Chapter 18 – Acid-Base Equilibria

1
Chapter 18 Acid-Base Equilibria

• 18.1 Acids and Bases in Water
• 18.2 Autoionization of Water and the pH Scale
• 18.3 Proton Transfer and the Brønsted-Lowry
  Acid-Base Definition
• 18.4 Solving Problems Involving Weak-Acid
  Equilibria
• 18.5 Weak Bases and Their Relation to Weak
  Acids
• 18.6 Molecular Properties and Acid Strength
• 18.7 Acid-Base Properties of Salt Solutions
• 18.8 Generalizing the Brønsted-Lowry Concept
  The Leveling Effect
• 18.9 Electron-Pair Donation and the Lewis
  Acid-Base Definition

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18.1 Acids and Bases in Water
3
18.1 Acids and Bases in Water
Zn with 1M HCl(aq) vs. 1M CH3COOH(aq)
4
18.1 Acids and Bases in Water
5
18.1 Acids and Bases in Water
SAMPLE PROBLEM 18.1
Classifying Acid and Base Strength from the
Chemical Formula
(a) H2SeO4
(b) (CH3)2CHCOOH
(c) KOH
(d) (CH3)2CHNH2
SOLUTION
6
18.2 Autoionization of Water and the pH Scale


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18.2 Autoionization of Water and the pH Scale
The Ion-Product Constant for Water, Kw
KcH2O2
H3OOH-
Kw
1.0 x 10-14 at 25 ?C
A change in H3O causes an inverse change in
OH-, and vice versa.
in an acidic solution, H3O gt OH-
in a basic solution, H3O lt OH-
in a neutral solution, H3O OH-
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18.2 Autoionization of Water and the pH Scale
H3O
OH-
ACIDIC SOLUTION
BASIC SOLUTION
NEUTRAL SOLUTION
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18.2 Autoionization of Water and the pH Scale
SAMPLE PROBLEM 18.2
Calculating H3O and OH- in an Aqueous
Solution
PROBLEM
A research chemist adds a measured amount of HCl
gas to pure water at 25 C and obtains a solution
with H3O 3.0x10-4M. Calculate OH-. Is
the solution neutral, acidic, or basic?
PLAN
Use the Kw at 250C and the H3O to find the
corresponding OH-.
SOLUTION
Kw 1.0x10-14 H3O OH- so
OH- Kw/ H3O 1.0x10-14/3.0x10-4
3.3x10-11M
H3O is gt OH- and the solution is acidic.
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18.2 Autoionization of Water and the pH Scale

• pH -log H3O
• of sig figs
• pH of a neutral soln 7.00
• pH of an acidic soln lt 7.00
• pH of a basic soln gt 7.00
• 1 pH unit 10x change
• H3O 10-pH
• p-scales
• pOH -log OH-
• pK -log K

