Chemistry 146 Lecture Problems
Acetic Acid/Acetate Buffer
First, what happens when a weak acid and the salt of the conjugate base are in a solution with
a solution that is 0.1 M acetic acid and 0.1 M sodium acetate:
Acetic acid is a weak acid so it is in an equlibrium:
CH3COOH + H2O <--> CH3COO1- + H3O1+
mole
M
liter
Sodium acetate is a salt, so when placed in water it will dissociate. This reaction goes to
completion:
NaCH3COO --> Na1+ + CH3COO1-
Since the dissociation of the salt, goes to completion, start by dealing with that reaction.
NaCH3COO
Initial
Change
Final
Na 1+
-->
0.1 . M
CH3COO1-
+
0 .M
0.1 . M
0 .M
0.1 .M
0 .M
0.1 .M
0.1 . M
0.1 . M
Now, look at the equlibrium expression for acetic acid.
CH3COOH
Initial
Change
Final
+
H2O
CH 3COO1-
<-->
0.1 . M
0.1 . M
X
0.1
+
H3O1+
0 .M
X
X
0.1
X
X
X
To solve for X, we need to look at the equlibrium expression:
Ka
C CH3COO .C H3O
1.8 . 10
C CH3COOH
5
( 0.1
( 0.1
acetate_buffer_a.mcd
X ) .( X )
X)
S.E. Van Bramer
X
1.7993523496892227 .10
5
.10003599352349689223
7/30/01
Simplifying the expression:
1.8 . 10
X ) .( X )
( 0.1
5
( 0.1
1.8 . 10
X)
( 0.1 ) .( X )
5
( 0.1 )
1.8 .10
X
5
Essentially the same as the exact solution.
Which gives final concentrations for everything as:
CH3COOH
0.1
X = 0.09998
CH3COO1-
0.1
X = 0.10002
H 3O1+
X = 1.8 . 10
Notice this value, it is not a
coincidence.
5
Finally, look at Kw (Since this is the smallest equlibrium constant):
2 H2O
H 3O1+
<---- >
1.8 . 10
Initial
Change
5
0
X
1.8 . 10
Final
OH1-
+
X
5
X
X
Solve for X using the equlibrium expression:
K w C H3O .C OH
1.0 . 10
X
1.8 . 10
14
5
X .( X )
1.8000555538409837511 . 10
5.555384098375108 .10
5
10
Or simplify the expression:
1.0 . 10
X
14
1.8 . 10
5.55556 .10
5 .
(X)
10
Which is identical to the exact solution (the meaningful root)
Which gives the final equlibrium concentrations as:
OH 1-
acetate_buffer_a.mcd
X = 5.55556 .10
10
S.E. Van Bramer
7/30/01