Acids, Bases, and Aqueous
Equilibria
Nature of Acids and Bases
Acids-Sour taste, corrosive to metals
Bases-Bitter taste, feel slippery, corrosive to fat
Don’t use these to identify acids/bases in lab!
Definitions of Acids and Bases
Arrhenius Concept: Acids produce H+ in
solution, bases produce OH ion.
Brønsted-Lowry: Acids are H+ donors, bases
are proton acceptors.
HCl + H2O  Cl + H3O+
acid base
Conjugate Acid/Base Pairs
HA(aq) + H2O(l)  H3O+(aq) + A(aq)
acid 1
base 2
conj
acid 2
conj
base 1
conjugate base: everything that remains of
the acid molecule after a proton is lost.
conjugate acid: formed when the proton is
transferred to the base.
Acid Dissociation Constant (Ka)
An equilibrium exists in water solutions of
acids :
HA(aq) + H2O(l)  H3O+(aq) + A(aq)
or HA(aq)  H+ (aq) + A- (aq)
Ka 
H3O

HA
A


H

A
HA

Example
Give dissociation reactions for these: HCl,
HC2H3O2, NH4+, C6H5NH3+
HCl  H+ + ClHC2H3O2  H+ + C2H3O2-
NH4+  H+ + NH3
C6H5NH3+  H+ + C6H5NH2-
Acid Strength
Strong Acid:
-
Its equilibrium position lies far to the right.
(HNO3, HCl, HBr, HI, HClO4, H2SO4) Ka
>> 1 – these are the ONLY strong acids
-
Yields a weak conjugate base. (NO3, or
others from above acids)
-
H2SO4 is only strong in its 1st H+
Acid Strength
(continued)
Weak Acid:
- Its equilibrium lies far to the left.
(CH3COOH, and other organic acids)
Ka << 1
-
Yields a much stronger (it is relatively
strong) conjugate base than water.
(CH3COO)
Types of Acids
Binary Acids– Hydrogen bonded to elements other
than water, which has acid characteristics– HCl,
HCN, H2S
Oxyacids– Hydrogen bonded to a polyatomic ion
containing oxygen—H2CO3, H3PO4
Organic Acids—contain the carboxyl group,
OH
-C=O which are all weak acids
14_322
Before dissociation
HA
After dissociation,
at equilibrium
H+ A–
Strong Acids
(a)
HA
HA
Weak Acids
H+ A–
(b)
14_1577
H+
- H+
A
A- H +
AH+
A(a)
AH+
H+
H+
A-
H+
HB
A-
HB
H+
HB
HB
H+
H+
H+
HB
HB
A-
AA-
A-
H+
HB
A(b)
HB
HB
B-
HB
HB
14_323
Relative
acid strength
Very
strong
Relative
conjugate
base strength
Very
weak
Strong
Weak
Weak
Strong
Very
weak
Very
strong
14_02T
Formula
HSO4
HClO2
HC2H2ClO2
HF
HNO2
HC2H3O2
[Al(H2O)6]3+
HOCl
HCN
NH4
HOC6H5
Values of Ka for Some Common Monoprotic Acids
Name
Hydrogen sulfate ion
Chlorous acid
Monochloracetic acid
Hydrofluoric acid
Nitrous acid
Acetic acid
Hydrated aluminum(III) ion
Hypochlorous acid
Hydrocyanic acid
Ammonium ion
Phenol
*The units of Ka are mol/L but are customarily omitted.
Value of K a*
1.2 x 102
1.2 x 102
1.35 x 103
7.2 x 104
4.0 x 104
1.8 x 105
1.4 x 105
3.5 x 108
6.2 x 1010
5.6 x 1010
1.6 x 1010
Increasing acid strength
Table 14.2
Example
From the previous slide, arrange these bases from
