Acid and Base Dissociation Constants
How do we calculate [H+] for a
weak acid?
 We know that strong acids dissociate 100% and that,
therefore, the [H+] equals that [acid] that we start with
 What about weak acids?
 Don’t ionize 100%, so the [H+] is NOT the same as our
starting concentration of our acid!
Recall Kw
 Autoionization of water:
H2O ↔ H+ + OHOr
H2O + H2O ↔ H3O+ + OH Keq =
 Kw = [H3O+] [OH-] = 1.0 x 10-14
Ka
 Similarly to Kw we can write an equilibrium expression for
the dissociation of a weak acid
 The equilibrium for a weak, monoprotic acid (HA) looks
like this:
HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq)
 So, we can write an equilibrium expression that looks like
this:
where Ka is the acid ionization constant
Ka
 Example: Write the equilibrium expression for the
ionization of acetic acid.
CH3COOH(aq) ↔ CH3COO-(aq) + H+(aq)
Ka and Acid Strength
 Ka values are typically between 1 – 1 x 10-16
 The higher the value of Ka, the more the acid
dissociates in water and, hence, the stronger the acid
What about weak bases?
 Weak bases also form an equilibrium in water:
B(aq) + H2O(l) ↔ HB+(aq) + OH-(aq)
 This can be represented by the base dissociation
constant:
 Like Ka, a higher Kb means that more B has dissociated
and, therefore, the stronger the base
Note: Coefficients and Equilibrium
Expressions
 If you have coefficients in your reaction equation, they
become subscripts in the equilibrium expression:
2AB → A2 + B2
Try it
 Write the equilibrium expression for the dissociation
of NH3 in water.
NH3(aq) + H2O(l) ↔ NH4(aq)+ + OH-(aq)
Try the Self Test 10.2
So how does Ka help us find the [H+]?
 The Ka’s for almost every weak acid you could think of
have been measured (at 25oC) and recorded
 If we know the value of Ka and the starting
concentration of our weak acid, we can solve for [H+]
Try It:
 What is the concentration of H+ in 0.50M HF at 25oC?
 From the acid table, Ka = 7.1 x 10-4, so:
HF(aq) ↔ H+(aq) + F-(aq)
 Now what? Now, we use ICE tables!
ICE table
HF(aq) ↔ H+(aq) + F-(aq)

Initial (M)
Change (M)
Equilibrium (M)
0.50
0
0
-x
+x
+x
0.50 – x
x
x
Solve for x
 In this case x is our [H+]
Short Cut:
 If
< 500, the change in the initial concentration
(x) is negligible and can be ignored.
HF(aq) ↔
Initial (M)
Change (M)
Equilibrium (M)
H+(aq)
+
F-(aq)
0.50
0
0
-x
+x
+x
0.50 – x
x
x
]
Percent Dissociation (aka. Percent
Ionization)
 The fraction of molecules that dissociate compared to
the initial concentration, expressed as a percent:
 Percent dissociation =
 Ex: If a 0.10 M solution of benzoic acid was found to
dissociate to give a [H+] = 1.1 x 10-3 M, the percent
dissociation would be: