Page 185 - Modern Analytical Chemistry

1400-CH06 9/9/99 7:41 AM Page 168

168 Modern Analytical Chemistry

A mixture of acetic acid and sodium acetate is one example of an acid/base
buffer. The equilibrium position of the buffer is governed by the reaction
+
–
CH 3 COOH(aq)+H 2 O(l) t H 3 O (aq)+CH 3 COO (aq)
and its acid dissociation constant
+
–
[ H O ][ CH COO ]
3
3
K a = = .175 ´10 –5 6.43
3
[ CH COOH]
The relationship between the pH of an acid–base buffer and the relative amounts of
–
CH 3 COOH and CH 3 COO is derived by taking the negative log of both sides of
equation 6.43 and solving for the pH
–
–
[ CH COO ] [ CH COO ]
3
3
pH = p a +log = .476 +log 6.44
K
3
3
[ CH COOH] [ CH COOH]
Buffering occurs because of the logarithmic relationship between pH and the ratio
of the weak base and weak acid concentrations. For example, if the equilibrium
–
concentrations of CH 3COOH and CH 3COO are equal, the pH of the buffer is 4.76.
If sufficient strong acid is added such that 10% of the acetate ion is converted to
–
acetic acid, the concentration ratio [CH 3 COO ]/[CH 3 COOH] changes to 0.818,
and the pH decreases to 4.67.
6H.1 Systematic Solution to Buffer Problems
Equation 6.44 is written in terms of the concentrations of CH 3 COOH and
–
CH 3 COO at equilibrium. A more useful relationship relates the buffer’s pH to the
initial concentrations of weak acid and weak base. A general buffer equation can be
derived by considering the following reactions for a weak acid, HA, and the salt of
its conjugate weak base, NaA.
+
–
NaA(aq) ® Na (aq)+A (aq)
+
–
HA(aq)+H 2 O(l) t H 3 O (aq)+A (aq)
–
+
2H 2 O(l) t H 3 O (aq)+OH (aq)
–
–
+
+
Since the concentrations of Na , A , HA, H 3 O , and OH are unknown, five equa-
tions are needed to uniquely define the solution’s composition. Two of these equa-
tions are given by the equilibrium constant expressions
–
+
[ HO ][ A ]
3
K a =
[ HA]
+
–
K w =[ HO ][ OH ]
3
The remaining three equations are given by mass balance equations on HA and
Na +
–
C HA + C NaA =[HA]+[A ] 6.45
+
C NaA = [Na ] 6.46
and a charge balance equation
–
+
–
+
[H 3 O ] + [Na ] = [OH ]+[A ]