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Ch 14 — Acids and Bases
1
Acids and Bases
Objectives
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Chapter 14
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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4
Acid and Bases
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Acid and Bases
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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Ch 14 — Acids and Bases
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ACID-BASE THEORIES
• The most general theory for common
aqueous acids and bases is the
BRØNSTED - LOWRY theory
ACID-BASE THEORIES
8
The Brønsted definition means NH3 is a
BASE in water — and water is itself an ACID
• ACIDS DONATE H+ IONS
• BASES ACCEPT H+ IONS
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Conjugate Pairs
ACID-BASE THEORIES
10
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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Acid/Base Conjugate Pair Practice
• Write the formula for the conjugate
acid of each of the following bases:
Base
Conj. Acid
HSO3
H2SO3
FHF
HPO42PO43CO
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 — Acids and Bases
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Strong and Weak Acids
Ch. 14.2 – Acid Strength
14
HNO3, HCl, HI, HBr, H2SO4, HClO3 and
HClO4 are the strong acids.
• 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
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• Weak acids are much less than 100% ionized in
water.
Water as an Acid and a Base
16
H2O can function as both an ACID and a BASE.
In pure water there can be AUTOIONIZATION
One of the best known is acetic acid = CH3CO2H
Also written as HC2H3O2 or HOAc
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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4
Ch 14 — Acids and Bases
Calculating [H3O+] & [OH-]
19
You add 0.0010 mol of NaOH to 1.0 L of
pure water. Calculate [H3O+] and [OH-].
Solution
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
H3O+(aq) + OH-(aq)
2 H2O(liq) 
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
This solution is _________
because
[H3O+] < [OH-]
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Ch. 14.3 – The pH Scale
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The pH Scale
A common way to express acidity and
basicity is with pH
What is the pH of the
0.0010 M NaOH solution?
pH = - log [H3O+]
[H3O+] = 1.0 x 10-11 M
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pH = - log (1.0 x 10-11) = 11.00
In a neutral solution,
[H3O+] = [OH-] =
1.00 x 10-7 at 25 oC
General conclusion —
Basic solution
Neutral
Acidic solution
pH = -log (1.00 x 10-7)
= - (-7) = 7
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pH > 7
pH = 7
pH < 7
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pH Calculations
The pH Scale
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
See Active Figure 17.1
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Ch 14 — Acids and Bases
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pH of Common
Substances
Other pX Scales
In general
pX = -log X
and so
pOH = - log [OH-]
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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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© 2009 Brooks/Cole - Cengage
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.
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pH of Strong Acid Solutions
Calculate the pH of a 1.5 × 10–11 M
solution of HCl.
What are the major species in solution?
H+, Cl–, H2O
pH = 7.00
What is the pH?
pH = 2.70
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Ch 14 — Acids and Bases
pH of Strong Acid Solutions
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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–
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
The reaction controlling the pH is: H2O + H2O  H3O+(aq) + OH‐(aq)
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Ch. 14.5 – Calculating pH of
Weak Acid Solutions
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Calculating pH of Weak Acid
Solutions
Acid
acetic, CH3CO2H
ammonium, NH4+
bicarbonate, HCO3-
Aspirin is a good
example of a
weak acid, where
most of the
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Conjugate Base
CH3CO2-, acetate
NH3, ammonia
CO32-, carbonate
A weak acid (or base) is one that ionizes
to a VERY small extent (< 5%).
species exist as
the neutral
molecule and not
as ions.
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Equilibrium Constants
for Weak Acids
Calculating pH of Weak Acid
Solutions
Consider acetic acid, CH3CO2H (HOAc)
 H3O+ + OAcHOAc + H2O
Acid
Conj. base
(K is designated Ka for ACID)
Because [H3O+] and [OAc-] are SMALL, Ka << 1.
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Weak acid has Ka < 1
Leads to small [H3O+] and a pH of 2 - 7
pKa = -log Ka
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Ch 14 — Acids and Bases
Ionization Constants for Acids/Bases
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You have 1.00 M HOAc. Calculate the
equilibrium concentrations of HOAc, H3O+,
OAc-, and the pH.
