Chapter 12
Acids and Bases
12.1 The Nature of Acids and Bases
12.2 Acid Strength
12.3 The pH Scale
12.4 Calculating the pH of Strong Acid Solutions
12.5 Calculating the pH of Weak Acid Solutions
12.6 Bases
12.7 Polyprotic Acids
12.8 Acid-Base Properties of Salts
12.9 Acid Solutions in Which Water Contributes to the H+
Concentration (skip)
12.10 Strong Acid Solutions in Which Water Contributes to
the H+ Concentration (skip)
12.11 Strategy for Solving Acid-Base Problems: A Summary
Acids and Bases and Their Reactions
Definitions
1. Arrhenius Acids and Bases Acids are
•
•
H+ donors
Bases are OH- donors
2. Arrhenius Broadened Definition
•
•
Acids increase H+ concentration or [H+]
increases
Bases increase OH- concentration or [OH-]
increases
3. Brønsted-Lowry Acids and Bases (1923)
•
•
Acids donate H+
Bases accept H+
Arrhenius
1903
Nobel Prize
Acids-Bases
Brønsted-Lowry Acids and Bases
A Brønsted-Lowry acid is a substance that can
donate a hydrogen ion (aka H+, proton).
A Brønsted-Lowry base is a substance that can
accept a hydrogen ion.
Acids and bases occur as conjugate acid base pairs.
Conjugate Base - subtract an H+ from the acid
Conjugate Acid add H+ to the base
Examples
1. OH– is the conjugate base of
2. H2O is the conjugated base of
3. H2O is the conjugated acid of
4. H3O+ (or often shown as H+) is the conjugate
acid of
CH3CO2H +
H 2O
H3O+ +
Acetic Acid
Pairs
CH3CO2Acetate Ion
1
2
acid
base
2
acid
1
base
Point of View #1
CH3CO2H +
acid
H 2O
H3O+ +
base
Conjugate acid
of H2O
CH3CO2Conjugate base
of CH3CO2H
Point of View #2
CH3CO2H
Conjugate acid
of CH3CO2-
+ H2O
Conjugate
base of H3O+
H3O+ + CH3CO2acid
base
Nomenclature
When H+ is hydrated, it is H3O+ and is called a
hydronium ion. A hydronium ion has the same
molecular geometry as NH3.
+
111.7°
HA
+ H2O
H3O+ + A –
HA = generic acid
• There is Competition for the proton between two bases
H2O and A–
• If H2O is a much stronger base than A– the equilibrium
lies far to the right.
• If A– is a much stronger base than H2O the equilibrium
lies far to the left.
Acid Strength: graphical representation of the behavior of
acids of different strengths in aqueous solution.
A strong acid: equilibrium lies
far to the right
HA + H2O
H3O+ + A –
A weak acid: equilibrium lies
far to the left
HA + H2O
H3O+ + A –
A weak acid yields a relatively
strong conjugate base
Relationship of acid strength and conjugate base
strength
HA + H2O
H3O+ + A-
[H3O  ][A ]
Ka 
[HA]
HA + H2O
H3O+ + A–
SAME AS
HA
H+ + A–
Ka is the Acidity constant
Autoionization of H2O
H2O
Pairs 1
+
H2O
2
base
acid
H3O+ + OH2
1
acid
base
Point of View #1
H2O
#1
acid
+ H2O
#2
base
H3O+
+
Conjugate acid
of H2O #2
OHConjugate base
of H2O #1
Point of View #2
H2O +
Conjugate acid
of OH-
H2O
Conjugate
base of H3O+
H3O+ +
acid
OHbase
Amphoterism - an ion or molecule can act as an
acid or base depending upon the reaction conditions
Amphoterism - an ion or molecule can act
as an acid or base depending upon the
reaction conditions
1.) Water in NH3 serves as an acid
H2O
+ NH3
acid
base
NH4+ +
acid
OHbase
2.) Water in acetic acid serves as a base
H2O +
base
CH3CO2H
acid
H3O+ +
acid
CH3CO2base
Conjugate acid
Conjugate base
