Strength of
Acids and Bases
Do they ionize 100%?
Strong Acids :
Give up H+ easily
Dissociate completely (100%) in water
HCl, HBr, HI, HNO3, H2SO4, HClO4, HClO3
Weak acids: (all others)
Hold onto H+
Few molecules dissociate
Ex: HC2H3O2
Strong/Weak Acid
Animation
http://educypedia.karadimov
.info/library/acid13.swf
Let’s examine the behavior of
an acid, HA, in aqueous solution.
HA
What happens to the HA molecules in solution?
100% dissociation of HA
HA
H+
Strong Acid
AWould the
solution be
conductive?
Oh yeah…
Partial dissociation of HA
HA
H+
Weak Acid
AWould the
solution be
conductive?
Not really…
HA  H+ + A-
HA
H+
A-
Weak Acid
At any one
time, only a
fraction of
the molecules
are
dissociated.
Strong Bases: Dissociate completely (100%) in water
- Group I metal hydroxides (NaOH, LiOH, etc.)
- Some Group II metal hydroxides
Ca(OH)2, Ba(OH)2,
Sr(OH)2
Weak Bases
Only a few ions dissociate
Ex: NH 3 (ammonia)
Strength and Reactivity

Acids/bases of the same initial molar concentration
can react differently and conduct electricity
differently if one is weak and the other strong.

Ex:
2M HCl
Strong Acid,
very conductive
very reactive
2M HC2H3O2 Weak Acid
Weak Conduction
Salad Dressing!!!
Conjugate Acid/Base Pairs
Strong acid will have a weak conjugate base
 Strong base will have a weak conjugate acid

Hydrolysis

Opposite reaction to neutralization
Salt + Water
Acid + Base
Parent Acid/Base

If you know the salt involved you should
be able to determine which acid and base
it would form if water is added.
Salt + Water
Acid + Base
Ex:
NaCl with water (HOH) would form HCl and NaOH
You Try It

Name the “parent” acid and base that
would be produced from these salts.

Ex:
Potassium chloride
Magnesium carbonate
pH and Hydrolysis

Salts can yield neutral, acidic or basic
solutions depending on what type of acid
or base they produce.
SA/SB = Neutral
SA/WB = Acidic
WA/SB = Basic
WA/WB = Undetermined
The Acid and Base
Dissociation Constant,
Ka & Kb
Setting up Ka and Kb Expressions
weak acid:
CH3COOH + H2O ↔ H3O+ + CH3COO-
Acid ionization constant: Ka = [H3O+][CH3COO-]
[CH3COOH]
weak base:
NH3 + H2O ↔ NH4+ + OH-
Base ionization constant: Kb = [NH4+][OH-]
[NH3]
Acid and base ionization constants are the measure of the
strengths of acids and bases.
The larger the Ka/Kb value the stronger the acid or base
Ka

Weak acids:
 only ionize to a small extent
 come to a state of chemical equilibrium.

Determine how much it ionizes by calculating the
equilibrium constant (Ka)

Larger Ka:
 stronger acid
 more ions found in solution
 more easily donate a proton
HCOOH (aq) + H2O (l) < -- > H3O+ (aq) + HCOO- (aq)

Ka = [H3O+][HCOO-]
[HCOOH]

Notice how the Ka ignores the water since
we are dealing with dilute solutions of
acids, water is considered a constant and
doesn’t have a concentration value
Practice:
1.
A solution of a weak acid, “HA”, is made
up to be 0.15 M. Its pH was found to be
2.96. Calculate the value of Ka.

Steps to follow:
1.
2.
3.
4.
Write balanced equation
Calculate [H+] using 10-pH
Set up chart for equilibrium (ICE)
Solve using Ka expression
Answer
1.
HA (aq) + H2O (l) < -- >H3O+ (aq) + A-(aq)
2.
[H3O+]= 10 -2.96 = 0.0011 M
HA
I
C
E
H2O
< -- >
H3O+
A-
0.0011
0.0011
0.15
-.0011
0.139
4.
Ka = [H3O+][ A-]
[HA]
= [0.0011][0.0011]
[0.139]
= 8.7 x 10 -6
Percent Ionization
The fraction of acid molecules that dissociate
compared with the initial concentration of the acid.
Percent Ionization =
[H3O+] x 100%
[HA i]
For the previous question:
Percent Ionization
= [0.0011] x 100% =0.73 %
[0.15]
Practice:

