Percent Dissociation
Chapter 16 part V
Percent Dissociation

This is the method to determine just how much
of a weak acid or weak base dissociates or
ionizes.

% dissociation=([dissociated]/[initial amount])X100

Previously we found in 1.00 HF
that [H+]=2.7 X10-2
What is % dissociate?
((2.7 X10-2)/1.00)X100=2.7%



Percent Dissociation




Example:
Find the % Dissociate of Acetic Acid
(Ka =1.8 X 10-5) in the following two
examples.
A. 1.00 M Acetic Acid
B. 0.10M Acetic Acid
First: Major species





Equations
Ka Expression
I
C
E
Answer









Acetic Acid Ka= 1.8 X 10-5
H2O Kw = 1.0 X 10-14
Acetic Acid wins!
HC2H3O2  H+ + C2H3O2I 1.00
0
0
C -X
+X +X
E 1.00-X X
X
Ka= 1.8X10-5 = [H+][C2H3O2-] =X2/(1.00-X)
[HC2H3O2]
Answer








X2/(1.00-X) ≈ X2/(1.00)= 1.8 X 10-5
Why? X is insignificant relative to the value 1,
But it IS significant relative to Zero, so we
keep the X2.
X2 = 1.8 X 10-5
X = 4.2 X 10-3
Valid?
(0.0042/1.00)X100 = 0.42 % 5% rule
% dissociation ([H+]/[HC2H3O2])X100=0.42%
Part B






Major species
Equations
Ka Expression
I
C
E
Answer






HC2H3O2  H+ + C2H3O2I 0.10
0
0
C -X
+X +X
E 0.10-X X
X
Ka= 1.8X10-5 = [H+][C2H3O2-] =X2/(0.10-X)
[HC2H3O2]
Answer







X2/(0.10-X) ≈ X2/(0.10)= 1.8 X 10-5
X = 1.3 X 10-3
Still Valid and % Dissociation
(1.3 X 10-3/0.10)X100= 1.3%
The duh Factor
As the initial []’s of a weak acid is smaller, the
% dissociation gets larger.
We can also find the Ka from % dissociation.
Ka from % dissociation