of H2O
of acetic acid
Water as an Acid and a Base
Autoionization of water:
[H3O+][OH-]
[H2
O(l)]2
2 H2O (l)
H3O+ (aq) + OH- (aq)
Kw = [H3O+][OH-] = [H+][OH-]
Kw = 1.0 x 10-14 (at 25oC)
In pure water [H+] = [OH-]
Kw = [H3O+][OH-]
Ka/Kb/Kw
NH4+
H+ + NH3
[H ][NH3 ]
Ka 
[NH+4 ]
NH3 + H2O
OH- + NH4+



+
4
[OH ][NH ]
Kb 
[NH3 ]
Ka/Kb/Kw
NH4+
H+ + NH3
[H ][NH3 ]
Ka 
[NH+4 ]
NH3 + H2O
OH- + NH4+


H 2O
+
4
[OH ][NH ]
Kb 
[NH3 ](1)
OH- + H+

K w = Ka K b 
[OH ][H+ ]
(1)
The pH Scale
pH = -log10[H3O+]
SAME AS
pH = -log10[H+]
pH < 7 acidic solution
[H3O+] > [OH-]
pH = 7 neutral solution
[H3O+] = [OH-]
pH > 7 basic solution
[H3O+] < [OH-]
Sig figs: for logs: the number of decimal
places in the log is equal to the number of sig
figs in the original number
Calculate the pH of Strong Acid-Base Solutions
EXAMPLE
pH = -log10[H3O+]
Calculate the pH (at 25oC) of an aqueous solution that has an
OH-(aq) concentration of 1.2 x 10-6 M (i.e., mol/liter)
Solution
The concentration of H+ (aq) is
[H+][OH-] = KW
[H+] = KW/[OH-] = 10-14/1.2x10-6 = 8.3 x 10-9
pH = -log[8.3 x 10-9] = 8.1
Strong Acids
A strong acid is one that dissociates completely in water to
produce H+(aq).
E.g., Hydrochloric acid (HCl) is a strong acid:
HCl (aq) → H+ (aq) + Cl- (aq)
(reaction essentially complete)
Dissolving 0.10 mol of HCl in enough water to make 1.0 L of
solution gives a final concentration of 0.10 M for H+(aq).
H2O (l)
H+ (aq) + OH- (aq)
[H+][OH-] = Kw
[OH-] = Kw / [H+] = 10-14/10-1 = 10-13
Weak Acids
Write the major species in solution and Ka
values. Which is dominant?
#1: HA
[H+ ][A  ]
Ka 
[HA]
H+ + A –

#2: H2O
–
H+ + OH
Kw  [H+ ][O H ]
Compare the value for Ka and Kw. Which is
+
larger? This is the dominate
source
of
H

Problem: (a) Calculate pH and (b) the fraction of
CH3CO2H ionized at equilibrium. The concentration of
CH3CO2H is 1 M (initial, or total). The Ka for acetic acid is
1.8 x 10-5
Estimate major species in solution CH3CO2H (a weak
acid) and H2O.
CH3CO2H
H2O
H+ + CH3CO2─
H+ + OH –
Ka = 1.8 x 10-5
Kw = 1.0 x 10-14
Problem: (a) Calculate pH and (b) the fraction of
CH3CO2H ionized at equilibrium.
H+ + CH3CO2─
CH3CO2H
Initial
1.0M
~0
0
Change
-y
+y
+y
Equilibrium
1.0 – y
y
y
Problem: (a) Calculate pH and (b) the fraction of
CH3CO2H ionized at equilibrium.
H+ + CH3CO2─
CH3CO2H
Initial
1.0M
~0
0
Change
-y
+y
+y
Equilibrium
1.0 – y
y
(y)(y)
Ka 
(1.0 - y)
assume y << 1.0
(y)(y)
Ka 
(1.0)
y
Problem: (a) Calculate pH and (b) the fraction of
CH3CO2H ionized at equilibrium.
H+ + CH3CO2─
CH3CO2H
Initial
1.0M
~0
0
Change
-y
+y
+y
Equilibrium
1.0 – y
y
y
(y)(y)
Ka 
(1.0 - y)
assume y << 1.0
(y)(y)
Ka  1.8x10 
(1.0)
(y)(y)
Ka 
(1.0)
y  4.3x103
check assumption: is y << 1.0?
5
y  1.8x105
Effect of dilution on the % dissociation and [H+] and pH
HA
H+ + A –
Le Chatelier’s Priniciple:
The more dilute the
weak acid solution,
the greater the
percent dissociation
If a chemical system at equilibrium
experiences a change in
concentration (or temperature,
volume, or partial pressure) then
the system shifts to counteract the
imposed change.
Strong Bases
A strong base reacts
water to
Strongcompletely
Acids andwith
Bases
produce OH-(aq) ions.