Conjugate
Bases
Acids
Calculating pH of Weak Acid Solutions
Increase
strength
Step 1. Define equilibrium concs. in ICE
table.
[HOAc]
[H3O+]
[OAc-]
initial
change
equilib
Increase
strength
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Calculating pH of Weak Acid Solutions
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Calculating pH of Weak Acid Solutions
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You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H3O+, OAc-, and the pH.
You have 1.00 M HOAc. Calc. the
equilibrium concs. of HOAc, H3O+, OAc-,
and the pH.
Step 2. Write Ka expression
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
This is a quadratic. However…
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 3. Solve Ka expression
First assume x is very small because
Ka is so small. Therefore,
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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
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
Now we can more easily solve this
approximate expression.
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Ch 14 — Acids and Bases
Calculating pH of Weak Acid Solutions
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Calculating pH of Weak Acid Solutions
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Calculate the pH of a 0.0010 M solution of
formic acid, HCO2H.
Consider the approximate expression

HCO2H + H2O  HCO2- + H3O+
Ka = 1.8 x 10-4
For many weak acids
Approximate solution
[H3O+] = [conj. base] = [Ka • Co]1/2
[H3O+] = [Ka • Co]1/2 = 4.2 x 10-4 M, pH = 3.37
where C0 = initial conc. of acid
Exact Solution
Useful Rule of Thumb:
[H3O+] = [HCO2-] = 3.4 x 10-4 M
If Co > 100*Ka then [H3O+] = [Ka•Co]1/2
[HCO2H] = 0.0010 - 3.4 x 10-4 = 0.0007 M
In other words, the amount of ionization is small relative to the initial
pH = 3.47
amount.
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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
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.
H2O(l) + H2O(l)  H3O+(aq) + OH–(aq) Kw=1.0 × 10‐14
Which reaction controls the pH? Why? © 2009 Brooks/Cole - Cengage
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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
2. Know Ka =
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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Ch 14 — Acids and Bases
Using Ka Practice
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50
a pH meter
0.0001 M
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-
Percent
Dissociation
0.003 M
pH of an acetic
acid solution.
0.06 M
What are your
observations?
2.0 M
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Percent Dissociation Example
Percent Dissociation (Ionization)
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.
• For a given weak acid, the percent
dissociation increases as the acid becomes
more dilute.
HCHO2(aq) + H2O  H3O+(aq) + CHO2‐(aq)
8.00
0
0
‐0.038
+0.038
+0.038
7.96 0.038
0.038
I
C
E
.
.
.
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= 1.8 x 10‐4
[H+]=0.038 so pH = ‐log(0.038) = 1.42
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Percent Dissociation
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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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 — Acids and Bases
Ch. 14.6 - Bases
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Bases
• Strong Base: 100% dissociated in
water.
• Weak base: less than 100% ionized
in water
One of the best known weak bases is
NaOH(aq)  Na+(aq) + OH-(aq)
ammonia
NH3(aq) + H2O(liq)  NH4+(aq) + OH-(aq)
Other common strong
bases include KOH and
Ca(OH)2.
CaO (lime) + H2O 
Ca(OH)2 (slaked lime)
CaO
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Weak Bases
Weak Bases
Weak base has Kb < 1
Leads to small [OH-] and a pH of 12 - 7
pKb = -log Kb
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Equilibria Involving A Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O 
Kb = 1.8 x
NH4+
+
Equilibria Involving A Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O  NH4+ +
OH-
10-5
Kb = 1.8 x
Step 1. Define equilibrium concs. in ICE table
[NH3]
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[NH4+]
OH-
10-5
Step 1. Define equilibrium concs. in ICE table
[OH-]
[NH3]
[NH4+]
[OH-]
initial
initial
0.010
0
0
change
change
-x
+x
+x
equilib
equilib
0.010 - x
x
x
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Ch 14 — Acids and Bases
Equilibria Involving A Weak Base
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You have 0.010 M NH3. Calc. the pH.
