Chapter 14
Acid and Base
Equilibria
pH of Weak Acids
Chapter 14
Table of Contents
14.5
14.6
14.7
14.8
Calculating the pH of Weak Acid Solutions
Bases
Polyprotic Acids
Acid–Base Properties of Salts
14.9
14.10
14.11
14.12
The Effect of Structure on Acid–Base Properties
Acid–Base Properties of Oxides
The Lewis Acid–Base Model
Strategy for Solving Acid–Base Problems: A Summary
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2
Section 14.2
Atomic
Acid
Strength
Masses
Objectives
1. To review how to calculate pH and pOH
for STRONG acids
2. To understand how equilibrium relates to
acid and base theory
3. To rank the strength of bases related to
the strength of their conjugate acids
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Section 14.2
Atomic
Acid
Strength
Masses
Exercise
Calculate the pH for each of the following
solutions.
a) 1.0 × 10–4 M H+
pH = 4.00
b) 0.040 M OH–
pH = 12.60
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4
Section 14.2
Atomic
Acid
Strength
Masses
Exercise
Calculate the pOH for each of the following
solutions.
a) 1.0 × 10–4 M H+
pOH = 10.00
b) 0.040 M OH–
pOH = 1.40
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5
Section 14.2
Atomic
Acid
Strength
Masses
Exercise
The pH of a solution is 5.85. What is the
[OH–] for this solution?
[OH–] = 7.1 × 10–9 M
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6
Section 14.2
Atomic
Acid
Strength
Masses
• Strong acid:
 Ionization equilibrium lies far to the right.
 Value of Ka?
 Yields a weak conjugate base
 (low ability to attract H+)
 HCl, H2SO4, HNO3 - recognize as strong
 HA + H2O  H3O+ + A CB (A-) is weak, won’t hold H+, [H+] = [HA]
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Section 14.2
Atomic
Acid
Strength
Masses
• Weak acid:
 Ionization equilibrium lies far to the left.
 Value of Ka?
 Weaker the acid, stronger its conjugate base.
 High ability to attract H+ and hold on to it!
 HCN (Ka = 6.2 x 10–10) Ka means weak acid
 HA + H2O  H3O+ + A HA weak, CB (A-) strong, will hold H+
 Can’t assume [H+] = [HA], ICE table for pH
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Section 14.2
Atomic
Acid
Strength
Masses
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Section 14.2
Atomic
Acid
Strength
Masses
Various Ways to Describe Acid Strength
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Section 14.2
Atomic
Acid
Strength
Masses
Water as an Acid and a Base
• Water is amphoteric:
 Behaves either as an acid or as a base.
• At 25°C:
Kw = [H+][OH–] = 1.0 × 10–14
• No matter what the solution contains, the
product of [H+] and [OH–] must always equal
1.0 × 10–14 at 25°C.
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Section 14.2
Atomic
Acid
Strength
Masses
Three Possible Situations
• [H+] = [OH–]; neutral solution
• [H+] > [OH–]; acidic solution
• [H+] < [OH–]; basic solution
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Section 14.2
Atomic
Acid
Strength
Masses
Self-Ionization of Water
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Section 14.2
Atomic
Acid
Strength
Masses
Concept Check
HA(aq) + H2O(l)
acid
base
H3O+(aq) + A-(aq)
conjugate conjugate
acid
base
What is the equilibrium constant expression
for an acid acting in water?


H3O   A 
K =
HA 
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Section 14.2
Atomic
Acid
Strength
Masses
Concept Check
If the equilibrium lies to the right, the value for
Ka is __________.
large (or >1)
If the equilibrium lies to the left, the value for Ka
is ___________.
small (or <1)
How do you calculate pH for each condition?
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Section 14.2
Atomic
Acid
Strength
Masses
Concept Check
HA(aq) + H2O(l)
H3O+(aq) + A–(aq)
If water is a better base than A–, do products
or reactants dominate at equilibrium?
Does this mean HA is a strong or weak acid?
Is the value for Ka greater or less than 1?
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Section 14.2
Atomic
Acid
Strength
Masses
Rank Acid Strength Relative to KA
• To page 628 ka values
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Section 14.2
Atomic
Acid
Strength
Masses
Concept Check
Consider a 1.0 M solution of HCl.
