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Acids and Bases
Chapter 14
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Objectives
•
•
•
•
•
•
•
•
•
•
Review acids and bases (Chem I & Ch. 4)
Bronstead acids and bases
Define autoionization of water, Kw and pH
Define Keq for acids (Ka) and bases (Kb)
Types of reactions (qualitative)
Calculations w/Keq’s (Ka, Kb and Kw)
Polyprotic acids
Acid/base properties of salts
Lewis acids and bases
Molecular structure and acid strength
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Ch. 14.1 – The Nature of Acids
and Bases
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Acid and Bases
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Acid and Bases
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Some Definitions
• Arrhenius acids and bases
– Acid: Substance that, when dissolved in
water, increases the concentration of hydrogen
ions (protons, H+).
HCl(aq)  H+(aq) + Cl-(aq)
– Base: Substance that, when dissolved in
water, increases the concentration of
hydroxide ions.
NaOH(aq)  Na+(aq) + OH-(aq)
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ACID-BASE THEORIES
• The most general theory for common
aqueous acids and bases is the
BRØNSTED - LOWRY theory
• ACIDS DONATE H+ IONS
• BASES ACCEPT H+ IONS
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ACID-BASE THEORIES
The Brønsted definition means NH3 is a
BASE in water — and water is itself an ACID
NH3
Base
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+
H2O
Acid
NH4+ + OHAcid
Base
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ACID-BASE THEORIES
NH3 is a BASE in water — and water is itself
an ACID
NH3 / NH4+ is a conjugate
by the gain or loss of H+
pair — related
Every acid has a conjugate base
- and vice-versa.
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Conjugate Pairs
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Acid/Base Conjugate Pair Practice
• Write the formula for the conjugate
acid of each of the following bases:
Base
Conj. Acid
HSO3H2SO3
FHF
PO43HPO42CO
HCO+
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Conjugate Acid-Base Pair
Examples
• H2S + NH3  NH4+ + HS• NH3 + HCO31-  NH4+ + CO32• H3PO4 + H2O  H2PO4- + H3O+
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Ch. 14.2 – Acid Strength
• Generally divide acids and bases into
STRONG or WEAK ones.
STRONG ACID: HNO3(aq) + H2O(liq) 
H3O+(aq) + NO3-(aq)
HNO3 is about 100% dissociated in water.
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Strong and Weak Acids
HNO3, HCl, HI, HBr, H2SO4, HClO3 and
HClO4 are the strong acids.
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Strong and Weak Acids
• Weak acids are much less than 100% ionized in
water.
One of the best known is acetic acid = CH3CO2H
Also written as HC2H3O2 or HOAc
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Water as an Acid and a Base
H2O can function as both an ACID and a BASE.
In pure water there can be AUTOIONIZATION
Dissociation Constant for Water= Kw
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
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Water as an Acid and as a Base
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
In a neutral solution [H3O+] = [OH-]
so Kw = [H3O+]2 = [OH-]2
and so [H3O+] = [OH-] = 1.00 x 10-7 M
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Calculating [H3O+] & [OH-]
You add 0.0010 mol of NaOH to 1.0 L of
pure water. Calculate [H3O+] and [OH-].
Solution
2 H2O(liq)  H3O+(aq) + OH-(aq)
Le Chatelier predicts equilibrium shifts to
the __left____.
[H3O+] < 10-7 at equilibrium.
Set up a ICE table.
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Calculating [H3O+] & [OH-]
You add 0.0010 mol of NaOH to 1.0 L of
pure water. Calculate [H3O+] and [OH-].
Solution
2 H2O(liq) 
H3O+(aq) + OH-(aq)
initial
0
0.0010
change
+x
+x
equilib
x
0.0010 + x
Kw = (x) (0.0010 + x)
Because x << 0.0010 M, assume
[OH-] = 0.0010 M
[H3O+] = x = Kw / 0.0010 = 1.0 x 10-11 M
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Calculating [H3O+] & [OH-]
You add 0.0010 mol of NaOH to 1.0 L of
pure water. Calculate [H3O+] and [OH-].