NH3 + H2O  NH4+ +
Equilibria Involving A Weak Base
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You have 0.010 M NH3. Calc. the pH.
NH3 + H2O  NH4+ +
OH-
Kb = 1.8 x 10-5
OH-
Kb = 1.8 x 10-5
Step 2. Solve the equilibrium expression
Step 3. Calculate pH
[OH-] = 4.2 x 10-4 M
so pOH = - log [OH-] = 3.37
Because pH + pOH = 14,
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 !
pH = 10.63
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Relationship between Ka and Kb
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Relationship between Ka and Kb
Ka and Kb are linked:
Ka and Kb are related in this way:
Ka  Kb = Kw
Therefore, if you know one of them, you can
calculate the other.
Combined reaction = ?
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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 — Acids and Bases
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Polyprotic Acids
Ch. 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.
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
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
• 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
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…
• 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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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
NH3
+ H3O+
NH4+ + H2O 
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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Ch 14 — Acids and Bases
Acid-Base Properties of Salts
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Acid-Base Properties of Salts
Example
74
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
75
Acid-Base Properties of Salts
Example
76
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 =
.
= 2.9 x 10-7
x = 0.025 ∗ 2.9 10
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
H2O  HCO3- +
OHCO32- +
base
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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14
Ch 14 — Acids and Bases
Acid-Base Example of a Polyprotic
Salt
79
Calculate the pH of a 0.10 M solution of Na2CO3.
Na+ + H2O is neutral
H2O  HCO3- +
OHCO32- +
base
acid
acid
base
Kb = 2.1 x 10-4
Step 2.
Acid-Base Example of a Polyprotic
Salt
80
Calculate the pH of a 0.10 M solution of Na2CO3.
Na+ + H2O is neutral
H2O  HCO3- +
OHCO32- +
base
acid
acid
base
Kb = 2.1 x 10-4
Solve the equilibrium expression
Step 3.
Calculate the pH
[OH-] = 0.0046 M
pOH = - log [OH-] = 2.34
Assume 0.10 - x ≈ 0.10, because 100•Kb < Co
x = [HCO3-] = [OH-] = 0.0046 M
pH + pOH = 14,
so pH = 11.66, and the solution is ___basic_.
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© 2009 Brooks/Cole - Cengage
Acid-Base Example of a Polyprotic
Salt
81
HCO3- + H2O  H2CO3 +
• Notice now that HCO3- can act as a base as
well. Does the Kb for this base affect pH?
+ OHHCO3- + H2O  H2CO3
base
Step 1.
initial
change
equilib
[H2CO3]
[OH-]
0.0046
-x
0.0046 - x
0
+x
x
0.0046
+x
0.0046 + x
Kb =
OH-
.
.
. 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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82
base
acid
acid
base
Kb = 2.4 x 10-8
Step 2. Solve the equilibrium expression
acid
acid
base
Kb = 2.4 x 10-8
Set up ICE table as before only…
[HCO31-]
Acid-Base Example of a Polyprotic
Salt
83
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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Ch 14 — Acids and Bases
85
86
Factors Affecting Acid Strength
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.
For a series of oxyacids, acidity increases
with the number of oxygens.
• This makes the O—H bond highly polar. The H of
O—H is very positive.
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Basicity of Oxoanions
NO3-
CO32-
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
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.
Another H2O
pulls this H
away as H+
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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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90
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 — Acids and Bases
Ch. 14.11 - Lewis Acid-Base Model
91
Lewis acid
a substance that accepts
an electron pair
92
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.
Lewis base
a substance that
donates an electron
pair
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Lewis Acids & Bases
93
94
Lewis Acid/Base Reaction
Formation of hydronium ion is also an excellent
example.
•Electron pair of the new O-H bond
originates on the Lewis base.
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© 2009 Brooks/Cole - Cengage
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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