Chemistry 12
 Keq
= equilibrium constant
 Ksp
= solubility product
 Kw
=
eqb exp for ionization of water
 Ka
=
acid ionization constant
 Kb
=
base ionization constant
Using Kw to calculate unknown [H3O+] and
[OH-] from a strong acid or a strong base.
Eg.) At 250C an HCl solution has a concentration
of 0.0100 M. What is the [OH-]?
 HCl
+ H20 ↔ H3O+ + Cl-
 [H3O+]
= 0.0100 M
 [H3O+][OH-]
 0.0100
 [OH-]
= Kw
M [OH-] = 1.00 x 10-14
= 1.00 x 10-12 M
What is the Ka expression for oxalic acid?
(Hint: use the table)
H2O(l) + H2C2O4(aq) ↔ H3O+(aq) + HC2O4-(aq)
Ka = [H3O+][HC2O4-]
[H2C2O4]
What is the equilibrium equation for HC2O4acting as a base and the corresponding Kb
expression?
 HC2O4-(aq)
 Kb
+ H2O(l) ↔H2C2O4(aq) +OH-(aq)
= [H2C2O4][OH-]
[HC2O4-]
What is Ka value for the following?
(simply use the table)
 H2C2O4
 HC2O4-
Ka = 5.9 x 10-2
Ka = 6.4 x 10 -5
…but solving for Kb is not that easy.
What is the Kb value of HPO42-?
… get out your B-L Table and follow along.
We must use the Ka
value of H2PO4to determine Kb of
HPO42-
HPO42- is
acting as an
acid
HPO42- is acting
as a base
Ka value of HPO42- so we
cannot use this to calculate Kb
Kb of HPO42- = Kw/ Ka(H2PO4-)
= 10-14/6.2 x 10-8
= 1.6 x10-7
You should notice that this is the
Ka value of the conjugate acid for
HPO42-
Kb for HC2O4-?
Find HC2O4- on right side of table
Ka = 5.9 x 10 -2
Kw = Ka x Kb
Actually Ka
of H2C2O4
1.0 x 10-14 = 5.9 x 10 -2 x Kb
Kb = 1.7 x 10-13
Why do we use the formula Kw = Ka x Kb?
H2O(l) + HC2O4-(aq) ↔ H3O+(aq) + C2O42-(aq)
Acid dissociation
Base dissociation
This can be explained using HC2O4- but
other weak acids could also be used
To get Kb we must figure out the base
dissociation equation of C2O42-.
H2O(l) + HC2O4-(aq) ↔ H3O+(aq) + C2O42-(aq)
Invert
H3O+(aq) + C2O42-(aq) ↔ H2O(l) + HC2O4-(aq)
+ OH- (aq)
↔ OH-(aq)
2 H2O(l)
H2O(l) + C2O42-(aq) ↔ OH-(aq) + HC2O4-(aq)
H2O(l) + C2O42-(aq) ↔ OH-(aq) + HC2O4-(aq)
H2O(l) + HC2O4-(aq) ↔ H3O+(aq) + C2O42-(aq)
Kb =
[HC2O4-][OH-]
[C2O42-]
Ka = [H3O+][C2O42-]
[HC2O4-]
 if
Kb = [HC2O4-][OH-]
[C2O42-]
Ka = [H3O+][C2O42-]
[HC2O4-]
Then Ka x Kb
[HC2O4-][OH-]
[C2O42-]
x [H3O+][C2O42-]
[HC2O4-]
 Ka x Kb = [OH-][H3O+] = Kw
 Can
you calculate [H3O+] and [OH-] of a
strong acid or base using Kw?
 Can you write equilibrium equations for
weak acids and bases?
 Can you write out Ka and Kb
expressions?
 Can you find Ka on the B-L table?
 Can you calculate Kb using the B-L table?
Read & Highlight
Self Notes p. 3 - 5
Hebden # 31-37