prof. T. L. Heise
33Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right:
b) NH4+(aq)+OH-(aq) NH3(aq)+H2O(l)
2 acids are: NH4+ and H2O
2 bases are: NH3 and OH-
shifts right
CHE116Prof. T. L. Heise
34The Autoionization of Water
One of the most important properties of water is its ability to ac as either a Bronsted acid or Bronsted base, depending on circumstances.One water molecule can donate a proton to another
water molecule Fig 16.10
CHE116Prof. T. L. Heise
35The Autoionization of Water
The autoionization of water, although rapid and weak, does exist as an equilibrium, and therefore has an equilibrium constant expression:
Keq = [H3O+][OH-] [H2O]2
* because water is a liquid, it can be excluded from the equation...
CHE116Prof. T. L. Heise
36The Autoionization of Water
Keq[H2O]2 = [H3O+][OH-]
Kw = [H3O+][OH-] = 1.0 x 10-14
* this equation is not only applicable to water, but to all aqueous solutions, and it is upon this fact that the pH scale was built.
CHE116Prof. T. L. Heise
37The Autoionization of Water
Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic:
a) [H+] = 2 x 10-5
b) [OH-] = 3 x 10-9
c) [OH-] = 1 x 10-7
CHE116Prof. T. L. Heise
38The Autoionization of Water
Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic:
a) [H+] = 2 x 10-5 then [OH-] must equal 1.0 x 10-14 which is 2 x10-5
[OH-] = 5.0 x 10-10
CHE116Prof. T. L. Heise
39The Autoionization of Water
Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic:
a) [H+] = 2 x 10-5 then [OH-] must equal 1.0 x 10-14 which is 2 x10-5
[OH-] = 5.0 x 10-10
[H+] > [OH-] so acidic
CHE116Prof. T. L. Heise
40The Autoionization of WaterSample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic:
b) [OH-] = 3 x 10-9 then [H+] must equal 1.0 x 10-14 which is 3 x10-9
[H+] = 3.3 x 10-6
CHE116Prof. T. L. Heise
41The Autoionization of WaterSample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic:
b) [OH-] = 3 x 10-9 then [H+] must equal 1.0 x 10-14 which is 3 x10-9
[H+] = 3.3 x 10-6
[H+] > [OH-] so acidic
CHE116Prof. T. L. Heise
42The Autoionization of Water
Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic:
c) [OH-] = 1 x 10-7 then [H+] must equal 1.0 x 10-14 which is 1 x10-7
[H+] = 1.0 x 10-7
CHE116Prof. T. L. Heise
43The Autoionization of Water
Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic:
c) [OH-] = 1 x 10-7 then [H+] must equal 1.0 x 10-14 which is 1 x10-7
[H+] = 1.0 x 10-7
[H+] = [OH-] so neutral
CHE116Prof. T. L. Heise
44The pH Scale
For convenience, we can use a logarithmic version of concentration to turn the very small concentrations of [H+] and [OH-] into whole numbers.
p(anything) = - log[anything]
p(H) = -log[H+]
p(OH) = - log[OH-]
** pH + pOH = 14
CHE116Prof. T. L. Heise
47The pH Scale
Sample exercise: In a sample of lemon juice [H+] is 3.8 x 10-4 M. What is the pH?
CHE116Prof. T. L. Heise
48The pH Scale
Sample exercise: In a sample of lemon juice [H+] is 3.8 x 10-4 M. What is the pH?
pH = -log[H+]
CHE116Prof. T. L. Heise
49The pH Scale
Sample exercise: In a sample of lemon juice [H+] is 3.8 x 10-4 M. What is the pH?
pH = -log[H+]
pH = -log[3.8 x 10-4]
CHE116Prof. T. L. Heise
50The pH Scale
Sample exercise: In a sample of lemon juice [H+] is 3.8 x 10-4 M. What is the pH?
pH = -log[H+]
pH = -log[3.8 x 10-4]
pH = 3.42
CHE116Prof. T. L. Heise
51The pH Scale
Sample exercise: A commonly available window-cleaning solution has a [H+] is 5.3 x 10-9 M. What is the pH?
