Chapter 14
Chapter 14
Table of Contents
14.1The Nature of Acids and Bases 14.2Acid Strength
14.3The pH Scale
14.4Calculating the pH of Strong Acid Solutions
14.5 Calculating the pH of Weak Acid Solutions
14.6 Bases
14.7 Polyprotic Acids
14.8 Acid–Base Properties of Salts
14.9 The Effect of Structure on Acid–Base Properties 14.10 Acid–Base Properties of Oxides
14.11 The Lewis Acid–Base Model
Section 14.1
The Nature of Acids and Bases
Models of Acids and Bases
• Arrhenius: Acids produce H+ ions in solution,
bases produce OH- ions.
• Brønsted–Lowry: Acids are proton (H+) donors,
bases are proton acceptors:
HCl + H2O Cl- + H
3O+
Section 14.1
The Nature of Acids and Bases
Brønsted–Lowry Reaction
acid base conjugate conjugate
acid base
Section 14.1
The Nature of Acids and Bases
Brønsted–Lowry Reaction
acid base conjugate conjugate
acid base
Section 14.2
Atomic Masses Acid Strength
• Strong acid:
Ionization equilibrium lies far to the right.
Yields a weak conjugate base.
(H
2O is a stronger base than A
-.)
Stronger the acid, the weaker its conjugate
base.
• Weak acid:
Section 14.2
Section 14.2
Atomic Masses Acid Strength
Section 14.2
Atomic Masses Acid Strength
Acid dissociation constant, K
a
or [H
+]
[A
-]
[HA]
High Ka = strong acid Write a dissociation equation for C5H6NH3+.
C5H6NH3+ + H
2O C5H6NH2 + H3O+
3
H O
A
Section 14.2
Atomic Masses Acid Strength
Acid dissociation constant, K
a H2SO4 is a strong, diprotic acid.
Oxyacids: The more oxygens in the formula, the stronger the acid.
Section 14.2
Atomic Masses Acid Strength
• Water is amphoteric:
Behaves either as an acid or as a base: 2 H2O(l) H3O+
(aq) + OH-(aq)
Or: H2O(l) H+
(aq) + OH-(aq)
Kw = [H+][OH–] = 1.0 × 10–14 at 25°C.
Section 14.2
Atomic Masses Acid Strength
• [H
+] = [OH
–] = 1.0 × 10
–7M;
neutral
solution
• [H
+] > [OH
–];
acidic
solution
Section 14.2
Atomic Masses Acid Strength
Section 14.2
Atomic Masses Acid Strength
Section 14.2
Atomic Masses Acid Strength
Review
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 ___________.
Section 14.2
Atomic Masses Acid Strength
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?
Section 14.2
Atomic Masses Acid Strength
Concept Check
Consider a 1.0 M solution of HCl.
Order the following from strongest to weakest
base and explain:
H2O(l)
Section 14.2
Atomic Masses Acid Strength
Let’s Think About It…
• How good is Cl–(aq) as a base?
• Is A–(aq) a good base?
The bases from strongest to weakest are:
A–, H
Section 14.2
Atomic Masses Acid Strength
Concept Check
Consider a solution of NaA where A– is the
anion from weak acid HA:
A–(aq) + H
2O(l) HA(aq) + OH–(aq)
base acid conjugate conjugate
Section 14.2
Atomic Masses Acid Strength
Concept Check
Consider a solution of NaA where A– is the
anion from weak acid HA:
A–(aq) + H
2O(l) HA(aq) + OH–
(aq)
base acid conjugate conjugate
acid base
b) Is the value for Kb greater than or less than
Section 14.2
Atomic Masses Acid Strength
Concept Check
Consider a solution of NaA where A– is the anion from
weak acid HA:
A–(aq) + H
2O(l) HA(aq) + OH–(aq)
base acid conjugate conjugate
acid base
Section 14.2
Atomic Masses Acid Strength
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– C
Section 14.2
Atomic Masses Acid Strength
Let’s Think About It…
• H2O(l) + H2O(l) H3O+(aq) + OH–(aq)
acid base conjugate conjugate
acid base
• At 25C, Kw = 1.0 x 10–14
The bases from weakest to strongest are:
Section 14.2
Atomic Masses Acid Strength
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; K
a for HF is 7.2×10–4.)
Section 14.2
Atomic Masses Acid Strength
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; K
a for HF is 7.2×10–4.)