Ka, for a hypothetical weak acid, HA, at 25°C
is 2.2 x 10-4.
a)
Calculate [H3O+] of a 0.20 M solution of HA.
b) Calculate the percent ionization of HA.
Answer
a) HA(aq) + H2O(l) < -- >H3O+(aq) + A-(aq)
Ka
= [H3O+] [A-] = 2.2 x 10-4
[HA]
2.2 x 10-4 = (x)(x)
(0.20M - x)
x2
10-4)
= (2.2 x
x = 0.0066 M
(0.20 M)
The [H3O+] is 0.0066 M.
Because it is a 1:1 ratio
they are both the same
concentration (x)
Since Ka is
rather small this
number can be
disregarded
b) % ionization = 0.0066 M x 100% = 3.3%
0.20 M
The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H2C2O4
I
C
E
0.100
⇄
H+
+
HC2O4-
The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H 2C 2O 4
I
C
E
0.100
pH =
1.28
⇄
H+
+
HC2O4-
The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H 2C 2O 4
I
C
E
⇄
0.100
pH =
1.28
[H+] =
10-1.28
H+
+
HC2O4-
The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H 2C 2O 4
I
C
E
⇄
H+
0.100
pH =
1.28
[H+] =
10-1.28
[H+] =
0.05248 M
+
HC2O4-
The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H 2C 2O 4
I
C
E
⇄
H+
+
HC2O4-
0.100
0.05248 0.05248
pH =
1.28
[H+] =
10-1.28
[H+] =
0.05248 M
The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H 2C 2O 4
⇄
I
0.100
C - 0.05248
E
0.04752
H+
+
HC2O4-
0.05248 0.05248
pH =
1.28
[H+] =
10-1.28
[H+] =
0.05248 M
1.
The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H 2C 2O 4
I
C
E
⇄
0.100
- 0.05248
0.04752
Ka =
H+
+
HC2O4-
0.05248 0.05248
[H+][HC2O4-] =(0.05248)2
[H2C2O4]
0.04752
Ka = 5.8 x 10-2
Kb

When using weak bases, the same rules
apply as with weak acids, except you are
solving for pOH and using [OH-]
If the pH of 0.40 M NH3 @ 25 oC is 11.427,
calculate the Kb.
NH3 + H2O
I
C
E
0.40
⇄
NH4+ +
OH-
If the pH of 0.40 M NH3 @ 25 oC is 11.427,
calculate the Kb.
NH3 + H2O
I
C
E
0.40
pH =
pOH =
11.427
2.573
⇄
NH4+ +
OH-
If the pH of 0.40 M NH3 @ 25 oC is 11.427,
calculate the Kb.
NH3 + H2O
I
C
E
⇄
0.40
pH =
pOH =
[OH-]
11.427
2.573
=10-2.573
[OH-]
=0.002673 M
NH4+ +
OH-
If the pH of 0.40 M NH3 @ 25 oC is 11.427,
calculate the Kb.
NH3 + H2O
⇄ NH4+ +
OHI
0.40
C - 0.002673
E
0.3973
[OH-]
=
0.002673
0.002673 M
0.002673
If the pH of 0.40 M NH3 @ 25 oC is 11.427,
calculate the Kb.
NH3 + H2O
⇄ NH4+ +
OHI
0.40
C - 0.002673
E
0.3973
0.002673
[OH-]
=
Kb =
[NH4+][OH-]
[NH3]
0.002673
0.002673 M
Kb = 1.8 x 10-5
=
(0.002673)2
0.3973
The pH of 0.20 M NaCN is 11.456,
calculate the Kb for CN-.
The pH of 0.20 M NaCN is 11.456,
calculate the Kb for CN-.
CN- + H2O
pH = 11.456
pOH = 2.55
[OH-] = 10-2.55
[OH-] = .002858M
⇄
HCN+
OH-
The pH of 0.20 M NaCN is 11.456,
calculate the Kb for CN-.
CN- + H2O
I
C
E
0.20
- 0.002858
0.1971
⇄
HCN
0.002858
+
OH-
0.002858
The pH of 0.20 M NaCN is 11.456,
calculate the Kb for CN-.
I
C
E
CN- + H2O
0.20
- 0.002858
0.1971
⇄
HCN
0.002858
+
OH-
0.002858
Kb = [HCN][OH-] = (0.002858)2= 4.1 x 10-5
[CN-]
0.1971
Calculate the pH of 0.020 M H3BO3 (Ka = 3.8 x 10-10)
(weak acid dissociates one H+ at a time)
H3BO3
⇄
H+
I
C
E
0.020 M
-x
0.020 - x
x2
+
disregard small Ka
x
= 3.8 x 10-10
0.020
x = [H+] =
H2BO3-
2.76 x 10-6 M
pH = -Log[2.76 x 10-6]
pH = 5.42
2 sig figs due to molarity and Ka
x