Solution
2 H2O(liq)  H3O+(aq) + OH-(aq)
[H3O+] = Kw / 0.0010 = 1.0 x 10-11 M
This solution is _________
because
[H3O+] < [OH-]
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Ch. 14.3 – The pH Scale
A common way to express acidity and
basicity is with pH
pH = - log [H3O+]
In a neutral solution,
[H3O+] = [OH-] =
1.00 x 10-7 at 25 oC
pH = -log (1.00 x 10-7)
= - (-7) = 7
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The pH Scale
What is the pH of the
0.0010 M NaOH solution?
[H3O+] = 1.0 x 10-11 M
pH = - log (1.0 x 10-11) = 11.00
General conclusion —
Basic solution
Neutral
Acidic solution
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pH > 7
pH = 7
pH < 7
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The pH Scale
See Active Figure 17.1
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pH Calculations
If the pH of Coke is 3.12, it is _acidic__.
Because pH = - log [H3O+] then
log [H3O+] = - pH
Take antilog and get
[H3O+] = 10-pH
[H3O+] = 10-3.12 =
7.6 x 10-4 M
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pH of Common
Substances
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Other pX Scales
In general
pX = -log X
and so
pOH = - log [OH-]
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
Take the log of both sides
-log (10-14) = - log [H3O+] + (-log [OH-])
pKw = 14 = pH + pOH
So if you know one, you can determine the
others.
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pH, pOH, [H+], [OH-]
Practice
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pH
pOH
[H+]
[OH-]
Acidic/Basic?
6.21
7.79
6.2 x 10-7
1.6 x 10-8
acidic
3.87
10.13
1.3 x 10-4
7.4 x 10-11
acidic
2.46
11.54
3.5 x 10-3
2.9 x 10-12
acidic
10.75
3.25
1.8 x 10-11
5.6 x 10-4
basic
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Ch. 14.4 – Calculating the pH of
Strong Acid Solutions
• 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.
• Solve for the pH if needed.
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pH of Strong Acid Solutions
Consider an aqueous solution of 2.0 × 10–3 M
HCl.
What are the major species in solution?
H+, Cl–, H2O
What is the pH?
pH = 2.70
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pH of Strong Acid Solutions
Calculate the pH of a 1.5 × 10–11 M
solution of HCl.
pH = 7.00
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pH of Strong Acid Solutions
Calculate the pH of a 1.5 × 10–2 M solution of
HNO3.
When HNO3 is added to water, a reaction
takes place immediately:
HNO3 + H2O  H3O+ + NO3–
The reaction controlling the pH is:
H2O + H2O  H3O+(aq) + OH-(aq)
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pH of Strong Acid Solutions
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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+ (H3O+)?
pH = - log (1.5x10-2) = 1.82
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Ch. 14.5 – Calculating pH of
Weak Acid Solutions
Aspirin is a good
example of a
weak acid, where
most of the
species exist as
the neutral
molecule and not
as ions.
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Calculating pH of Weak Acid
Solutions
Acid
acetic, CH3CO2H
ammonium, NH4+
bicarbonate, HCO3-
Conjugate Base
CH3CO2-, acetate
NH3, ammonia
CO32-, carbonate
A weak acid (or base) is one that ionizes
to a VERY small extent (< 5%).
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Calculating pH of Weak Acid
Solutions
Consider acetic acid, CH3CO2H (HOAc)
HOAc + H2O
 H3O+ + OAcAcid
Conj. base
Ka =
[H3O+ ][OAc - ]
[HOAc]
= 1.8 x 10-5
(K is designated Ka for ACID)
Because [H3O+] and [OAc-] are SMALL, Ka << 1.
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Equilibrium Constants
for Weak Acids
Weak acid has Ka < 1
Leads to small [H3O+] and a pH of 2 - 7
pKa = -log Ka
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Ionization Constants for Acids/Bases
Acids
Conjugate
Bases
Increase
strength
Increase
strength
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Calculating pH of Weak Acid Solutions
You have 1.00 M HOAc. Calculate the
equilibrium concentrations of HOAc, H3O+,
OAc-, and the pH.
Step 1. Define equilibrium concs. in ICE
table.
[HOAc]
initial
change
equilib
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[H3O+]
[OAc-]
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Calculating pH of Weak Acid Solutions
You have 1.00 M HOAc. Calc. the
equilibrium concs. of HOAc, H3O+, OAc-,
and the pH.
Step 1. Define equilibrium concs. in ICE
table. HOAC(aq) + H2O  OAC- + H3O+
[HOAc]
[H3O+]
[OAc-]
initial
1.00
0
0
change
-x
+x
+x
equilib
1.00-x
x
x
Note that we neglect [H3O+] from H2O.
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Calculating pH of Weak Acid Solutions
You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H3O+, OAc-, and the pH.