CHE116Prof. T. L. Heise
52The pH Scale
Sample exercise: A commonly available window-cleaning solution has a [H+] is 5.3 x 10-9 M. What is the pH?
pH = -log[H+]
CHE116Prof. T. L. Heise
53The pH Scale
Sample exercise: A commonly available window-cleaning solution has a [H+] is 5.3 x 10-9 M. What is the pH?
pH = -log[H+]
pH = -log[5.3 x 10-9]
CHE116Prof. T. L. Heise
54The pH Scale
Sample exercise: A commonly available window-cleaning solution has a [H+] is 5.3 x 10-9 M. What is the pH?
pH = -log[H+]
pH = -log[5.3 x 10-9]
pH = 8.28
CHE116Prof. T. L. Heise
55The pH Scale
Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H+].
CHE116Prof. T. L. Heise
56The pH Scale
Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H+].
pH = -log[H+]
CHE116Prof. T. L. Heise
57The pH Scale
Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H+].
pH = -log[H+]
9.18 = -log[H+]
CHE116Prof. T. L. Heise
58The pH Scale
Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H+].
pH = -log[H+]
9.18 = -log[H+]
10-9.18 = [H+]
CHE116Prof. T. L. Heise
59The pH ScaleSample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18.
Calculate [H+].
pH = -log[H+]
9.18 = -log[H+]
10-9.18 = [H+]
6.61 x 10-10 = [H+]
CHE116Prof. T. L. Heise
60The pH Scale
Measuring pH: a pH can be measured quickly and accurately using a pH meter.A pair of electrodes connected to a meter capable of measuring small voltagesa voltage which varies with pH is generated when the electrodes are placed in
a solutioncalibrated to give pH
CHE116Prof. T. L. Heise
61The pH Scale
Measuring pH: a pH can be measured quickly and accurately using a pH meter.Electrodes come in a variety of shapes and sizesa set of electrodes exists that can be placed inside a human cellacid base indicators can be used, but are much less precise
CHE116Prof. T. L. Heise
63Strong Acids and Bases
Strong acids and bases are strong electrolytes, existing in aqueous solution entirely as ions. Strong Acids HCl
HBr HImonoprotic HNO3 HClO3
HClO4 H2SO4
diprotic
CHE116Prof. T. L. Heise
64Strong Acids and Bases
Strong acids and bases are strong electrolytes, existing in aqueous solution entirely as ions.
Calculating the pH of a solution made up entirely of ions means the [H+] is proportional to [acid]
CHE116Prof. T. L. Heise
65Strong Acids and Bases
An aqueous solution of HNO3 has a pH of 2.66. What is the concentration of the acid?
CHE116Prof. T. L. Heise
66Strong Acids and Bases
An aqueous solution of HNO3 has a pH of 2.66. What is the concentration of the acid?
pH = -log[H+]
CHE116Prof. T. L. Heise
67Strong Acids and Bases
An aqueous solution of HNO3 has a pH of 2.66. What is the concentration of the acid?
pH = -log[H+] 2.66 = -log[H+]
CHE116Prof. T. L. Heise
68Strong Acids and Bases
An aqueous solution of HNO3 has a pH of 2.66. What is the concentration of the acid?
pH = -log[H+] 2.66 = -log[H+] 10-2.66 = [H+]
2.2 x 10-3 = [H+]
CHE116Prof. T. L. Heise
69Strong Acids and Bases
An aqueous solution of HNO3 has a pH of 2.66. What is the concentration of the acid?