Order the following from strongest to weakest base
and explain:
A–(aq) (from weak acid HA)
Cl–(aq)
H2O…………………..next slide
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Section 14.2
Atomic
Acid
Strength
Masses
Let’s Think About It…
A–(aq) (from weak acid HA)
Cl–(aq)
H2O
• Is A–(aq) a good base? (From weak acid HA)
• How good is Cl–(aq) as a base? HCl – strong or
weak?..............VERY WEAK! From Very Strong Acid.
• Water? Weak Base, but stronger than ClThe bases from strongest to weakest are:
A–, H2O, Cl–
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19
Section 14.2
Atomic
Acid
Strength
Masses
Concept Check
Acetic acid (HC2H3O2) and HCN are both weak
acids. Acetic acid is a stronger acid than HCN.
Arrange these bases from weakest to strongest
and explain your answer:
H2O
Cl–
CN–
C2H3O2–
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Section 14.2
Atomic
Acid
Strength
Masses
Let’s Think About It…
•
Cl–
CN–
C2H3O2–
The bases from weakest to strongest are:
Cl–, H2O, C2H3O2–, CN–
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Section 14.3
The pH
Mole
Scale
Objectives Review
1. To review how to calculate pH and pOH
for STRONG acids
2. To understand how equilibrium relates to
acid and base theory
3. To rank the strength of bases related to
the strength of their conjugate acids
4. Work Session: page 674 # 33, 35
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Section 14.5
Calculating the pH of Weak Acid Solutions
Objectives
• To realize what we were missing when we
calculated pH of strong acids
• To use the ICE table and equilibrium to
calculate pH of weak acids solutions
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Section 14.5
Calculating the pH of Weak Acid Solutions
Concept Check
Calculate the pH of a 1.5 x 10–2 M solution of
HNO3.
Name HNO3
Is it a strong or weak acid?
pH = 1.8
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Section 14.5
Calculating the pH of Weak Acid Solutions
Calculate the pH of a 1.5 x 10–2 M solution of HNO3.
• When HNO3 is added to water, write the reaction that
takes place immediately (dissociation of the acid):
HNO3 + H2O  H3O+ + NO3–
• Why is this reaction not likely?
NO3–(aq) + H2O(l)
HNO3(aq) + OH–(aq)
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Section 14.5
Calculating the pH of Weak Acid Solutions
• What else is happening in aq?
• H2O(l) + H2O(l)
H3O+(aq) + OH–(aq)
• In aqueous solutions, this reaction is always taking
place.
• But is water the major contributor of H+ (H3O+)?
• Which reaction controls the pH?
• HNO3 + H2O  H3O+ + NO3– or
• H2O(l) + H2O(l)
H3O+(aq) + OH–(aq)
• We assumed that [H+] = M of sol’n for strong acids
• We assumed that the autoionization of water was a
small effect of the overall H+ (aq) for strong acids.
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Section 14.5
Calculating the pH of Weak Acid Solutions
Solving Weak Acid Equilibrium Problems
1. List the major species in the solution.
2. Choose the species that can produce H+, and
write balanced equations for the reactions
producing H+.
3. Using the values of the equilibrium constants for
the reactions you have written, decide which
equilibrium will dominate in producing H+.
4. Write the equilibrium expression for the
dominant equilibrium.
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Section 14.5
Calculating the pH of Weak Acid Solutions
Solving Weak Acid Equilibrium Problems
5. List the initial concentrations of the species
participating in the dominant equilibrium.
6. Define the change needed to achieve
equilibrium; that is, define x.
7. Write the equilibrium concentrations in terms
of x.
8. Substitute the equilibrium concentrations into
the equilibrium expression.
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Section 14.5
Calculating the pH of Weak Acid Solutions
Solving Weak Acid Equilibrium Problems
9. Solve for x the “easy” way, that is, by
assuming that [HA]0 – x about equals [HA]0.
10.Use the 5% rule to verify whether the
approximation is valid.
11.Calculate [H+] and pH.
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Section 14.5
Calculating the pH of Weak Acid Solutions
Solving Weak Acid Equilibrium Problems
1. List the major species in the solution that can
produce H+, and write balanced equations for
the reactions producing H+.
2. Using the values of the equilibrium constants for
the reactions you have written, decide which
equilibrium will dominate in producing H+.