weak to stronger: H2O, F-, Cl-, NO2-, CNCl- is from strong acid, as is H2O (from H3O+), so
both are very weak. CN- is from the weakest
acid and is therefore the strongest. HF is a
stonger acid than HNO2, so NO2- is stronger than
F-, so the ranking from weak to strong is:
Cl- < H2O < F- < NO2- < CN-
Water as an Acid and a Base
Water is amphoteric (it can behave either as an
acid or a base).
H2O + H2O  H3O+ + OH
acid
base
conj
acid
conj
base
Kw = 1  1014 at 25°C= [H+] [OH-]
Must always be a balance between H+ and OH-
Example
Calculate [H+] and [OH-] in these solutions:
a. 1.0 x 10-5 M OHb. 1.0 x 10-7 M OH-
c. 10.0 M H+
a.
[OH-]= 1.0 x 10-5 M
= 1.0 x 10-9 M
[H+]= Kw / [OH-] =1 x 10-14/ 1.0 x 10-5
b.
[OH-]= 1.0 x 10-7 M
= 1.0 x 10-7 M
[H+]= Kw / [OH-] = 1 x 10-14/ 1.0 x 10-7
c.
[H+]= 10.0 M
[OH-]= Kw / [H+] = 1 x 10-14/10.0 =1.0 x 10-15 M
The pH Scale
There is a more convenient way to indicate [H+]
pH  log[H+]
pH in water normally ranges from 0 to 14, but can
extend to negative or >14 values
Kw = 1.00  1014 = [H+] [OH]
pKw = 14.00 = pH + pOH
As pH rises, pOH falls (sum = 14.00).
Example
Calculate pH and pOH for each of these
solutions: a. 1.0 x 10-3 M OH- b. 1.0 M H+
a. -log [OH-] = 3 = pOH pH = 14 – pOH = 11
b. -log [H+] = 0= pH
pOH = 14 – pH = 14
14_324
[H+] pH
10–14 14
1 M NaOH
10–13 13
Basic
10–12 12
10–11 11
Ammonia
(Household
cleaner)
10–10 10
10–9
9
10–8
8
–7
7
10–6
6
10–5
5
10–4
4
10–3
3
10–2
2
10–1
1
1
0
Neutral 10
Acidic
Blood
Pure water
Milk
Vinegar
Lemon juice
Stomach acid
1 M HCl
Example
If pH of blood is 7.41, find pOH, [H+], and [OH-]
pOH= 14-pH= 6.59 [H+] = 10-pH = 10-7.41 = 3.9 x 10-8 M
[OH-]= 10-6.59 = 2.6 x 10-7 M
Calculate pH for 0.10 M HNO3 and 1 x 10-10 M HCl
Since both are strong acids, they are totally dissociated, and
[H+] = acid strength of the major species. pH of HNO3 is
therefore –log(0.10) = 1. But the HCl solution is so dilute
that the water provides most of the [H+], and so the pH=7.
Solving Weak Acid Equilibrium
Problems
List major species in solution.
Choose species that can produce H+ and write reactions.
Based on K values, decide on dominant equilibrium.
Write equilibrium expression for dominant equilibrium.
List initial concentrations in dominant equilibrium. (I)
Solving Weak Acid Equilibrium
Problems (continued)
Define change at equilibrium (as “x”). (C)
Write equilibrium concentrations in terms of x.
(E)
Substitute equilibrium concentrations into equilibrium
expression.
Solve for x the “easy way.”
Verify assumptions using 5% rule.
Calculate [H+] and pH.
Example
Calculate the pH of a
0.100 M solution of
HOCl (Ka=3.5 x 10-8)
HOCl
H+
I 0.1
0
0
C -x
+x
+x
x
x
E 0.1-x
OCl-
( x)( x) x 2
Ka 