pH = -log[H+] 2.66 = -log[H+] 10-2.66 = [H+] 2.2 x 10-3 = [H+] 2.2 x 10-3 M H+ 1 mole HNO3 1 mole H+
CHE116Prof. T. L. Heise
70Strong Acids and Bases
An aqueous solution of HNO3 has a pH of 2.66. What is the concentration of the acid?
pH = -log[H+] 2.66 = -log[H+] 10-2.66 = [H+]2.2 x 10-3 = [H+] 2.2 x 10-3 M H+ 1 mole HNO3
1 mole H+ 2.2 x 10-3 HNO3
CHE116Prof. T. L. Heise
71Strong Acids and Bases
The most soluble common bases are the ionic hydroxides of the alkali and alkaline earth metals.
Due to the complete dissociation of the base into its ion components makes the pH calculation equally straightforward
CHE116Prof. T. L. Heise
72Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
CHE116Prof. T. L. Heise
73Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
pH = -log[H+]
CHE116Prof. T. L. Heise
74Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
pH = -log[H+]11.89 = -log[H+]
CHE116Prof. T. L. Heise
75Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
pH = -log[H+] 11.89 = -log[H+]10-11.89 = [H+]
CHE116Prof. T. L. Heise
76Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
pH = -log[H+]11.89 = -log[H+] 10-11.89 = [H+]
1.29 x 10-12 = [H+]
CHE116Prof. T. L. Heise
77Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
pH = -log[H+] 11.89 = -log[H+]10-11.89 = [H+] 1.29 x 10-12 = [H+]1.0 x 10-14 = [OH-] 1.29 x 10-12
CHE116Prof. T. L. Heise
78Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
pH = -log[H+] 11.89 = -log[H+]10-11.89 = [H+] 1.29 x 10-12 = [H+]1.0 x 10-14 = [OH-] = 7.8 x 10-3 1.29 x 10-12
CHE116Prof. T. L. Heise
79Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
1.0 x 10-14 = [OH-] = 7.8 x 10-3 1.29 x 10-12
7.8 x 10-3 M OH- 1 mol KOH 1 mol OH-
CHE116Prof. T. L. Heise
80Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
1.0 x 10-14 = [OH-] = 7.8 x 10-3 1.29 x 10-12
7.8 x 10-3 M OH- 1 mol KOH 1 mol OH- 7.8 x 10-3 M KOH
CHE116Prof. T. L. Heise
81Weak Acids
Most acids are weak acids and only partially dissociate in aqueous solution.
The extent to which a weak acid dissociates can be expressed using an equilibrium constant for the ionization reaction HX + H2O H3O+ + X- Ka = [H3O+][X-] [HX] *the larger the Ka the stronger the acid
CHE116Prof. T. L. Heise
82Weak Acids
Calculating Ka from pH: a much more complicated calculation is required for the determinations of weak acids, in many cases, due to the extremely small magnitude of the values, some simpler approximations can be made.
CHE116Prof. T. L. Heise
83Weak Acids
Sample exercise: Niacin, one of the B vitamins, has the following molecular structure: C O H O N
A 0.020 M solution of niacin has a pH of 3.26
a) What % of the acid is ionized in this solution?
b) What is the acid-dissociation constant?
CHE116Prof. T. L. Heise
84Weak Acids
Sample exercise: A 0.020 M solution of niacin has a pH of 3.26
a) What % of the acid is ionized in this solution? pH = -log[H+] 3.26 = -log[H+] e-3.26 = [H+] 5.5 x 10-4 = [H+]
CHE116Prof. T. L. Heise
85Weak Acids
Sample exercise: A 0.020 M solution of niacin has a pH of 3.26
a) What % of the acid is ionized in this solution?C6H4NOOH C6H4NOO- + H+
initial 0.020 0 0
change -5.5 x 10-4 +5.5 x 10-4 +5.5 x 10-4
equil 0.01945 5.5 x 10-4 5.5 x 10-4
CHE116Prof. T. L. Heise
86Weak Acids
Sample exercise: A 0.020 M solution of niacin has a pH of 3.26
a) What % of the acid is ionized in this solution?% = part/total x 100
= 5.5 x 10-4/0.01945 x 100
= 2.8%
CHE116Prof. T. L. Heise
87Weak Acids
Sample exercise: A 0.020 M solution of niacin has a pH of 3.26
b) What is the acid-dissociation constant?