(bigger K)
3. Write the equilibrium expression for the
dominant equilibrium and use the ICE table to
calculate pH of weak acids
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Section 14.5
Calculating the pH of Weak Acid Solutions
Concept Check
Calculate the pH of a 0.80 M aqueous solution
of the acid HCN (Ka = 6.2 x 10–10).
Is this a strong or weak acid? Write Keq = Ka =
What are the major species in solution?
HCN, H2O
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31
Section 14.5
Calculating the pH of Weak Acid Solutions
Let’s Think About It…
• Why aren’t H+ or CN– major species?
• Write the balanced equations for the reactions
producing H+.
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Section 14.5
Calculating the pH of Weak Acid Solutions
Consider This
HCN(aq) + H2O(l)
H3O+(aq) + CN–(aq)
Ka = 6.2 x 10-10
H3O+(aq) + OH–(aq)
Kw = ?
Kw = 1.0 x 10-14
H2O(l) + H2O(l)
• Which reaction controls the pH? Explain.
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Section 14.5
Calculating the pH of Weak Acid Solutions
HCN(aq) + H2O(l)
H3O+(aq) + CN–(aq)
Ka = 6.2 x 10-10
Original Question?
Calculate the pH of a 0.80 M aqueous solution of
the acid HCN (Ka = 6.2 x 10–10).
I
C
-----------------------------------------------------------------E
[H+] = 2.2 X10-5
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pH = 4.7
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34
Section 14.5
Calculating the pH of Weak Acid Solutions
HCN(aq) + H2O(l)
H3O+(aq) + CN–(aq)
Ka = 6.2 x 10-10
Original Question?
Calculate the pH of a 0.80 M aqueous solution of
the acid HCN (Ka = 6.2 x 10–10).
I
C
-----------------------------------------------------------------E
[H+] = 2.2 X10-5
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pH = 4.7
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35
Section 14.5
Calculating the pH of Weak Acid Solutions
Exercise
Calculate the pH of a 0.50 M aqueous solution
of the weak acid HF.
(Ka = 7.2 x 10–4)
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Section 14.5
Calculating the pH of Weak Acid Solutions
Let’s Think About It…
• What are the major species in solution?
HF, H2O
• Why aren’t H+ and F– major species?
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Section 14.5
Calculating the pH of Weak Acid Solutions
Let’s Think About It…
• What are the possibilities for the dominant
reaction?
HF(aq) + H2O(l)
H3O+(aq) + F–(aq)
Ka=7.2 x 10-4
H2O(l) + H2O(l)
H3O+(aq) + OH–(aq)
Kw=1.0 x 10-14
• Which reaction controls the pH? Why?
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Section 14.5
Calculating the pH of Weak Acid Solutions
Steps Toward Solving for pH
H3O+(aq) +
F–(aq)
0.50 M
~0
~0
–x
+x
+x
0.50–x
x
x
HF(aq) + H2O
Initial
Change
Equilibrium
Ka = 7.2 x 10–4
pH = 1.72
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Section 14.5
Calculating the pH of Weak Acid Solutions
Objectives
• To realize what we were missing when we
calculated pH of strong acids
• To use the ICE table and equilibrium to
calculate pH of weak acids solutions
• Work Session: Page 674 #56, #55 (CH3COOH
= Acetic Acid), 58, 59, 60, 61, 69, 70
• pH = -log[H+]
must find [H+] first through ICE
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Section 14.5
Calculating the pH of Weak Acid Solutions
Objectives
• To calculate percent dissociation of weak acids
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Section 14.5
Calculating the pH of Weak Acid Solutions
Percent Dissociation (Ionization)
Percent dissociation =
amount dissociated (mol/L)
 100%
initial concentration (mol/L)
• For a given weak acid, the percent dissociation
increases as the acid becomes more dilute.
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Section 14.5
Calculating the pH of Weak Acid Solutions
Exercise
A solution of 8.00 M formic acid (HCHO2) is 0.47%
ionized in water.
Calculate the Ka value for formic acid. ICE Table
Ka = 1.8 x 10–4
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Section 14.5
Calculating the pH of Weak Acid Solutions
Exercise
Calculate the pH of an 8.00 M solution of formic
acid. Use the data from the previous slide to
help you solve this problem.
pH = 1.42
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Section 14.5
Calculating the pH of Weak Acid Solutions
Concept Check
The value of Ka for a 4.00 M formic acid
solution should be:
higher than
lower than
the same as
the value of Ka of an 8.00 M formic acid
solution.
Explain.
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45
Section 14.5
Calculating the pH of Weak Acid Solutions
Concept Check
The percent ionization of a 4.00 M formic acid
solution should be:
higher than
lower than
the same as
the percent ionization of an 8.00 M formic acid
solution.
Write the ionization reaction of formic acid and
explain in terms of equilibrium shift.
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46
Section 14.5
Calculating the pH of Weak Acid Solutions
Concept Check
The pH of a 4.00 M formic acid solution should
be:
higher than
lower than
the same as
the pH of an 8.00 M formic acid solution.
Explain. Calculate…
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Section 14.5
Calculating the pH of Weak Acid Solutions
Exercise
Calculate the percent ionization of a 4.00 M
formic acid solution in water. Same Ka…
% Ionization = 0.67%
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Section 14.5
Calculating the pH of Weak Acid Solutions
Exercise
Calculate the pH of a 4.00 M solution of formic
acid.
pH = 1.57
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Section 14.5
Calculating the pH of Weak Acid Solutions
Objectives Review
• To calculate percent dissociation of weak acids
• Work Session: 63, 64, 65, 66, 67, 69, 70
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Section 14.6
Bases
Objectives
• To calculate pH of strong and weak bases
• To name a few bases that aren’t hydroxides (OH-)
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Section 14.6
Bases
Arrhenius: bases produce OH– ions.
Brønsted–Lowry: bases are proton acceptors.
In a basic solution at 25°C, pH > 7.
Ionic compounds containing OH are generally
considered strong bases.
 LiOH, NaOH, KOH, Ca(OH)2
• pOH = –log[OH–]
• pH = 14.00 – pOH
•
•
•
•
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Section 14.6
Bases
Concept Check
Calculate the pH of a 1.0 x 10–3 M solution of
sodium hydroxide.
pH = 11.00
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Section 14.6
Bases
Concept Check
Calculate the pH of a 1.0 x 10–3 M solution of
calcium hydroxide.
pH = 11.30
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Section 14.6
Bases
• Equilibrium expression for weak bases uses Kb.
CN–(aq) + H2O(l)
Kb =
HCN(aq) + OH–(aq)



HCN
OH


CN 
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Section 14.6
Bases
• pH calculations for solutions of weak bases are
very similar to those for weak acids.
• Kw = [H+][OH–] = 1.0 × 10–14
• pOH = –log[OH–]
• pH = 14.00 – pOH
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Section 14.6
Bases
Concept Check
Calculate the pH of a 2.0 M solution of the
weak base ammonia (NH3) .
(Kb = 1.8 × 10–5)
pH = 11.78
How about a 4.0 M solution?
8.0 M solution?
12.0 M solution?
11.93, 12.08, 12.17
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Section 14.6
Bases
Objectives Review
• To calculate pH of strong and weak bases
• To name a few bases that aren’t hydroxides (OH-)
• Work Session: 71, 73, 75, 77, 79, 81, 83, 85
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Section 14.7
Polyprotic Acids
Objectives
• To calculate pH of polyprotic acids
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Section 14.7
Polyprotic Acids
• Acids that can furnish more than one proton.
• Always dissociates in a stepwise manner, one
proton at a time.
• The conjugate base of the first dissociation
equilibrium becomes the acid in the second
step.
• For a typical weak polyprotic acid like H3PO4 :
Ka1 > Ka2 > Ka3
• For a typical polyprotic acid in water, only the
first dissociation step is important to pH.
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Section 14.7
Polyprotic Acids
Exercise
Calculate the pH of a 1.00 M solution of H3PO4.
Ka1 = 7.5 x 10-3
Ka2 = 6.2 x 10-8
Ka3 = 4.8 x 10-13
pH = 1.08
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Section 14.7
Polyprotic Acids
Concept Check
Calculate the equilibrium concentration of PO43in a 1.00 M solution of H3PO4.