Step 2. Write Ka expression
K a = 1.8 x 10-5 =
[H3O+ ][OAc - ]
[HOAc]
x2
=
1.00 - x
This is a quadratic. However…
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Calculating pH of Weak Acid Solutions
You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H3O+, OAc-, and the pH.
Step 3. Solve Ka expression
K a = 1.8 x 10-5 =
[H3O+ ][OAc - ]
[HOAc]
x2
=
1.00 - x
First assume x is very small because
Ka is so small. Therefore,
Ka = 1.8 x 10-5
x2
=
1.00
Now we can more easily solve this
approximate expression.
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Calculating pH of Weak Acid Solutions
You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H3O+, OAc-, and the pH.
Step 3. Solve Ka approximate expression
Ka = 1.8 x 10-5
x2
=
1.00
x = [H3O+] = [OAc-] = [Ka • 1.00]1/2
x = [H3O+] = [OAc-] = 4.2 x 10-3 M
pH = - log [H3O+] = -log (4.2 x 10-3) = 2.37
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Calculating pH of Weak Acid Solutions
Consider the approximate expression
Ka = 1.8 x 10-5
x2
=
1.00
x = [H3O+ ] = [Ka
1/2
 1.00]
For many weak acids
[H3O+] = [conj. base] = [Ka • Co]1/2
where C0 = initial conc. of acid
Useful Rule of Thumb:
If Co > 100*Ka then [H3O+] = [Ka•Co]1/2
In other words, the amount of ionization is small relative to the initial
amount.
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Calculating pH of Weak Acid Solutions
Calculate the pH of a 0.0010 M solution of
formic acid, HCO2H.
HCO2H + H2O  HCO2- + H3O+
Ka = 1.8 x 10-4
Approximate solution
[H3O+] = [Ka • Co]1/2 = 4.2 x 10-4 M, pH = 3.37
Exact Solution
[H3O+] = [HCO2-] = 3.4 x 10-4 M
[HCO2H] = 0.0010 - 3.4 x 10-4 = 0.0007 M
pH = 3.47
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Calculating pH of Weak Acid Solutions
Calculate the pH of a 0.50 M aqueous
solution of HF. (Ka = 7.2 x 10-4)
What are the major species in solution?
HF, H2O
Why aren’t H+ and F– major species?
Only a few are formed from the
dissociation of HF.
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Calculating pH of Weak Acid
Solutions
What are the possibilities for the dominant
reaction?
HF(aq) + H2O(l)  H3O+(aq) + F–(aq)
Ka=7.2 × 10-4
H2O(l) + H2O(l)  H3O+(aq) + OH–(aq)
Kw=1.0 × 10-14
Which reaction controls the pH? Why?
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Calculating pH of Weak Acid
Solutions
Steps Toward Solving for pH
Initial
Change
Equilibrium

H3O+(aq) +
F–(aq)
0.50 M
~0
~0
–x
+x
+x
0.50–x
x
x
HF(aq) + H2O
Ka = 7.2 × 10–4
[H3O+] =(Ka*Co)1/2= (7.2x10-4*0.05)1/2=0.0190
pH = -log(0.0190) = 1.72
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Using Ka Example
A 0.10 M aqueous solution of lactic acid,
CH3CHOHCOOH, has a pH of 2.43. What is the
Ka of lactic acid?
1. Write the equilibrium expression:
HLac(aq) + H2O(l) ⇌ H3O+(aq) + Lac-(aq)
0.10 – x
x
x
+
−
𝑯𝟑𝑶 [𝑳𝒂𝒄 ]
2. Know Ka =
[𝑯𝑳𝒂𝒄]
3. Also know [H3O+] = [Lac-] = 10-2.43 =0.00372
and [HLac] = 0.10 – 0.00372 = 0.0963
4. Substitute: [0.00372][0.00372] = 1.4 x 10-4
[0.0963]
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Using Ka Practice
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A solution prepared from 0.55 mol of butanoic acid is dissolved
in water to give 1.0 L of solution with a pH of 2.21. Determine
the Ka for butanoic acid. The acid ionizes according to the
equation below.
CH3(CH2)2COOH(aq) + H2O(l)  H3O+(aq) + CH3(CH2)2COO-
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a pH meter
0.0001 M
Percent
Dissociation
0.003 M
0.06 M
pH of an acetic
acid solution.
What are your
observations?