Section 14.3
The Mole
The pH Scale
• pH = –log[H+] (Also: pOH = –
log[OH-] )
• An easy way to represent acidity. • pH decreases as [H+] increases.
• Significant figures: The number of places after the decimal in the log is equal to the number of sig figs in the original number.
Section 14.3
The Mole
The pH Scale
pH Range
• pH = 7; neutral
• pH > 7; basic
Higher pH, more basic. • pH < 7; acidic
Section 14.3
The Mole
The pH Scale
The pH scale and pH values of some
Section 14.3
The Mole
The pH Scale
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–
Section 14.3
The Mole
The pH Scale
Exercise
The pH of a solution is 5.85. What is the [H+]
for this solution?
Section 14.3
The Mole
The pH Scale
pH and pOH
• Recall:
Kw = [H+][OH–]
–log Kw = –log[H+] – log[OH–]
Section 14.3
The Mole
The pH Scale
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–
Section 14.3
The Mole
The pH Scale
Exercise
The pH of a solution is 5.85. What is the [OH–] for this solution?
Section 14.4
Calculating the pH of Strong Acid Solutions
Thinking about acid–base problems
• What are the major species in solution?
• What is the dominant reaction that will take place?
Is it an equilibrium reaction or a reaction that will go essentially to completion?
React all major species until you are left with an equilibrium reaction.
Section 14.4
Calculating the pH of Strong Acid Solutions
Concept check
Consider an aqueous solution of 2.0 x 10–3 M
HCl.
What are the major species in solution?
H+, Cl–, H 2O
What is the pH?
Section 14.4
Calculating the pH of Strong Acid Solutions
Calculate the pH of a 1.5 x 10–11 M solution of
HCl.
pH = 7.00
Section 14.4
Calculating the pH of Strong Acid Solutions
Calculate the pH of a 1.5 x 10–2 M solution of
HNO3.
Section 14.4
Calculating the pH of Strong Acid Solutions
Let’s Think About It…
• When HNO3 is added to water, a reaction takes place immediately:
Section 14.4
Calculating the pH of Strong Acid Solutions
Let’s Think About It…
• Why is this reaction not likely?
NO3–(aq) + H
Section 14.4
Calculating the pH of Strong Acid Solutions
Let’s Think About It…
• What reaction controls the pH?
• 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+ (H
3O+)?
Section 14.4
Calculating the pH of Strong Acid Solutions
End of lesson
Section 14.4
Calculating the pH of Strong Acid Solutions
14.5-6 pH of a weak acid solution
Objective
:•
Students will be able to calculate the
pH and the concentrations of all
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+.
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.
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
Section 14.5
Calculating the pH of Weak Acid Solutions
The easy way to find pH of a weak acid…
Ex. 1 Find the [H3O+], [HCO
3-], [H2CO3], and pH of a
0.20 M solution of H2CO3 (p. A23: Ka = 4.3 x 10-7).
Note: H2CO3 gives off more H+ than H
2O does… So:
H2CO3 + H2O H3O+ + HCO 3
-Initial 0.20 M ~ 0 ~ 0
Change –x +x +x
Section 14.5
Calculating the pH of Weak Acid Solutions
Ka = [H3O+] [HCO 3-]
[H2CO3]
4.3 x 10–7 = x2
H2CO3 + H2O H3O+ + HCO 3
-Initial 0.20 M ~ 0 ~ 0
Change –x +x +x
Equilibrium 0.20–x x x
4.3 x 10–7 = x2
0.20
x = √(4.3 x 10–7)(0.20)
Very small (because Ka is so small) so we ignore x in this case.
Section 14.5
Calculating the pH of Weak Acid Solutions
[HCO
3-] = [H
3
O
+] = 2.9 x 10
-4M
[H
2CO
3] = 0.20 - 2.9 x 10
-4M ≈ 0.20 M
Section 14.5
Calculating the pH of Weak Acid Solutions
Concept Check
Consider a 0.80
M
aqueous solution of the
weak acid HCN (
K
a= 6.2 x 10
–10).
Section 14.5
Calculating the pH of Weak Acid Solutions
Let’s Think About It…
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
H2O(l) + H2O(l) H3O+(aq) + OH–(aq)
Kw = 1.0 x 10-14
Section 14.5
Calculating the pH of Weak Acid Solutions
Exercise
Section 14.5
Calculating the pH of Weak Acid Solutions
Let’s think about it…
• What are the major species in solution?