0.1  x 0.1
x2 = 3.5 x 10-8(0.1)
=3.5 x 10-9
x= 5.9 x 10-5 M = [H+]
pH = -log(5.9 x 10-5)
=4.23
Percent Dissociation
(Ionization)
Calculate this from [H+]
amount dissociated( M )
% dissociation 
 100%
initial concentration( M )
This becomes greater as the acid concentration
becomes more dilute—Example 14.10
Example
Calculate percent dissociation for a. 1.00 M HC2H3O2 and b. 0.100 M
HC2H3O2
Ka= 1.8 x 10-5
a.
H+
Acid
A-
b. Acid
H+
A-
0
0
I 1.00
0
0
0.100
C -x
+x
+x
-x
+x
+x
E 1-x
x
x
0.1-x
x
x
1.8 x 10-5 = x2 / 1
1.8 x 10-5 = x2 / 0.1
x= [H+] = 4.2 x 10-3 M
x= 1.3 x 10-3 M
% diss= 4.2 x 10-3 / 1.0
= 0.42 %
% diss = 1.3 x 10-3 / 0.100
= 1.3 %
14_325
More concentrated
More dilute
Acid concentration
Percent dissociation
H+ concentration
Bases
“Strong” and “weak” are used in the same
sense for bases as for acids.
strong = complete dissociation (hydroxide ion
supplied to solution) Most common are metal
hydroxides. Kb is very large.
NaOH(s)  Na+(aq) + OH(aq)
Example
Calculate pH of 5.0 x 10-2 M NaOH
Strong base means [OH-] = [NaOH]= 0.05 M
pOH = -log(0.05) = 1.30
pH = 14- pOH = 12.70
Bases
(continued)
weak = very little dissociation (or reaction
with water) Usually contain an -NHn group
H3CNH2(aq) + H2O(l)  H3CNH3+(aq) + OH(aq)
Kb for weak bases is usually very small < 10-3
Calcualtions Involving Weak
Bases
Use the same ICE method as with weak acids
Notice that x will equal [OH-] rather than [H+]
Calculate pOH and from that calculate pH
Example
Calculate the pH of a 15.0 M solution of NH3
Kb = 1.8 x 10-5
NH4+ OH-
NH3
I
15.0
0
C -x +x
E 15-x
0
+x
x
x
1.8 x 10-5 = x2 / 15 x2= 1.8 x 10-5 (15)=2.7 x 10-4
x= 1.6 x 10-2 M = [OH-]
pH= 14-pOH=12.2
pOH= 1.80
Polyprotic Acids
. . . can furnish more than one proton (H+) to
the solution. Each one comes off
separately.
H 2CO3  H   HCO3
( Ka1 )
HCO3  H   CO32 
( Ka 2 )
Acid-Base Properties of Salts
Cation
neutral
neutral
Acidic
or Basic
neutral
basic
Anion
neutral
conj base of
weak acid
conj acid of
neutral
acidic
weak base
conj acid of conj base of depends on
weak base weak acid
Ka & Kb
values
Example
NaCl
NaF
NH4Cl
Al2(SO4)3
Relationship of Ka to Kb
For acidic/basic salts, Ka or Kb must be
calculated from the parent acid/base value
K b = Kw
K a = Kw
Ka
Kb
Example:
Ka of HC2H3O2 = 1.8 x 10-5
Kb = 1 x 10-14 / 1.8 x 10-5 = 5.6 x 10-10
Structure and Acid-Base
Properties
Two factors for acidity in binary compounds:
-
Bond Polarity (high is good)
-
Bond Strength (low is good)
14_326
Cl
O
H
Electron density
O
Cl
O
H
Electron density
O
Cl
O
H
O
Electron density
O
O Cl
O
O
H
Electron density
Oxides
Acidic Oxides (Acid Anhydrides):
-
OX bond is strong and covalent.
SO2, NO2, CrO3
Basic Oxides (Basic Anhydrides):
-
OX bond is ionic.
K2O, CaO
Lewis Acids and Bases
Lewis Acid: electron pair acceptor
Lewis Base: electron pair donor
Al3+ + 6 O
H
H
Al
H
3+
O
H
6