C6H4NOOH C6H4NOO- + H+
Ka = [C6H4NOO- ][H+] [C6H4NOOH] = (5.5 x 10-4 )(5.5 x 10-4 ) 0.01945 = 1.55 x10-5
CHE116Prof. T. L. Heise
88Weak Acids
Using Ka to calculate pH:
Using the value of Ka and knowing the initial concentration of the weak acid, we can calculate the concentration of H+(aq).
Example: Calculate the pH of a 0.30 M solution of acetic acid.
HC2H3O2(aq) H+(aq) + C2H3O2-(aq)
CHE116Prof. T. L. Heise
89Weak Acids
Example: Calculate the pH of a 0.30 M solution of acetic acid.
HC2H3O2(aq) H+(aq) + C2H3O2-(aq)
From Table 16.2, Ka = 1.8 x 10-5
Ka = 1.8 x 10-5 = [H+][C2H3O2-] [HC2H3O2]
set up data table of concentrations involved...
CHE116Prof. T. L. Heise
90Weak Acids
Ka = 1.8 x 10-5 = [H+][C2H3O2-] [HC2H3O2]
set up data table of concentrations involved…
[HC2H3O2] [H+] [C2H3O2-]
Initial 0.30 M 0 0
Change -x +x +x
Equilibrium 0.30 - x x x
CHE116Prof. T. L. Heise
91Weak Acids
Input concentrations in formula
Ka = 1.8 x 10-5 = [x][x] [0.30 -x]
This will lead to a quadratic equation, but we can simplify the problem a bit. All weak acids dissociate so little that we can assume the initial concentration of the acid remains essentially the same, that means we can rewrite the formula to read...
CHE116Prof. T. L. Heise
92Weak AcidsKa = 1.8 x 10-5 = [x][x] [0.30]
1.8 x 10-5 (0.30) = x2
1.8 x 10-5 (0.30) = x
0.0023 = x
0.0023 = [H+] pH = -log [H+]
= -log[0.0023]
= 2.64
CHE116Prof. T. L. Heise
93Weak Acids
Sample Exercise: The Ka for niacin is 1.5 x 10-5. What is the pH of a 0.010 M solution of niacin?
CHE116Prof. T. L. Heise
94Weak Acids
Sample Exercise: The Ka for niacin is 1.5 x 10-5. What is the pH of a 0.010 M solution of niacin?
C6H4NOOH C6H4NOO- + H+
Ka = 1.5 x10-5 = [C6H4NOO-][H+] [C6H4NOOH]
CHE116Prof. T. L. Heise
95Weak Acids
Sample Exercise: The Ka for niacin is 1.5 x 10-5. What is the pH of a 0.010 M solution of niacin?
C6H4NOOH C6H4NOO- H+
initial 0.010 0 0 change -x +x +x equilibrium 0.010 -x x x
CHE116Prof. T. L. Heise
96Weak Acids
Sample Exercise: The Ka for niacin is 1.5 x 10-5. What is the pH of a 0.010 M solution of niacin?
C6H4NOOH C6H4NOO- + H+
Ka = 1.5 x10-5 = [x][x] [0.010]
CHE116Prof. T. L. Heise
97Weak Acids
Sample Exercise: The Ka for niacin is 1.5 x 10-5. What is the pH of a 0.010 M solution of niacin?
1.5 x10-5 [0.010] = x2
1.5 x10-5 [0.010] = x
3.87 x 10-4 = x
CHE116Prof. T. L. Heise
98Weak Acids
Sample Exercise: The Ka for niacin is 1.5 x 10-5. What is the pH of a 0.010 M solution of niacin?
3.87 x 10-4 = x
3.87 x 10-4 = [H+] pH = -log[H+] pH = -log[3.87 x 10-4] pH = 3.41
CHE116Prof. T. L. Heise
99Weak Acids
The results seen in the two previous examples are typical of weak acids. The lower number of ions produced during the partial dissociation causes less electrical conductivity and a slower reaction rate with metals.
Percent ionization is a good way to discover the actual conductivity, however...
CHE116Prof. T. L. Heise
100Weak Acids
As the concentration of a weak acid increases, the % ionized decreases.
CHE116Prof. T. L. Heise
101Weak Acids
As the concentration of a weak acid increases, the % ionized decreases.
CHE116Prof. T. L. Heise
102Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in
a) the previous exerciseb) a 1.0 x 10-3 M solution
CHE116Prof. T. L. Heise
103Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in
a) the previous exercise
Our approximation was good so% = part x 100 total
= 3.87 x 10-4 x 100 = 3.9% 0.010
CHE116Prof. T. L. Heise
104Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in
b) a 1.0 x 10-3 M solution
Ka = 1.5 x10-5 = [x][x] [1.0 x 10-3]
1.5 x10-5 (1.0 x 10-3) = x2
1.2 x 10-4
CHE116Prof. T. L. Heise
105Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in
b) a 1.0 x 10-3 M solution
Ka = 1.5 x10-5 = [x][x] [1.0 x 10-3] but, 1.2 x 10-3 x 100
1.5 x10-5 (1.0 x 10-3) = x2 1.0 x 10-3
1.2 x 10-3 is greater than 5% so use quadratic...
CHE116Prof. T. L. Heise
106Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in
b) a 1.0 x 10-3 M solution
Ka = 1.5 x10-5 = [x][x] [1.0 x 10-3 - x]
-1.5 x 10-5(1.0 x10-3) + 1.5 x 10-5x + x2 = 0
CHE116Prof. T. L. Heise
107Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in
b) a 1.0 x 10-3 M solution
-1.5 x 10-5(1.0 x10-3) + 1.5 x 10-5x + x2 = 0
x = -b ± b2 - 4ac 2a
x =
CHE116Prof. T. L. Heise
108Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in
b) a 1.0 x 10-3 M solution
-1.5 x 10-5(1.0 x10-3) + 1.5 x 10-5x + x2 = 0
x = -b ± b2 - 4ac 2a
x = 1.1 x 10-4 or -1.3 x 10-4
CHE116Prof. T. L. Heise
109Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in
b) a 1.0 x 10-3 M solution
x = 1.1 x 10-4
1.1 x 10-4 x 1001.0 x 10-3
CHE116Prof. T. L. Heise
110Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in
b) a 1.0 x 10-3 M solution
x = 1.1 x 10-4
1.1 x 10-4 x 100 = 11%1.0 x 10-3
CHE116Prof. T. L. Heise
111Weak Acids
Polyprotic Acids: many acids have more than one ionizable H atom.
H2SO3(aq) H+(aq) + HSO3-(aq) Ka1
HSO3-(aq) H+(aq) + SO3
-2(aq) Ka2
The Ka are labeled according to which proton is dissociating.
-it is always easier to remove the first proton than the second
CHE116Prof. T. L. Heise
113Weak Acids
If Ka values differ by 103 or more, only use Ka1 to determine calculations.
CHE116Prof. T. L. Heise
114Weak Acids
Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O4
2-] in a 0.020 M solution of oxalic acid.
CHE116Prof. T. L. Heise
115Weak Acids
Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O4
2-] in a 0.020 M solution of oxalic acid.
H2C2O4 HC2O4- + H+
Ka = 5.9 x 10-2 = [HC2O4- ][H+] [H2C2O4]
CHE116Prof. T. L. Heise
116Weak Acids
Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O4
2-] in a 0.020 M solution of oxalic acid.
H2C2O4 HC2O4- + H+
strong acid so 100% dissociation
0.020 M H+
CHE116Prof. T. L. Heise
117Weak Acids
Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O4
2-] in a 0.020 M solution of oxalic acid.
HC2O4- C2O4
-2 + H+
I 0.020 0 0.020 -x +x +x E 0.020 - x x 0.020 + x
CHE116Prof. T. L. Heise
118Weak Acids
Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O4
2-] in a 0.020 M solution of oxalic acid.
Ka = 6.4 x 10-5 = [0.020 + x ][x] [0.020 - x]
use your assumption
CHE116Prof. T. L. Heise
119Weak Acids
Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O4
2-] in a 0.020 M solution of oxalic acid.
Ka = 6.4 x 10-5 = [0.020 ][x] [0.020]
6.4 x 10-5 = x = [C2O42-]
CHE116Prof. T. L. Heise
120Weak Acids
Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O4
2-] in a 0.020 M solution of oxalic acid.
[H+] = 0.020
CHE116Prof. T. L. Heise
121Weak Acids
Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O4
2-] in a 0.020 M solution of oxalic acid.
pH = -log[H+]
pH = - log[0.020]
pH = 1.70
CHE116Prof. T. L. Heise
122Weak Bases
Many substances behave as weak bases in water. Such substances react with water, removing protons from water, leaving the OH- ion behind.
NH3 + H2O NH4+ + OH-
Kb = [NH4+][OH-] [NH3]
* Kb is the base dissociation constant utilizing the [OH-]
CHE116Prof. T. L. Heise
123Weak Bases
* Kb is the base dissociation constant utilizing the [OH-]bases must contain one or more lone pair to bond with the H+
from water.as before, the larger the Kb the stronger the basestronger base have low pOH, but high pH
CHE116Prof. T. L. Heise
124Weak Bases
Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution?A) pyridine
B) methylamine
C) nitrous acid
CHE116Prof. T. L. Heise
125Weak Bases
Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution?A) pyridine Kb = 1.7 x 10-9
B) methylamine Kb = 4.4 x 10-4
C) nitrous acid Kb = 2.2 x 10-11
CHE116Prof. T. L. Heise
126Weak Bases
Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? A) pyridine Kb = 1.7 x 10-9
Kb = 1.7 x 10-9 = [x][x] [0.05]
x = [OH-] = 9.2 x 10-6
pOH = 5.0 so pH = 9.0
CHE116Prof. T. L. Heise
127Weak Bases
Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? B) methylamine Kb = 4.4 x 10-4
Kb = 4.4 x 10-4 = [x][x] [0.05]
x = [OH-] = 4.6 x 10-3
pOH = 2.32 so pH = 11.68
CHE116Prof. T. L. Heise
128Weak Bases
Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? C) nitrous acid Kb = 2.2 x 10-11
Kb = 2.2 x 10-11 = [x][x] [0.05]
x = [OH-] = 1.0 x 10-6
pOH = 5.97 so pH = 8.02
CHE116Prof. T. L. Heise
129Weak Bases
Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution?A) pyridine pH = 9.0
B) methylamine pH = 11.68
C) nitrous acid pH = 8.02
CHE116Prof. T. L. Heise
130Weak Bases
Identifying a Weak BaseNeutral substances that have an atom with a nonbonding pair
of electrons that can serve as a proton acceptor.Most of these are nitrogen atoms
Anions of weak acids
CHE116Prof. T. L. Heise
131Weak Bases
Sample exercise: A solution of NH3 in water has a pH of 10.50. What is the molarity of the solution?