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18.2 Autoionization of Water and the pH Scale
Table 18.3 The Relationship Between Ka and pKa
Acid Name (Formula)
Ka at 25 ?C
pKa
1.02x10-2
Hydrogen sulfate ion (HSO4-)
1.991
3.15
7.1x10-4
Nitrous acid (HNO2)
4.74
1.8x10-5
Acetic acid (CH3COOH)
8.64
2.3x10-9
Hypobromous acid (HBrO)
1.0x10-10
Phenol (C6H5OH)
10.00
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18.2 Autoionization of Water and the pH Scale
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18.2 Autoionization of Water and the pH Scale
SAMPLE PROBLEM 18.3
Calculating H3O, pH, OH-, and pOH
PROBLEM
In an art restoration project, a conservator
prepares copper-plate etching solutions by
diluting concentrated HNO3 to 2.0M, 0.30M, and
0.0063M HNO3. Calculate H3O, pH, OH-, and
pOH of the three solutions at 25 C.
SOLUTION
For 2.0M HNO3, H3O 2.0M and -log H3O
-0.30 pH
OH- Kw/ H3O 1.0x10-14/2.0 5.0x10-15M
pOH 14.30
For 0.3M HNO3, H3O 0.30M and -log H3O
0.52 pH
OH- Kw/ H3O 1.0x10-14/0.30 3.3x10-14M
pOH 13.48
For 0.0063M HNO3, H3O 0.0063M and -log
H3O 2.20 pH
OH- Kw/ H3O 1.0x10-14/6.3x10-3
1.6x10-12M pOH 11.80
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18.3 Proton Transfer and the Brønsted-Lowry
Acid-Base Definition
(acid, H donor)
(base, H acceptor)
(base, H acceptor)
(acid, H donor)
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18.3 Proton Transfer and the Brønsted-Lowry
Acid-Base Definition
Table 18.4 The Conjugate Pairs in Some
Acid-Base Reactions
Conjugate Pair
Conjugate Pair
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18.3 Proton Transfer and the Brønsted-Lowry
Acid-Base Definition
SAMPLE PROBLEM 18.4
Identifying Conjugate Acid-Base Pairs
conjugate pair2
conjugate pair1
SOLUTION
proton donor
proton acceptor
proton acceptor
proton donor
conjugate pair2
conjugate pair1
proton donor
proton acceptor
proton acceptor
proton donor
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18.3 Proton Transfer and the Brønsted-Lowry
Acid-Base Definition
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18.3 Proton Transfer and the Brønsted-Lowry
Acid-Base Definition
SAMPLE PROBLEM 18.5
Predicting the Net Direction of an Acid-Base
Reaction
SOLUTION
Net direction is to the right with Kc gt 1.
Net direction is to the left with Kc lt 1.
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18.4 Solving Problems Involving Weak-Acid
Equilibria
SAMPLE PROBLEM 18.6
Finding the Ka of a Weak Acid from the pH of Its
Solution
With a pH of 2.62, the H3OHPAc gtgt H3Owater.
Assumptions
PAc- H3O since HPAc is weak,
HPAcinitial HPAcinitial -
HPAcdissociation
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18.4 Solving Problems Involving Weak-Acid
Equilibria
SAMPLE PROBLEM 18.6
Finding the Ka of a Weak Acid from the pH of Its
Solution
continued
Concentration(M)
HPAc(aq) H2O(l) H3O(aq)
PAc-(aq)
H3O 10-pH 2.4x10-3 M which is gtgt 10-7 (the
H3O from water)
x 2.4x10-3 M H3O PAc-
HPAcequilibrium 0.12-x 0.12 M
So Ka
4.8 x 10-5
Be sure to check for error.
4x10-3
2.0
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18.4 Solving Problems Involving Weak-Acid
Equilibria
SAMPLE PROBLEM 18.7
Determining Concentrations from Ka and Initial
HA
PROBLEM
Propanoic acid (CH3CH2COOH, which we simplify and
HPr) is an organic acid whose salts are used to
retard mold growth in foods. What is the H3O
of 0.10M HPr (Ka 1.3x10-5)?
x HPrdiss H3Ofrom HPr Pr-
SOLUTION
Concentration(M)
Since Ka is small, we will assume that x ltlt 0.10
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18.4 Solving Problems Involving Weak-Acid
Equilibria
SAMPLE PROBLEM 18.7
Determining Concentrations from Ka and Initial
HA
continued
1.1x10-3 M H3O
Check HPrdiss 1.1x10-3M/0.10 M x 100 1.1
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18.4 Solving Problems Involving Weak-Acid
Equilibria
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18.4 Solving Problems Involving Weak-Acid
Equilibria
Polyprotic acids
acids with more than more ionizable proton
7.2x10-3
6.3x10-8
4.2x10-13
Ka1 gt Ka2 gt Ka3
25
18.4 Solving Problems Involving Weak-Acid
Equilibria
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18.4 Solving Problems Involving Weak-Acid
Equilibria
SAMPLE PROBLEM 18.8
Calculating Equilibrium Concentrations for a
Polyprotic Acid
PROBLEM
Ascorbic acid (H2C6H6O6 H2Asc for this
problem), known as vitamin C, is a diprotic acid