Sodium hydroxide (NaOH) is a strong base:
Others are KOH–, NH2– (amide ion) and H– (Hydride ion)
NaOH (s)
→
Na+ (aq) + OH- (aq)
(reaction essentially complete)
Dissolving 0.10 mol of NaOH in enough water to make 1.0 L
of solution gives a final concentration of 0.10 M for OH- (aq).
From this you can calculate pH and pOH.
[H+][OH-] = Kw
[H+] = Kw/[OH-] =10-14/10-1 = 10-13
[OH-] = 10-1
[H+] = 10-13
pOH = 1
pH = 13
Weak Bases
B (aq) + H2O (l)
BH+ (aq) + OH– (aq)
B = Base
Kb is the basicity constant
Kb = [BH+][OH–]/[B]
Calculations for solutions of weak bases are similar
to those for weak acids. Bases (B) compete with
OH–, a very strong base, for H+ ions.
B (aq) + H2O (l)
BH+ (aq) + OH– (aq)
Acid-Base Equilibria
Base Strength
– strong acids have weak
conjugate bases
– weak acids have strong
conjugate bases
The strength of a base is
inversely related to the strength
of its conjugate acid; the weaker
the acid, the stronger its
conjugate base, and vice versa
pKa + pKb = pKw
This equation applies to an acid and its conjugate base.
Weak Bases
Kb = 1.8 x 10-5
NH3 is a base
NH3 + H20 NH4+ + OHacid1 base2 acid2 base1
NH4+ is an acid
Ka = 5.6 x 10-10
NH4+
acid1
NH3 + H+
base1
Weak Bases
NH3 is a base
NH3 + H20
Kb = 1.8 x 10-5
NH4+ + OH-
NH4+ is an acid
Ka = 5.6 x 10-10
NH3 + H+
NH4+
Ka Kb = Kw = (5.6 x 10-10) (1.8 x 10-5) = 10-14
pKa + pKb = pKw = (-9.25) + (-4.75) = -14
Weak Bases with Weak Acids
NH4+ is an acid
NH4+
Ka(am) = 5.6 x 10-10
NH3 + H+
Ka(aa) = 1.8 x 10-5
CH3COOH is an acid
CH3COOH
CH3COO- + H+
change the direction of the aa reaction and add them together
NH4+ + CH3COO-
NH3 + CH3COOH
K = Ka(am)/Ka(aa) = 5.6 x 10-10/1.8 x 10-5
=3.1 x 10-5
Assuming 0.1M NH3 initial, calculate the pH of
the resulting solution
H20 + NH3
NH4+ + OH-
Assuming 0.1M NH3 initial, calculate the pH of
the resulting solution
H20 + NH3
NH4+ + OHInit. conc.
0.1M
0
~0
Change
-y
+y
+y
Equil. conc.
0.1– y
y
y
Assuming 0.1M NH3 initial, calculate the pH of
the resulting solution
H20 + NH3
NH4+ + OH-
Kb
Init. conc.
0.1M
0
~0
Change
-y
+y
+y
Equil. conc.
0.1– y
y
[NH 4  ][OH - ]
y2


[NH 3 ][1]
0.1  y
K b for NH 3  1.8x10 5
assume y is small
y2
5
Kb 
 1.8x10
 0.1
y  [OH - ]  [NH 4  ]  1.3x10-3
y
[H+] =
Kw/[OH-] =
10-14/y
pH=-log[H+]
Polyprotic
Acids
It is always easier to remove the first proton in a
polyprotic acid than the second. That is, Ka1 > Ka2 > Ka3
Note: Sulfuric acid is a strong acid in its first dissociation step and a weak
acid in its second step.
H2SO4 (aq) → H+ (aq) + HSO4– (aq)
HSO4– (aq)
H+ (aq) + SO42– (aq)
Ka1 ≈ 1.0x102
Ka2 ≈ 1.2x10-2
Polyprotic Acids
1.
2.
3.
4.
Polyprotic acids have more than one ionizable proton.
The protons are removed in steps, not all at once.
The extent of the steps can be calculated sequentially.
From the successive acidity constants, the equilibrium concentrations
of all species present can be calculated at any value of the pH.
H2SO3 (aq) + H2O (l) → H3O+ (aq) + HSO3- (aq)
Ka1 = 103
HSO3- (aq) + H2O (l)
H3O+ (aq) + SO32- (aq)
Ka2 = 1.2 x 10-2
Polyprotic Acids
• pH calculations for solutions of Polyprotic acids
appear complicated, most common cases (with
weak acids) are surprising straightforward.