CHE116Prof. T. L. Heise
132Weak Bases
Sample exercise: A solution of NH3 in water has a pH of 10.50. What is the molarity of the solution?
pH = 10.50 so pOH = 3.50
CHE116Prof. T. L. Heise
133Weak Bases
Sample exercise: A solution of NH3 in water has a pH of 10.50. What is the molarity of the solution?
pOH = 3.50NH3 + H20 NH4+ + OH-
[OH-] = 3.16 x 10-4
CHE116Prof. T. L. Heise
134Weak Bases
Sample exercise: A solution of NH3 in water has a pH of 10.50. What is the molarity of the solution?
NH3 + H20 NH4+ + OH-
x 0 0 -3.16 x 10-4 +3.16 x 10-4 +3.16 x 10-4
x - 3.16 x 10-4 3.16 x 10-4 3.16 x 10-4
CHE116Prof. T. L. Heise
135Weak Bases
Sample exercise: A solution of NH3 in water has a pH of 10.50. What is the molarity of the solution?
NH3 + H20 NH4+ + OH-
Kb = 1.8 x 10-5 = [3.16 x 10-4][3.16 x 10-4] [x - 3.16 x 10-4]
CHE116Prof. T. L. Heise
136Weak Bases
Sample exercise: A solution of NH3 in water has a pH of 10.50. What is the molarity of the solution?
NH3 + H20 NH4+ + OH-
Kb = 1.8 x 10-5 = [3.16 x 10-4][3.16 x 10-4] [x - 3.16 x 10-4]
x = 0.0058 M
CHE116Prof. T. L. Heise
137Relationship Between Ka and Kb
When two reactions are added to give a third reaction, the equilibrium constant for the third reaction is equal to the product of the equilibrium constants for the two added reactants.
NH4+(aq) NH3(aq) + H+(aq)
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
H2O(l) H+(aq) + OH-(aq)
CHE116Prof. T. L. Heise
138Relationship Between Ka and Kb
NH4+(aq) NH3(aq) + H+(aq)
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
H2O(l) H+(aq) + OH-(aq)
Ka x Kb = Kw
pKa + pKb = pKw = 14
CHE116Prof. T. L. Heise
139Relationship Between Ka and Kb
Sample exercise: Which of the following anions has the largest base-dissociation constant? A) NO2
-
B) PO43-
C) N3-
CHE116Prof. T. L. Heise
140Relationship Between Ka and Kb
Sample exercise: Which of the following anions has the largest base-dissociation constant? A) NO2
- Ka = 4.5 x 10-4
Ka x Kb = 1.0 x 10-14
Kb = 1.0 x 10-14 = 2.2 x 10-11
4.5 x 10-4
CHE116Prof. T. L. Heise
141Relationship Between Ka and Kb
Sample exercise: Which of the following anions has the largest base-dissociation constant? B) PO43- Ka = 4.2 x
10-13
Ka x Kb = 1.0 x 10-14
Kb = 1.0 x 10-14 = 2.4 x 10-2
4.2 x 10-13
CHE116Prof. T. L. Heise
142Relationship Between Ka and Kb
Sample exercise: Which of the following anions has the largest base-dissociation constant? C) N3
- Ka = 1.9 x 10-5
Ka x Kb = 1.0 x 10-14
Kb = 1.0 x 10-14 = 5.2 x 10-10
1.9 x 10-5
CHE116Prof. T. L. Heise
143Relationship Between Ka and Kb
Sample exercise: Which of the following anions has the largest base-dissociation constant? A) NO2
- Kb = 2.2 x 10-11
B) PO43- Kb = 2.4 x 10-2
C) N3- Kb = 5.2 x 10-10
CHE116Prof. T. L. Heise
144Relationship Between Ka and Kb
Sample exercise: The base quinoline has a pKa of 4.90. What is the base dissociation constant for quinoline?
