Acids and Bases
Chapter 14
Classifying Acids
Organic acids contain a carboxyl group or
-- HC2H3O2 & citric acid.
-COOH
Inorganic acids -- HCl, H2SO4, HNO3.
Oxyacids -- acid proton attached to oxygen–
H3PO4. more oxygen the stronger the acid
Monoprotic -- HCl & HC2H3O2
Diprotic -- H2SO4
Triprotic -- H3PO4
Models 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
Bronsted-Lowry Model
The Bronsted-Lowry Model is not limited to
aqueous solutions like the Arrhenius
Model.
NH3(g) + HCl(g) ----> NH4Cl(s)
This is an acid-base reaction according to
Bronsted-Lowry, but not according to
Arrhenius!
Hydronium Ion
Hydronium ion is a hydrated proton -H+.H2O.
The H+ ion is simply a proton. It has a very
high charge density, so it strongly is
attracted to the very electronegative
oxygen of the polar water molecule.
Conjugate Acid/Base Pairs
HA(aq) + H2O(l)  H3O+(aq) + A(aq)
conj
acid 1
conj
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.
Which is the stronger base--H2O or A-?
HF(aq) + OH-(aq) -> F-(aq) + H2O(l)
HCl + OH-  Cl- + H2O
HCl + NH3  Cl- + NH4+
14_323
Relative
acid strength
Very
strong
Relative
conjugate
base strength
Very
weak
Strong
Weak
The relationship of
acid strength and
conjugate base
strength for acidbase reactions.
Weak
Strong
Very
weak
Very
strong
Acid Strength
Strong Acid:
-
Its equilibrium position lies far to the right.
(HNO3)
-
Yields a weak conjugate base. (NO3)
Acid Strength
(continued)
Weak Acid:
-
Its equilibrium lies far to the left.
(CH3COOH)
-
Yields a much stronger (water is relatively
strong) conjugate base than water.
(CH3COO)
Acid Dissociation Constant (Ka)
HA(aq) + H2O(l)  H3O+(aq) + A(aq)
Ka 
H3O

HA
A


H

A

HA
Ka values for common acids are found in Table 14.2
on page 663.
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
HCOOH(aq) + H20 (l)  COOH- (aq) + H30+ (aq)
HClO4(aq) + H20 (l)  ClO4- (aq) + H3O+(aq)
14_1577
H+
+
H
A- H + A
AH+
A(a)
A- H +
H+
H+
A-
H+
HB
A-
HB
H+
HB
HB
H+
H+
H+
HB
HB
A-
AA-
A-
H+
HB
A-
HB
HB
B-
HB
HB
(b)
A strong acid is nearly 100 % ionized, while a weak acid
is only slightly ionized.
14_322
Before dissociation
HA
After dissociation,
at equilibrium
H+ A–
(a)
HA
HA
H+ A–
(b)
Diagram a represents a strong acid, while b represents a weak
acid which remains mostly in the molecular form.
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 1 base 2
conj
acid 2
conj
base 1
Ion product Constant, Kw
Kw is called the ion-product constant or
dissociation constant.
Kw = 1  1014 M2 at 25°C
neutral solution [H+] = [OH-] = 1.0 x 10 -7 M
acidic solution [H+] > [OH-] [H+] > 1.0 x 10-7 M
basic solution [H+] < [OH-] [OH-] > 1.0 x 10-7 M
No matter what the concentration of H+ or OH- in an
aqueous solution, the product, Kw, will remain
the same for that Temp.
[H+] & [OH-] Calculations
Calculate the [H+] for a 1.0 x 10-5 M OH-.
Kw = [H+][OH-]
[H+] = Kw/[OH-]
[H+] = 1.0 x 10-14 M2/1.0 x 10-5 M
[H+] = 1.0 x 10-9 M
[H+] & [OH-] Calculations
Continued
Calculate the [OH-] for a 10.0 M H+.
Kw = [H+][OH-]
[OH-] = Kw/[H+]
[OH-] = 1.0 x 10-14 M2/10.0 M
[OH-] = 1.0 x 10-15 M
The pH Scale
pH = log[H+]
pH in water usually ranges from 0 to 14.
Kw = 1.00  1014 = [H+] [OH]
pKw = 14.00 = pH + pOH
As pH rises, pOH falls (sum = 14.00).
14_324
[H+] pH
10–14 14
–13
10
Basic
1 M NaOH
13
10–12 12
10–11 11
Ammonia
(Household
cleaner)
10–10 10
pH scale and pH values for
common substances. A pH of
1 is 100 times more acidic
than a pH of 3.
10–9
9
10–8
8
–7
7
10–6
6
10–5
5
10–4
4
10–3
3
Acidic 10–2
2
10–1
1
1
0
Neutral 10
Blood
Pure water
Milk
Vinegar
Lemon juice
Stomach acid
1 M HCl
Logarithms
-log 1.00 x 10-7 = 7.000
7.000
characteristic
mantissa
The number of significant digits in 1.00 x 10-7
is three, therefore, the log has three
decimal places. The mantissa represents
the log of 1.00 and the characteristic
represents the exponent 7.
pH & Significant Figures
# decimal places pH -------> # Significant Figure in [H+]
# Significant Figures [H+] -------> #
pH = - log [H+]
[H+] = 10(-pH)
[H+] = 1.0 x 10-5 M
pH = 5.00
decimal places pH
pH Calculations
What is the pOH, [H+], & [OH-] for human
blood with a pH of 7.41?
pH + pOH = 14.00
pOH = 14.00 - pH
pOH = 14.00 - 7.41
pOH = 6.59
pH Calculations
Continued
What is the pOH, [H+], & [OH-]
for human blood with a pH of
7.41?
pH = - log [H+]
[H+] = antilog (-pH)
[H+] = antilog (-7.41)
[H+] = 3.9 x 10-8 M
Note: The number of
significant figures in the
antilog is equal to the number
of decimal places in the pH.
pH Calculations
Continued
What is the pOH, [H+], & [OH-]
for human blood with a pH of
7.41?
pOH = - log [OH-]
[OH-] = antilog (-pOH)
[OH-] = antilog (-6.59)
[OH-] = 2.6 x 10-7 M
Note: The number of
significant figures in the
antilog is equal to the number
of decimal places in the pOH.
Review Book 338-340
1-22
pH of Strong Acid Solutions
Calculate the pH of a 0.10 M HNO3 solution.
Major species are: H+, NO3-,
Sources of H+ are from HNO3 and H2O -amount from water is insignificant.
[H+] = 0.10 M
Note: The number of
significant figures in
the [H+] is the same as
the decimal places in
the pH.
pH = - log [H+]
pH = - log [0.10]
pH = 1.00
A solution is prepared by adding 15.8g of HCl
to enough water to make a total volume of
400.ml What is the pH of the solution ?
Solving Weak Acid Equilibrium
Problems
- Write equilibrium expression for
dominant equilibrium.
- Use an ice table .
pH of Weak Acid Solutions
Calculate the pH of a 0.100 M HOCl solution.
Ka HOCl = 3.5 x 10-8
Major species: HOCl and HOH
 HOCl will be only significant source of [H+].
Ka = 3.5 x 10-8 = [H+][OCl-]/[HOCl]
pH of Weak Acid Solutions
Continued
ICE
[HOCl]
[OCl-]
[H+]
Initial (mol/L)
0.100
0
0
Change (mol/L)
-x
+x
+x
0+x
0+x
Equil. (mol/L)
0.100 - x
pH of Weak Acid Solutions
Continued
Ka = 3.5 x 10-8 = [H+][OCl-]/[HOCl]
3.5 x 10-8 = [x][x]/[0.100 - x]
Ka is more than 100 x smaller than concentration,
x can be neglected in the denominator.
Ka = 3.5 x 10-8 = [x][x]/[0.100]
x2 = 3.5 x 10-9
x = 5.9 x 10-5 M
pH of Weak Acid Solutions
Continued
Approximation check:
% dissociation = (x/[HOCl]o) (100%)
% dissociation = (5.9 x 10-5/0.100)(100%)
% dissociation = 0.059 %
This is much less than 5 % and therefore the
approximation was valid.
Percent Dissociation
(Ionization)
amount dissociated( M )
% dissociation 
(100%)
initial concentration( M )
The percent dissociation calculation is
exactly the same as the one to check the
5 % approximation.
pH of a weak acid practice
Calculate the Ph of a .500M aqueous solution
of formic acid,HCOOH (Ka=1.77x10-4)
Rb324
pH of a weak acid practice II
Calculate the pH of a .200 M HCA Solution
with a Ka=7.45x10-4
Rb324-325
Bases
Bases are often called alkalis because they
often contain alkali or alkaline earth metals.
“Strong” and “weak” are used in the same
sense for bases as for acids.
strong = complete dissociation (hydroxide ion
supplied to solution)
NaOH(s)  Na+(aq) + OH(aq)
Bases
(continued)
weak = very little dissociation (or reaction
with water)
H3CNH2(aq) + H2O(l)  H3CNH3+(aq) + OH(aq)
Kb calculations are identical to Ka calculations.
Calculate the pH of a solution made by adding
4.63 g of LiOH into water for a total
volume of 400 ml
Calculate the pH of a .350M solution of
CH3NH2 (Kb=4.38x10-4)
What is the pH of a 0.100 M solution of
ammonia (NH3) (Kb = 1.8x10-5)?
Kw =Ka x Kb
Calculate the pH of a .500M KF solution at
25dc
Kb for = 1.4.0x10-11
text689
pH =8.31
pH of a Salt
Calculate the pH of a .500M NaNO2
solution at 25dc
(Ka for HNO2= 4.0x10-4
Rb334
8.55
Ph of a salt practice acid
Calculate the pH of a .010M AlCl3 solution .
Ka Value for Al(H2O6)3+ is 1.4x10-5
Calculate the pH of a .010M NH4Cl solution .
Kb Value for NH3 is 5.6x10-10
pH=5.13
Polyprotic Acids
. . . can furnish more than one proton (H+) to
the solution.
H 2CO3  H   HCO3
( Ka1 )
HCO3  H   CO32 
( Ka 2 )
Calculate the pH of a 5.0M H3PO4 solution
and the equilibrium concentrations of
H3PO4, H2PO4-, HPO4-2, PO3-3
Ka1= 7.5 x10-3
Ka2= 6.2 x10-8
Ka3= 4.8 x10-13
Rb
331-332