2.0 M
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Percent Dissociation (Ionization)
amount dissociated (mol/L)
Percent dissociation =
 100%
initial concentration (mol/L)
• For a given weak acid, the percent
dissociation increases as the acid becomes
more dilute.
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Percent Dissociation Example
A solution of 8.00 M formic acid (HCHO2) is 0.47%
ionized in water. Calculate the Ka value for formic
acid. Determine the pH of the solution.
If 8.00 M of the acid is 0.47% ionized, then 0.038 M
dissociates.
I
C
E
HCHO2(aq) + H2O  H3O+(aq) + CHO2-(aq)
8.00
0
0
-0.038
+0.038
+0.038
7.96
0.038
0.038
𝑲𝒂 =
(𝟎.𝟎𝟑𝟖)(𝟎.𝟎𝟑𝟖)
𝟕.𝟗𝟔
= 1.8 x 10-4
[H+]=0.038 so pH = -log(0.038) = 1.42
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Percent Dissociation
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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Percent Dissociation
However, the % Dissociation and the pH are
higher!!
Dominant rxn: HCHO2(aq) + H2O  H3O+(aq) + CHO2-(aq)
HCHO2(aq) + H2O  H3O+(aq) + CHO2-(aq)
I
4.00
0
0
C
-x
+x
+x
E
4.00-x
x
x
Ka = 1.8 x 10-4 = (x)2/(4.00-x)
x = 0.027; (0.027/4.00) x 100 = 0.67%
pH=-log(0.027) = 1.57
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Ch. 14.6 - Bases
• Strong Base: 100% dissociated in
water.
NaOH(aq)  Na+(aq) + OH-(aq)
Other common strong
bases include KOH and
Ca(OH)2.
CaO (lime) + H2O 
CaO
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Ca(OH)2 (slaked lime)
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Bases
• Weak base: less than 100% ionized
in water
One of the best known weak bases is
ammonia
NH3(aq) + H2O(liq)  NH4+(aq) + OH-(aq)
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Weak Bases
Weak base has Kb < 1
Leads to small [OH-] and a pH of 12 - 7
pKb = -log Kb
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Weak Bases
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Equilibria Involving A Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O  NH4+ +
OH-
Kb = 1.8 x 10-5
Step 1. Define equilibrium concs. in ICE table
[NH3]
initial
change
equilib
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[NH4+]
[OH-]
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Equilibria Involving A Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O  NH4+ +
OH-
Kb = 1.8 x 10-5
Step 1. Define equilibrium concs. in ICE table
[NH3]
[NH4+]
[OH-]
initial
0.010
0
0
change
-x
+x
+x
equilib
0.010 - x
x
x
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Equilibria Involving A Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O  NH4+ +
OH-
Kb = 1.8 x 10-5
Step 2. Solve the equilibrium expression
-5
Kb = 1.8 x 10
=
[NH4+ ][OH- ]
[NH3 ]
x2
=
0.010 - x
Assume x is small (Co>100•Kb), so
x = [OH-] = [NH4+] = 4.2 x 10-4 M
and [NH3] = 0.010 - 4.2 x 10-4 ≈ 0.010 M
The approximation is valid !
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Equilibria Involving A Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O  NH4+ +
OH-
Kb = 1.8 x 10-5
Step 3. Calculate pH
[OH-] = 4.2 x 10-4 M
so pOH = - log [OH-] = 3.37
Because pH + pOH = 14,
pH = 10.63
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Relationship between Ka and Kb
Ka and Kb are linked:
Combined reaction = ?
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Relationship between Ka and Kb
Ka and Kb are related in this way:
Ka  Kb = Kw
Therefore, if you know one of them, you can
calculate the other.
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Relationship between Ka and Kb
Ka and Kb are linked:
Combined reaction = ?
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Ka and Kb Comparison
Several acids are listed here with their
respective Ka’s.
C6H6OH(aq) + H2O(l)  H3O+(aq) + C6H5O-(aq)
Ka = 1.3 x 10-10
HCO2H(aq) + H2O(l)  H3O+(aq) + HCO2-(aq)
Ka = 1.8 x 10-4
HC2O4-(aq) + H2O(l)  H3O+(aq) + C2O42-(aq)
Ka = 6.4 x 10-5
a) Which acid is the strongest? Weakest?
b) Which acid has the weakest conjugate base?
Strongest?
c) What are the pKa values of each acid?
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Ch. 14.7 - Polyprotic Acids
Have more than one acidic proton.