CHE116Prof. T. L. Heise
145Relationship Between Ka and Kb
Sample exercise: The base quinoline has a pKa of 4.90. What is the base dissociation constant for quinoline?
pKa + pKb = 14
4.90 + x = 14
x = 9.1
CHE116Prof. T. L. Heise
146Relationship Between Ka and Kb
Sample exercise: The base quinoline has a pKa of 4.90. What is the base dissociation constant for quinoline?
pKb = 9.1
pKb = -log[Kb]
9.1 = -log[Kb] [Kb] = 7.9 x 10-10
CHE116Prof. T. L. Heise
147Acid Base Properties of Salt Soln’s
Salt solutions have the potential to be acidic or basic.Hydrolysis of a saltacid base properties are due to the behavior of their cations and anionsperform the necessary double replacement reaction and examine the
products using your strength rules
CHE116Prof. T. L. Heise
148Acid Base Properties of Salt Soln’s
If a strong acid and strong base are produced, the resultant solution will be neutral.
If a strong acid and weak base are produced, the resultant solution will be acidic.
If a strong base and a weak acid are produced, the resultant solution will be basic.
CHE116Prof. T. L. Heise
149Acid Base Properties of Salt Soln’s
If a weak acid and weak base are produced, the resultant solution will be dependent on the Ka values.
CHE116Prof. T. L. Heise
150Acid Base Properties of Salt Soln’s
Sample exercise: In each of following, which salt will form the more acidic solution.
A) NaNO3, Fe(NO3)3
NaNO3 + H2O NaOH + HNO3
SB SA
CHE116Prof. T. L. Heise
151Acid Base Properties of Salt Soln’sSample exercise: In each of following, which salt will form the more acidic solution.
A) NaNO3, Fe(NO3)3
Fe(NO3)3 + 3H2O Fe(OH)3 + 3HNO3
WB SA
CHE116Prof. T. L. Heise
152Acid Base Properties of Salt Soln’sSample exercise: In each of following, which salt will form the more acidic solution.
A) NaNO3, Fe(NO3)3
Fe(NO3)3 + 3H2O Fe(OH)3 + 3HNO3
WB SA
CHE116Prof. T. L. Heise
153Acid Base Properties of Salt Soln’s
Sample exercise: In each of following, which salt will form the more acidic solution.
B) KBr, KBrO
CHE116Prof. T. L. Heise
154Acid Base Properties of Salt Soln’sSample exercise: In each of following, which salt will form the more acidic solution.
B) KBr, KBrO
KBr + H2O KOH + HBr
SB SA
CHE116Prof. T. L. Heise
155Acid Base Properties of Salt Soln’sSample exercise: In each of following, which salt will form the more acidic solution.
B) KBr, KBrO
KBrO + H2O KOH + HBrO
SB WA
CHE116Prof. T. L. Heise
156Acid Base Properties of Salt Soln’sSample exercise: In each of following, which salt will form the more acidic solution.
B) KBr, KBrO
CHE116Prof. T. L. Heise
157
Acid-Base Behavior & Chem Structure
How does the chemical structure determine which of the behaviors will be exhibited?
Acidica molecule containing H will transfer a proton only if the H-X bond is polarized like
H -- X the stronger the bond the weaker the acid and vice versa
CHE116Prof. T. L. Heise
158
Acid-Base Behavior & Chem Structure
How does the chemical structure determine which of the behaviors will be exhibited?
Acidicoxyacids exist when the H is attached to an oxygen bonded to a central atom if
the OH’s attached are equal in number to the O’s present, acid strength increases with electronegativity
CHE116Prof. T. L. Heise
159
Acid-Base Behavior & Chem Structure
How does the chemical structure determine which of the behaviors will be exhibited?
Acidicoxyacids exist when the H is attached to an oxygen bonded to a central atom
the more O’s present compared to the OH’s, the more polarized the OH bond becomes and the stronger the acid is
CHE116Prof. T. L. Heise
160
Acid-Base Behavior & Chem Structure
How does the chemical structure determine which of the behaviors will be exhibited?
Acidiccarboxylic acids exist when the functional group COOH is present the
strength also increases as the number of electronegative atoms in the molecule increase