Ka1 = 7.5 x 10-3
Ka2 = 6.2 x 10-8
Ka3 = 4.8 x 10-13
[PO43-] = 3.6 x 10-19 M
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Section 14.7
Polyprotic Acids
Objectives Review
• To calculate pH of polyprotic acids
• Work Session: 93, 95, 97
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Section 14.8
Acid–Base Properties of Salts
Salts
• Ionic compounds.
• When dissolved in water, break up into its ions
(which can behave as acids or bases).
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Section 14.8
Acid–Base Properties of Salts
Salts
• The salt of a strong acid and a strong base
gives a neutral solution.
 KCl, NaNO3
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Section 14.8
Acid–Base Properties of Salts
Salts
• A basic solution is formed if the anion of the salt
is the conjugate base of a weak acid.
 NaF, KC2H3O2
 Kw = Ka × Kb
 Use Kb when starting with base.
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Section 14.8
Acid–Base Properties of Salts
Salts
• An acidic solution is formed if the cation of the
salt is the conjugate acid of a weak base.
 NH4Cl
 Kw = Ka × Kb
 Use Ka when starting with acid.
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Section 14.8
Acid–Base Properties of Salts
Cation
neutral
neutral
Anion
neutral
conjugate
base of
weak acid
neutral
Acidic
or Basic
neutral
basic
conjugate
acidic
acid of
weak base
conjugate conjugate depends
acid of
base of on Ka & Kb
weak base weak acid
values
Example
NaCl
NaF
NH4Cl
Al2(SO4)3
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Section 14.8
Acid–Base Properties of Salts
Qualitative Prediction of pH of Salt Solutions (from Weak Parents)
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Section 14.8
Acid–Base Properties of Salts
Exercise
HC2H3O2
HCN
Ka = 1.8 x 10-5
Ka = 6.2 x 10-10
Calculate the Kb values for: C2H3O2− and CN−
Kb (C2H3O2-) = 5.6 x 10-10
Kb (CN-) = 1.6 x 10-5
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Section 14.8
Acid–Base Properties of Salts
Concept Check
Arrange the following 1.0 M solutions from
lowest to highest pH.
HBr
NaCN
NaCl
NaOH
NH3
HF
NH4Cl
HCN
Justify your answer.
HBr, HF, HCN, NH4Cl, NaCl, NaCN, NH3, NaOH
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71
Section 14.8
Acid–Base Properties of Salts
Concept Check
Consider a 0.30 M solution of NaF. The Ka for
HF is 7.2 x 10-4.
What are the major species?
Na+, F-, H2O
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Section 14.8
Acid–Base Properties of Salts
Let’s Think About It…
• Why isn’t NaF considered a major species?
• What are the possibilities for the dominant
reactions?
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Section 14.8
Acid–Base Properties of Salts
Let’s Think About It…
The possibilities for the dominant reactions are:
1.
2.
3.
4.
F–(aq) + H2O(l)
H2O(l) + H2O(l)
Na+(aq) + H2O(l)
Na+(aq) + F–(aq)
HF(aq) + OH–(aq)
H3O+(aq) + OH–(aq)
NaOH + H+(aq)
NaF
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Section 14.8
Acid–Base Properties of Salts
Let’s Think About It…
• How do we decide which reaction controls the
pH?
F–(aq) + H2O(l)
H2O(l) + H2O(l)
HF(aq) + OH–(aq)
H3O+(aq) + OH–(aq)
• Determine the equilibrium constant for each
reaction.
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Section 14.8
Acid–Base Properties of Salts
Exercise
Calculate the pH of a 0.75 M aqueous solution
of NaCN.
Ka for HCN is 6.2 x 10–10.
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Section 14.8
Acid–Base Properties of Salts
Let’s Think About It…
• What are the major species in solution?
Na+, CN–, H2O
• Why isn’t NaCN considered a major species?
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Section 14.8
Acid–Base Properties of Salts
Let’s Think About It…
• What are all possibilities for the dominant
reaction?
• The possibilities for the dominant reaction are:
1. CN–(aq) + H2O(l)
HCN(aq) + OH–(aq)
2. H2O(l) + H2O(l)
H3O+(aq) + OH–(aq)
3. Na+(aq) + H2O(l)
NaOH + H+(aq)
4. Na+(aq) + CN–(aq)
NaCN
• Which of these reactions really occur?