If the difference between the Ka for the first
dissociation and subsequent Ka values is
103 or more, the pH generally depends only
on the first dissociation.
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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.
Example: Citric Acid Dissociation:
H3C6H5O7 + H2O  H2C6H5O7- + H3O+ Ka1 = 7.4x10-4
H2C6H5O7- + H2O  HC6H5O72- + H3O+ Ka2 = 1.7x10-5
HC6H5O72- + H2O  C6H5O73- + H3O+ Ka3 = 4.0x10-7
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Polyprotic Acids
• For a typical weak polyprotic acid:
Ka1 > Ka2 > Ka3
• For a typical polyprotic acid in water, only the
first dissociation step is important to pH,
especially if Ka1 >1000Ka2 and so on.
• Example Ascorbic Acid:
H2C6H6O6 + H2O  HC6H6O6- + H3O+ Ka1 = 8.0 x10-5
HC6H6O6- + H2O  C6H6O62- + H3O+ Ka2 = 1.6x10-12
• In other words, the amount of H3O+ that HC6H6O6- is
contributing to the solution is negligible compared
to the H2C6H6O6. (It’s negligible compared to water
too!)
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Ch. 14.8 - Acid-Base Properties of
Salts
• Salts are ionic compounds.
• Salts of the cation of strong bases and
the anion of strong acids are neutral.
For example: NaCl, KNO3
• There is no equilibrium for strong
acids and bases.
• We ignore the reverse reaction.
• So…
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Acid-Base Properties of Salts
MX + H2O  acidic or basic solution?
Consider NH4Cl
NH4Cl(aq)  NH4+(aq) + Cl-(aq)
(a) Reaction of Cl- with H2O
Cl- + H2O 
HCl
+ OHbase
acid
acid
base
Cl- ion is a VERY weak base because its
conjugate acid is strong.
Therefore, [Cl-] does not affect pH and
above reaction lies completely to the left.
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Acid-Base Properties of Salts
NH4Cl(aq)  NH4+(aq) + Cl-(aq)
(b) Reaction of NH4+ with H2O
NH4+ + H2O 
NH3
+ H3O+
acid
base
base
acid
NH4+ ion is a moderate acid because its
conjugate base is weak.
Therefore, NH4+ is an acidic solution
See TABLE 14.6 for a summary of
acid-base properties of ions.
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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
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Example
NaCl
NaF
NH4Cl
Al2(SO4)3
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Acid-Base Properties of Salts
Example
What is the pH of a 0.015 M solution of
sodium acetate?
C2H3O2-(aq) + H2O(l)  C2H3O2H(aq) + OH-(aq)
I 0.015
C -x
E 0.015-x
0
+x
x
0
+x
+x
Since the acetate ion is the conjugate base of a weak acid,
it’s going to act as a base and take H+ away from the ‘acid’
water.
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Acid-Base Properties of Salts
Example
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Acid-Base Properties of Salts
Example
pH = 14 – 5.54 = 8.46
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Acid-Base Properties of Salts Practice
Sodium hypochlorite, NaClO, is used as a
disinfectant in swimming pools and water
treatment plants. What are the concentrations of
HClO and OH- and the pH of a 0.025 M solution of
NaClO?
ClO-(aq) + H2O(l) ⇌ HClO(aq) + OH-(aq)
I 0.025
0
0
C -x
+x
+x
E 0.025-x
x
x
Kb =
X2
0.025−x
= 2.9 x 10-7
x = 0.025 ∗ 2.9 𝑥10−7 = 8.51x10-5 = [OH-]
pOH = -log(8.51x10-5) = 4.07
So, the pH = 14.00 – 4.07 = 9.93
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Acid-Base Example of a Polyprotic
Salt
Calculate the pH of a 0.10 M solution of Na2CO3.
Na+ + H2O  neutral
CO32- +
H2O  HCO3- +
OHbase
acid
acid
base
Kb = 2.1 x 10-4
Step 1.
Set up ICE table
[CO32-]
initial 0.10
change -x
equilib 0.10 - x
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[HCO3-]
[OH-]
0
+x
x
0
+x
x
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Acid-Base Example of a Polyprotic
Salt
Calculate the pH of a 0.10 M solution of Na2CO3.
Na+ + H2O is neutral
CO32- +
H2O  HCO3- +
OHbase
acid
acid
base
Kb = 2.1 x 10-4
Step 2.