(Ka1 1.0x10-5 and Ka2 5x10-12) found in
citrus fruit. Calculate H2Asc, HAsc-,
Asc2-, and the pH of 0.050M H2Asc.
Ka1 gtgt Ka2 so the first dissociation produces
virtually all of the H3O.
Ka1 is small so H2Ascinitial H2Ascdiss
After finding the concentrations of various
species for the first dissociation, we can use
them as initial concentrations for the second
dissociation.
SOLUTION
1.0x10-5
5x10-12
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18.4 Solving Problems Involving Weak-Acid
Equilibria
continued
Concentration(M)
Ka1 HAsc-H3O/H2Asc 1.0x10-5
(x)(x)/0.050 M
pH -log(7.1x10-4) 3.15
x 7.1x10-4 M
6x10-8 M
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18.5 Weak Bases and Their Relation to Weak Acids
Kb
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18.5 Weak Bases and Their Relation to Weak Acids
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18.5 Weak Bases and Their Relation to Weak Acids
SAMPLE PROBLEM 18.9
Determining pH from Kb and Initial B
Assumptions
Kb gtgt Kw so OH-from water is neglible
(CH3)2NH2 OH- x (CH3)2NH2 - x
(CH3)2NHinitial
SOLUTION
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18.5 Weak Bases and Their Relation to Weak Acids
x 3.0x10-2M OH-
Check assumption
3.0x10-2M/1.5M x 100 2
H3O Kw/OH- 1.0x10-14/3.0x10-2
3.3x10-13M
pH -log 3.3x10-13 12.48
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18.5 Weak Bases and Their Relation to Weak Acids
SAMPLE PROBLEM 18.10
Determining the pH of a Solution of A-
Write the association equation for acetic acid
use the Ka to find the Kb.
SOLUTION
5.6x10-10M
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18.5 Weak Bases and Their Relation to Weak Acids
Ac- 0.25M-x 0.25M
5.6x10-10 x2/0.25M
x 1.2x10-5M OH-
pH -log 8.3x10-10M 9.08
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18.6 Molecular Properties and Acid Strength
Figure 18.12 The effect of atomic and molecular
properties on nonmetal hydride acidity.
Electronegativity increases, acidity increases
Bond strength decreases, acidity increases
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18.6 Molecular Properties and Acid Strength
The relative strengths of oxoacids.
Figure 18.13
ltlt
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18.6 Molecular Properties and Acid Strength
Table 18.7 Ka Values of Some Hydrated Metal
Ions at 250C
Free Ion
Hydrated Ion
Ka
37
18.6 Molecular Properties and Acid Strength
Figure 18.13
The acidic behavior of the hydrated Al3 ion.
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18.7 Acid-Base Properties of Salt Solutions
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18.7 Acid-Base Properties of Salt Solutions
SAMPLE PROBLEM 18.11
Predicting Relative Acidity of Salt Solutions
(a) Potassium perchlorate, KClO4
(b) Sodium benzoate, C6H5COONa
(c) Chromium trichloride, CrCl3
(d) Sodium hydrogen sulfate, NaHSO4
SOLUTION
(a) The ions are K and ClO4- , both of which
come from a strong base(KOH) and a strong
acid(HClO4). Therefore the solution will be
neutral.
(b) Na comes from the strong base NaOH while
C6H5COO- is the anion of a weak organic acid.
The salt solution will be basic.
(c) Cr3 is a small cation with a large
charge, so its hydrated form will react with
water to produce H3O. Cl- comes from the strong
acid HCl. Acidic solution.
(d) Na comes from a strong base. HSO4- can
react with water to form H3O. So the salt
solution will be acidic.
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18.7 Acid-Base Properties of Salt Solutions
SAMPLE PROBLEM 18.12
Predicting the Relative Acidity of Salt
Solutions from Ka and Kb of the Ions
SOLUTION
Ka Zn(H2O)62 1x10-9
Ka HCOO- 1.8x10-4 Kb Kw/Ka
1.0x10-14/1.8x10-4 5.6x10-11
Ka for Zn(H2O)62 gtgtgt Kb HCOO-, therefore the
solution is acidic.
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18.9 Electron-Pair Donation and the Lewis
Acid-Base Definition
An acid is an electron-pair acceptor.
A base is an electron-pair donor.
acid
base
adduct
M(H2O)42(aq)
adduct
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18.9 Electron-Pair Donation and the Lewis
Acid-Base Definition
Figure 18.15
The Mg2 ion as a Lewis acid in the chlorophyll
molecule.
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18.9 Electron-Pair Donation and the Lewis
Acid-Base Definition
SAMPLE PROBLEM 18.13
Identifying Lewis Acids and Bases
SOLUTION
acceptor
donor
donor
acceptor
acceptor
donor