HF, H
2O
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
Section 14.5
Calculating the pH of Weak Acid Solutions
Solving for pH
Ka = [H3O+] [F–]
[HF]
7.2 x 10–4 = x2
0.20 - x
HF(aq) + H2O H3O+(aq) + F–(aq)
Initial 0.20 M ~ 0 ~ 0
Change –x +x +x
Equilibrium 0.20–x x x
This time we cannot
ignore x. Ka is too large… > 1% is ionized.
7.2 x 10-4 (0.20 – x) = x2
1.44 x 10-4 - 7.2 x 10-4 x - x2 = 0
x2 + 7.2 x 10-4 x - 1.44 x 10-4 = 0
Section 14.5
Calculating the pH of Weak Acid Solutions
Percent dissociation
Ex. 3 Find percent dissociation for a 0.20 M HF solution. (See example 2.)
Note: % dissociation = amount dissoc (M) x 100% initial conc (M)
• % dissociation = amount dissoc (M) x 100% initial conc (M)
Section 14.5
Calculating the pH of Weak Acid Solutions
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.
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.
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.
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.
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.
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.
Section 14.5
Calculating the pH of Weak Acid Solutions
Exercise
Calculate the pH of a 4.00 M solution of formic acid.
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–]
Section 14.6
Bases
Concept Check
Calculate the pH of a 1.0 x 10–3 M solution of
sodium hydroxide.
Section 14.6
Bases
Concept Check
Calculate the pH of a 1.0 x 10–3 M solution of
calcium hydroxide.
Section 14.6
Bases
• Equilibrium expression for weak bases uses Kb.
CN–(aq) + H
2O(l) HCN(aq) + OH–(aq)
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–]
Section 14.6
Bases
Concept Check
Calculate the pH of a 2.0 M solution of ammonia (NH3). (Kb = 1.8 × 10–5)
NH3 + H2O
NH4+ + OH
-Initial 2.0 M ~ 0 ~ 0
Change –x +x +x
Section 14.6
Bases
Kb = [NH4+] [OH-]
[NH3]
1.8 x 10 = x
1.8 x 10–5 = x2
2.0
x = √(1.8 x 10–5)(2.0)
NH3 + H2O NH4+ + OH
-Initial 2.0 M ~ 0 ~ 0
Change –x +x +x
Section 14.6
Bases
[OH-] = 6.0 x 10-3 M
= 2.22 pOH = antilog (6.0 x 10-3 M)
Section 14.6
Bases
End of lesson.
Section 14.7
Polyprotic Acids
• Acids that can furnish more than one proton.
• Always dissociate 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:
K
a1>
K
a2>
K
a3Section 14.7
Polyprotic Acids
Exercise
Calculate the pH of a 1.00 M solution of H3PO4, given:
Ka1 = 7.5 x 10-3
Ka2 = 6.2 x 10-8
Ka3 = 4.8 x 10-13
Section 14.7
Polyprotic Acids
Extremely long, challenging question:
Calculate the equilibrium concentration of PO43- in
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
Section 14.8
Acid–Base Properties of Salts
Salts
•
Ionic compounds, when dissolved in
Section 14.8
Acid–Base Properties of Salts
Salts
•
The salt of a strong acid and a strong
base gives a neutral solution.
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, KC
2H
3O
2
K
b=
K
w/
K
aSection 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.
NH
4Cl
K
a=
K
w/
K
bSection 14.8
Acid–Base Properties of Salts
Cation
Anion
Acidic
or basic
Example
from
strong
base
from strong
acid
neutral
NaCl
from
strong
base
from weak
acid
basic
NaF
from weak
base
from strong
acid
acidic
NH
4Cl
from weak
base
from weak
acid
depends
on
K
a&
K
bSection 14.8
Acid–Base Properties of Salts
Qualitative prediction of pH of salt solutions (from weak parents)
Let’s say a salt “came from” an acid and a base, like NaF “came from” HF and NaOH.
Section 14.8
Acid–Base Properties of Salts
pH of a salt of a weak acid
Ex. 4 Find the pH of a 0.30 M solution of NaClO.
Note: NaClO is not an acid or base, but it is the salt of a weak acid, HClO. A hydrolysis reaction occurs
when it is placed in water, with the salt dissolving in water and leading to changes in [H+] and [OH-].
