Page 186 - Modern Analytical Chemistry

1400-CH06 9/9/99 7:41 AM Page 169

Chapter 6 Equilibrium Chemistry 169

–
Substituting equation 6.46 into the charge balance equation and solving for [A ]
gives
–
–
+
[A ]= C NaA – [OH ]+[H 3 O ] 6.47
which is substituted into equation 6.45 to give the concentration of HA
–
+
[HA] = C HA + [OH ]–[H 3 O ] 6.48
Finally, substituting equations 6.47 and 6.48 into the K a equation for HA and solv-
ing for pH gives the general buffer equation
–
+
C NaA –[ OH ] +[ H O ]
3
pH = p a +log – +
K
C HA +[ OH ] – [ H O ]
3
+
If the initial concentrations of weak acid and weak base are greater than [H 3 O ] and
–
[OH ], the general equation simplifies to the Henderson–Hasselbalch equation. Henderson–Hasselbalch equation
Equation showing the relationship
C NaA between a buffer’s pH and the relative
pH = pK a +log 6.49
C HA amounts of the buffer’s conjugate weak
acid and weak base.
As in Example 6.13, the Henderson–Hasselbalch equation provides a simple way to
calculate the pH of a buffer and to determine the change in pH upon adding a
strong acid or strong base.
EXAMPLE 6.1 3
Calculate the pH of a buffer that is 0.020 M in NH 3 and 0.030 M in NH 4 Cl.
What is the pH after adding 1.00 mL of 0.10 M NaOH to 0.10 L of this buffer?
SOLUTION

+
The acid dissociation constant for NH 4 is 5.70 ´10 –10 ; thus the initial pH of
the buffer is
.
C 0 020
pH = 924 +log NH 3 =9 24 +log =906
.
.
.
C + 0 030
.
NH 4
+
Adding NaOH converts a portion of the NH 4 to NH 3 due to the following
reaction
–
+
NH 4 (aq)+OH (aq) t NH 3 (aq)+H 2 O(l)
Since the equilibrium constant for this reaction is large, we may treat the
+
reaction as if it went to completion. The new concentrations of NH 4 and NH 3
are therefore
moles NH + – moles OH –
4
C + =
NH 4
V tot
.
( 0 030 M 0 10 L) – ( 0 10 M 1 00 ´ 10 3 – L)
.
)(
(
.
.
)
= = 0 029 M
.
.
0 101 L
moles NH 3 + moles OH –
=
C NH 3
V tot
0 10 )( .
0 020
)
(. M 0 10)( . L +(. M 1 00 ´ 10 3 – L)
.
= = 0 021 M
.
0 101 L