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Section 14.8
Acid–Base Properties of Salts
Let’s Think About It…
• How do we decide which reaction controls the
pH?
CN–(aq) + H2O(l)
H2O(l) + H2O(l)
HCN(aq) + OH–(aq)
H3O+(aq) + OH–(aq)
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Section 14.8
Acid–Base Properties of Salts
Steps Toward Solving for pH
CN–(aq) + H2O
Initial
0.75 M
0
~0
–x
+x
+x
0.75–x
x
x
Change
Equilibrium
HCN(aq) + OH–(aq)
Kb = 1.6 x 10–5
pH = 11.54
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80
Section 14.9
The Effect of Structure on Acid–Base Properties
Models of Acids and Bases
• Two factors for acidity in binary compounds:
 Bond Polarity (high is good)
 Bond Strength (low is good)
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81
Section 14.9
The Effect of Structure on Acid–Base Properties
Bond Strengths and Acid Strengths for Hydrogen Halides
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Section 14.9
The Effect of Structure on Acid–Base Properties
Oxyacids
• Contains the group H–O–X.
• For a given series the acid strength increases
with an increase in the number of oxygen atoms
attached to the central atom.
• The greater the ability of X to draw electrons
toward itself, the greater the acidity of the
molecule.
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83
Section 14.9
The Effect of Structure on Acid–Base Properties
Several Series of Oxyacids and Their Ka Values
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84
Section 14.9
The Effect of Structure on Acid–Base Properties
Comparison of Electronegativity of X and Ka Value
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85
Section 14.10
Acid–Base Properties of Oxides
Oxides
• Acidic Oxides (Acid Anhydrides):
 O X bond is strong and covalent.
SO2, NO2, CO2
• When H O X grouping is dissolved in water,
the O X bond will remain intact. It will be the
polar and relatively weak H O bond that will
tend to break, releasing a proton.
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86
Section 14.10
Acid–Base Properties of Oxides
Oxides
• Basic Oxides (Basic Anhydrides):
 O X bond is ionic.
K2O, CaO
• If X has a very low electronegativity, the O X
bond will be ionic and subject to being broken in
polar water, producing a basic solution.
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87
Section 14.11
The Lewis Acid–Base Model
Lewis Acids and Bases
• Lewis acid: electron pair acceptor
• Lewis base: electron pair donor
3+
Al3+ + 6 O
Al
H
Lewis acid
H
H
O
H
6
Lewis base
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88
Section 14.11
The Lewis Acid–Base Model
Three Models for Acids and Bases
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89
Section 14.2
Atomic
Acid
Strength
Masses
Concept Check SKIP>>>>1,2,3
Consider a solution of NaA where A– is the
anion from weak acid HA:
A–(aq) + H2O(l)
base
acid
HA(aq)
conjugate
acid
+
OH–(aq)
conjugate
base
a) Which way will equilibrium lie? Write Ka…
left
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90
Section 14.2
Atomic
Acid
Strength
Masses
Concept Check SKIP>>>>
Consider a solution of NaA where A– is the
anion from weak acid HA:
A–(aq) + H2O(l)
base
acid
HA(aq) +
OH–(aq)
conjugate
conjugate
acid
base
b) Is the value for Kb greater than or less
than 1? SKIP>>>>>
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91
Section 14.2
Atomic
Acid
Strength
Masses
Concept Check SKIP>>>>>>>
Consider a solution of NaA where A– is the anion from
weak acid HA:
A–(aq) + H2O(l)
base
acid
HA(aq) +
conjugate
acid
OH–(aq)
conjugate
base
c) Does this mean A– is a strong or weak base?
weak base
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92
Section 14.2
Atomic
Acid
Strength
Masses
Concept Check
Discuss whether the value of K for the reaction:
HCN(aq) + F–(aq)
CN–(aq) + HF(aq)
is 1
1
=1
(Ka for HCN is 6.2×10–10; Ka for HF is 7.2×10–4.)
Explain your answer.
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93
Section 14.2
Atomic
Acid
Strength
Masses
Concept Check
Calculate the value for K for the reaction:
HCN(aq) + F–(aq)
CN–(aq) + HF(aq)
(Ka for HCN is 6.2×10–10; Ka for HF is 7.2×10–4.)
K = 8.6 x 10–7
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94