Solve the equilibrium expression
Kb = 2.1 x 10-4 =
[HCO3- ][OH- ]
[CO3 2- ]
x2
=
0.10 - x
Assume 0.10 - x ≈ 0.10, because 100•Kb < Co
x = [HCO3-] = [OH-] = 0.0046 M
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Acid-Base Example of a Polyprotic
Salt
Calculate the pH of a 0.10 M solution of Na2CO3.
Na+ + H2O is neutral
CO32- +
H2O  HCO3- +
OHbase
acid
acid
base
Kb = 2.1 x 10-4
Step 3.
Calculate the pH
[OH-] = 0.0046 M
pOH = - log [OH-] = 2.34
pH + pOH = 14,
so pH = 11.66, and the solution is ___basic_.
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Acid-Base Example of a Polyprotic
Salt
• Notice now that HCO3- can act as a base as
well. Does the Kb for this base affect pH?
HCO3- + H2O  H2CO3
+ OHbase
Step 1.
initial
change
equilib
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acid
acid
base
Kb = 2.4 x 10-8
Set up ICE table as before only…
[HCO31-]
[H2CO3]
[OH-]
0.0046
-x
0.0046 - x
0
+x
x
0.0046
+x
0.0046 + x
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Acid-Base Example of a Polyprotic
Salt
HCO3- + H2O  H2CO3 +
82
OH-
base
acid
acid
base
Kb = 2.4 x 10-8
Step 2. Solve the equilibrium expression
Kb =
[𝐱][𝟎.𝟎𝟎𝟒𝟔+𝐱]
[𝟎.𝟎𝟎𝟒𝟔 −𝐱]
= 𝟐. 𝟒 𝐱 𝟏𝟎−𝟖
Step 3. Solve. Again assume x is small relative to
0.0046.
Then x = 2.4 x 10-8, which really doesn’t add any
measurable difference. Since the Ka’s for the two
acids differ by >103, all of the OH- are made by
CO32-.
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Ch. 14.9 – Effect of
Structure on Acid/Base
Properties
• Why are some compounds
acids?
• Why are some compounds
bases?
• Why do acids and bases vary in
strength?
• Can we predict variations in
acidity or basicity?
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84
Why is CH3CO2H an Acid?
+0.12
+0.24
See Figure 17.12
–0.32
1. The electronegativity of the O atoms causes the H
attached to O to be highly positive.
2. The O—H bond is highly polar.
3. The H atom of O—H is readily attracted to polar H2O.
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Acetic acid
Ka = 1.8 x 10-5
Trichloroacetic acid
Ka = 0.3
• Trichloroacetic acid is a much stronger acid owing
to the high electronegativity of Cl.
• Cl withdraws electrons from the rest of the
molecule.
• This makes the O—H bond highly polar. The H of
O—H is very positive.
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Factors Affecting Acid Strength
For a series of oxyacids, acidity increases
with the number of oxygens.
86
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Basicity of Oxoanions
NO3-
CO32-
PO43-
These ions are BASES.
They become more and more basic as the
negative charge increases.
As the charge goes up, they interact more
strongly with polar water molecules.
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Acid Properties of Metal Ion
Solutions
Electron withdrawing explains why water
solutions of Fe3+, Al3+, Cu2+, Pb2+, etc. are
acidic.
This interaction weakens this bond
Another H2O
pulls this H
away as H+
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89
Ch. 14.10 - Acid/Base Properties of
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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Acid/Base Properties of Oxides
• Basic Oxides (Basic Anhydrides):
– O—X bond is ionic.
K2O, CaO
• If X has a very low electronegativity
(i.e. metal), the O—X bond will be
ionic and subject to being broken in
polar water, producing a basic
solution.
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Ch. 14.11 - Lewis Acid-Base Model
91
Lewis acid
a substance that accepts
an electron pair
Lewis base
a substance that
donates an electron
pair
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Reaction of a Lewis Acid and
Lewis Base
• New bond formed
using electron
pair from the
Lewis base.
• Coordinate
covalent bond
• Notice geometry
change on
reaction.
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Lewis Acids & Bases
Formation of hydronium ion is also an excellent
example.
H
+
ACID
•• ••
O—H
H
BASE
••
H O—H
H
•Electron pair of the new O-H bond
originates on the Lewis base.
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Lewis Acid/Base Reaction
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AP Exam Practice
• 2009B AP Exam #1
• 2009 AP Exam #1
• 2007 AP Exam #1
• 2005 AP Exam #1(a,b,d,e)
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