Acid-Base Strength:
Ka, Kb, Kw
Mrs. Kay
Chemistry 12
Chapter 15 Pages: 583-584,587-597
Relative Strengths Of Binary Acids
H –X
The greater the tendency for the transfer of a proton
from HX to H2O, the more the forward reaction is
favored and the stronger the acid.
in a periodic group:
• The weaker the bond, the stronger the acid.
• The larger the resultant anion’s radius, the stronger is
the acid.
• The strengths of binary acids increase from top to
bottom in a group of the periodic table.
Relative Strengths Of Binary Acids
H –X
in a periodic group:
Bond dissociation energy: the weaker the bond, the stronger the acid.
Bond dissociation energy
569
> 431 > 368 > 297
(kJ/mol)
HF
HCl
HBr
HI
Acid strength Ka
6.6x10-4 < ~106 <
~108 < ~109
Anion radius: the larger the anion’s radius, the stronger the acid.
Anion radius (ppm)
136
< 181 < 195 < 216
(kJ/mol)
HF
HCl
HBr
HI
Acid strength Ka
6.6x10-4 < ~106 <
~108 < ~109
The strength of binary acids increase from top to bottom in a
group of the periodic table.
Relative Strengths Of Binary Acids
H –X
in a period:
• The larger the electronegativity difference between H and X,
the more easily the proton is removed and the stronger is the
acid.
 EN
Acid strength
0.4
CH4
<
0.9
NH3
<
1.4
H2O
<
1.9
HF
• The strengths of binary acids increase from left to right
across a period of the periodic table.
Representative Trends
In Strengths of Binary Acids
The Acid dissociation
constant, Ka
Ka
• A weak acid only ionizes to a small extent and
comes to a state of chemical equilibrium.
• We can determine how much it ionizes by
calculating the equilibrium constant for this
reaction, the ionization constant, Ka.
• The larger the Ka the more acid ions are found
in solution and the stronger the acid because the
more easily it donates a proton.
• The reverse is true for the smaller the Ka
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
when multiplied by both sides it is
cancelled out.
Equilibrium In Solutions Of
Weak Acids And Weak Bases
weak acid:
HA + H2O  H3O+ + A[H3O+][A-]
Ka =
[HA]
weak base:
B + H2O  HB+ + OH[HB+][OH-]
Kb =
[B]
You need to be able to write acid and base ionization
equations!!!
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 or i Δ f)
Solve using Ka expression
Answer
1. HA (aq) + H2O (l) < -- >H3O+ (aq) + A-(aq)
2. [H3O+]= 10 -2.96 = 0.0011 M
3.
Set up table:
H3O+
A-
0.15
0
0
-.0011
+0.0011
+0.0011
0.139
0.0011
0.0011
HA
i
Δ
f
H2O
< -- >
4. Ka = [H3O+][ A-]
[HA]
= [0.0011][0.0011]
[0.139]
= 8.7 x 10 -6
Percent Dissociation
The fraction of acid molecules that dissociate
compared with the initial concentration of the
acid.
Percent Dissociation = [H3O+] x 100%
[HA i]
For the previous question:
Percent Dissociation = [0.0011] x 100% =0.73 %
[0.15]
Practice:
• The ionization constant, Ka, for a
hypothetical weak acid, HA, at 25°C is 2.2
x 10-4.
a) Calculate the [H3O+] of a 0.20 M solution
of HA.
b) Calculate the percent ionization of HA.
c) Calculate the [A-].
d) What initial concentration of HA is needed
to produce a [H3O+] of 5.0 x 10-3 M?
Answer
a) HA(aq) + H2O(l) < -- >H3O+(aq) + A-(aq)
= [H3O+][A-] = 2.2 x 10-4
[HA]
2.2 x 10-4 = (x)(x)
0.20 M
x2 = (2.2 x 10-4) (0.20 M)
x = 0.0066 M
Ka
The [H3O+] is 0.0066 M.
Because it is a
1:1 ratio they
are both the
same
concentration
(x)
b) % ionization = 0.0066 M x 100% = 3.3%
0.20 M
c) From the stoichiometry of the reaction,
[H3O+] = [A-]
Therefore, [A-] = 0.0066 M
Found d) from table set up
d) 2.2 x 10-4 = (5.0 x 10-3 M)(5.0 x 10-3 M)
x – 5.0 x 10-3 M
• 2.2 x 10-4(x – 5.0 x 10-3M) = 2.5 x 10-5
• 2.2 x 10-4x – 1.1 x 10-6 = 2.5 x 10-5
• x = (2.5 x 10-5 + 1.1 x 10-6) / 2.2 x 10-4
• x= 0.12 M
• The initial concentration of HA required is
0.12 M.
Acid And Base Ionization Constants
CH3COOH + H2O  H3O+ + CH3COO[H3O+][CH3COO-]
Acid ionization constant: Ka =
[CH3COOH]
weak acid:
NH3 + H2O  NH4+ + OH[NH4+][OH-]
Base ionization constant: Kb =
[NH3]
weak base:
Acid and base ionization constants are the measure of the
strengths of acids and bases.
Kb
• When using weak bases, the same rules
apply as with weak acids, except you are
solving for pOH and using [OH-]
Another relationship
• Useful to know:
Ka x Kb = Kw = 1.0 x 10-14
Buffer Solutions
• A buffer solution is a solution that changes
pH only slightly when small amounts of a
strong acid or a strong base are added.
• A buffer contains
a weak acid with its salt (conjugate base) or
a weak base with its salt (conjugate acid)
CH3COOH/CH3COONa
NH3/NH4Cl
Depicting Buffer Action
How A Buffer Solution Works
• The acid component of the buffer can neutralize
small added amounts of OH-, and the basic
component can neutralize small added amounts
of H3O+.
• Pure water does not buffer at all.