Step 1: NaClO(s) Na+
Section 14.8
Acid–Base Properties of Salts
Kb = [HClO] [OH–]
[ClO-]
Step 2: ClO- + H
2O HClO + OH–
Initial 0.30 M 0 ~ 0
Change - x + x + x
Equilibrium 0.30 - x x x
2.9 x 10–7 = x2
0.30 - x
Use Kb = 1 x 10-14
Ka for HClO
See p. A22.
Section 14.8
Acid–Base Properties of Salts
2.9 x 10–7 = x2
0.30 - x
x2 = (0.30) (2.9 x 10–7)
x = √(0.30) (2.9 x 10–7)
x = √8.7 x 10–8
x = 2.9 x 10–4
[OH-] = 2.9 x 10–4
Section 14.8
Acid–Base Properties of Salts
Exercise… depending on time, let’s skip ahead to slide #90.
HC2H3O2 Ka = 1.8 x 10-5
HCN Ka = 6.2 x 10-10
Calculate the Kb values for: C2H3O2− and CN−
Section 14.8
Acid–Base Properties of Salts
Concept Check
Arrange the following 1.0 M solutions from lowest to highest pH.
HBr NaOH NH4Cl NaCN NH3 HCN NaCl HF
Justify your answer.
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?
Section 14.8
Acid–Base Properties of Salts
• Why isn’t NaF considered a major species? • What are the possibilities for the dominant
Section 14.8
Acid–Base Properties of Salts
The possibilities for the dominant reactions are:
1. F–(aq) + H
2O(l) HF(aq) + OH–(aq)
2. H2O(l) + H2O(l) H3O+(aq) + OH–(aq)
3. Na+(aq) + H
2O(l) NaOH + H+(aq)
Section 14.8
Acid–Base Properties of Salts
Let’s Think About It…
• How do we decide which reaction controls the pH?
F–(aq) + H
2O(l) HF(aq) + OH–(aq)
H2O(l) + H2O(l) H3O+(aq) + OH–(aq)
Section 14.8
Acid–Base Properties of Salts
Ex. 1
Calculate the pH of a 0.30 M solution of NaC2H3O2.
Section 14.8
Acid–Base Properties of Salts
(Notice it’s like example 4 yesterday…)
1. NaC
2H
3O
2(s)
Na
+(aq)+ C
2H
3O
2-(aq)2. C
2H
3O
2-+ H
2O
HC
2H
3O
2+ OH
-3. K
b= K
w/K
a= 1.00 x 10
-14/1.8 x 10
-5= 5.6 x 10
-10Section 14.8
Acid–Base Properties of Salts
Find the pH and the conc of all species at
equilibrium for a 0.50 M solution of CH3NH3Cl. Kb for CH3NH2 = 4.4 x 10-4
-11
Section 14.9
The Effect of Structure on Acid–Base Properties
Models of acids and bases
• Two factors for acidity in binary compounds:
Section 14.9
The Effect of Structure on Acid–Base Properties
Section 14.9
The Effect of Structure on Acid–Base Properties
Oxyacids
• … contain 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
Section 14.9
The Effect of Structure on Acid–Base Properties
Several series of
oxyacids and
Section 14.9
The Effect of Structure on Acid–Base Properties
Comparison of electronegativity of X and Ka
Section 14.9
The Effect of Structure on Acid–Base Properties
End of lesson
Section 14.10
Acid–Base Properties of Oxides
Oxides
•
Objectives: Students will learn about
acid-base properties of oxides, and use a
Section 14.10
Acid–Base Properties of Oxides
Oxides
• Acidic oxides (acid anhydrides):
• The O - X bond is strong and covalent. (X
is a nonmetal.)
SO
2, NO
2, CO
2Section 14.10
Acid–Base Properties of Oxides
Oxides
• Basic oxides (basic anhydrides):
• O - X bond is ionic. (X is a metal.)
K
2O, CaO
Section 14.11
The Lewis Acid–Base Model
Lewis acids and bases (Not on AP Exam)
• Lewis acid: electron pair
acceptor
• Lewis base: electron pair
donor
O
H
H
O
H
H
Al
Al
3++ 6
6
Section 14.11
The Lewis Acid–Base Model
Section 14.11
The Lewis Acid–Base Model
Review the main steps for solving
acid-base equilibrium problems, p. 666 – 667.
Section 14.11
The Lewis Acid–Base Model
End of lesson
• See homework on board.