• Setup equations, K values, set up ICE type
table, assume dissociation is small (check
assumption) and solve.
• For typical weak Polyprotic acids.
Ka1 >> Ka2 >> Ka3
• The first dissociation step dominates the H+
concentration and thus the pH
A plot of the fractions of H2CO3, HCO-3 and CO32At pH = 9.00 H2CO3 ≈ 0%, HCO-3 = 95% and CO32- = 5%
At pH = 10.00 H2CO3 ≈ 0%, HCO-3 = 68% and CO32- = 32%
Acid-Base Properties of Salts
“Hydrolysis”
Hydrolysis is a Brønsted-Lowry Acid
and Base Reaction
Anion: A─ + H2O ↔ HA + OH─
Cation: M+2 + H2O ↔ H3O+ + M(OH)+
E.g., Cations Ni+2 and Fe+2
• Hydrolysis is a term applied to reactions of
aquated ions that change the pH from 7
• When NaCl is placed in water, the resulting solution
is observed to be neutral (pH = 7)
• However when sodium acetate (NaC2H3O2) is
dissolved in water the resulting solution is basic
• Other salts behave similarly, NH4Cl and AlCl3 give
acid solutions.
• These interactions between salts and water are
called hydrolysis
Example problem:
Suppose a 0.1 mole solution sodium acetate is
dissolved in 1 liter of water. What is the pH of the
solution?
CH3CO2– + H2O ↔ CH3CO2H + OH–
base
acid
acid
Init. conc.
0.1M
0
∆ conc.
-y
+y
Equil. conc.
0.1– y
y
1. Find Kb
2. Find [OH-]
3. Find [H+]
4. Find pH
base
~0
+y
y
1. Find Kb
2. Find [OH-]
Example problem:
3. Find [H+]
What is the pH of the solution?
CH3CO2– + H2O ↔ CH3CO2H
Init. conc.
0.1M
∆ conc.
-y
0
+y
Equil. conc. 0.1– y
y
Ka x Kb = Kw
4. Find pH
+ OH–
~0
+y
y
• Hydrolysis is a term applied to reactions of aquated ions
that change the pH from 7
• When NaCl is placed in water, the resulting solution is
observed to be neutral (pH = 7)
• However when sodium acetate (NaC2H3O2) is dissolved in
water the resulting solution is basic (Problem: found at 0.1M
NaAc, pH =8.89)
• a non-hydrolyzed cation (Na+)
• a hydrolyzed anion (acetate ion)
• Other salts behave similarly, NH4Cl and AlCl3 give acid
solutions.
• a non-hydrolyzed anion (Cl-)
• a hydrolyzed cation (NH4+ or Al+3)
• These interactions between salts and water are called
hydrolysis
Hydrolysis of
Result
Anions
Cations
Raise pH
Lower pH
Non-Hyrolyzed Ions (a few)
7 Anions, not hydrolyzed
Cl –, Br –, I –, HSO4–, NO3–, ClO3–, ClO4–
10 Cations, not hydrolyzed
Li+, Na+, K+, Rb+, Sc+, Mg++, Ca++, Sr++,
Ba++, Ag+
Hydrolysis of
Result
Anions
Cations
Raise pH
Lower pH
Predict pH of salts in water (relative pH)
Na33PO44? is basic (raised pH)
(a non hydrolyzed cation and a hydrolyzed anion)
FeCl33? is acidic (lowers pH)
(a hydrolyzed cation and a non hydrolyzed anion)
Summary
Acid-Base Properties of Salts
“Hydrolysis”
Summary
• Acid-Base Equilibrium Problems
– Which major species are present
– Does a reaction occur that can be assumed to go to
completion?
– Which equilibrium dominates the solution?
– Set up ICE type table
– Solve for equilibrium concentrations using know K
values
– Check any simplifying assumptions
– Typically solve for pH or % species in solution
Chapter 7
Acids and Bases
7.1 The Nature of Acids and Bases
7.2 Acid Strength
7.3 The pH Scale
7.4 Calculating the pH of Strong Acid Solutions
7.5 Calculating the pH of Weak Acid Solutions
7.6 Bases
7.7 Polyprotic Acids
7.8 Acid-Base Properties of Salts
7.9 Acid Solutions in Which Water Contributes to the H+
Concentration (skip)
7.10 Strong Acid Solutions in Which Water Contributes to the
H+ Concentration (skip)
7.11 Strategy for Solving Acid-Base Problems: A Summary