Page 160 - Modern Analytical Chemistry

1400-CH06 9/9/99 7:40 AM Page 143

Chapter 6 Equilibrium Chemistry 143

+
Solving for [H 3 O ] gives 1
Gastric juice
2
–14
–7
+
10
3
[HO ] = . 1 00 ´ 10 = . 1 00 ´
3 Vinegar
–7
A neutral solution has a hydronium ion concentration of 1.00 ´10 M and a pH of 4
+
7.00.* For a solution to be acidic, the concentration of H 3 O must be greater than
–
that for OH , or 5
“Pure” rain
6
+
–7
[H 3 O ] > 1.00 ´10 M Milk
7 Neutral
The pH of an acidic solution, therefore, must be less than 7.00. A basic solution, on pH
8 Blood
the other hand, will have a pH greater than 7.00. Figure 6.3 shows the pH scale Seawater
along with pH values for some representative solutions. 9
10 Milk of magnesia
Tabulating Values for K a and K b A useful observation about acids and bases is that
11
the strength of a base is inversely proportional to the strength of its conjugate acid.
12
Consider, for example, the dissociation reactions of acetic acid and acetate.
Household bleach
13
–
+
CH 3 COOH(aq)+H 2 O(l) t H 3 O (aq)+CH 3 COO (aq) 6.11
14
–
–
CH 3 COO (aq)+H 2 O(l) t CH 3 COOH(aq)+OH (aq) 6.12 Figure 6.3
pH scale showing values for representative
Adding together these two reactions gives
solutions.
–
+
2H 2 O(l) t H 3 O (aq)+OH (aq) 6.13
The equilibrium constant for equation 6.13 is K w . Since equation 6.13 is obtained
by adding together reactions 6.11 and 6.12, K w may also be expressed as the product
–
of K a for CH 3 COOH and K b for CH 3 COO . Thus, for a weak acid, HA, and its con-
–
jugate weak base, A ,
K w = K a ´K b 6.14
This relationship between K a and K b simplifies the tabulation of acid and base dis-
sociation constants. Acid dissociation constants for a variety of weak acids are listed
in Appendix 3B. The corresponding values of K b for their conjugate weak bases are
determined using equation 6.14.
EXAMPLE 6. 3
Using Appendix 3B, calculate the following equilibrium constants
(a) K b for pyridine, C 5 H 5 N
–
(b) K b for dihydrogen phosphate, H 2 PO 4
SOLUTION
K w . 100 ´ 10 –14
() K , bC H N = = = . 169 ´ 10 –9
a
5
5
K , aC H NH + . 590 ´ 10 –6
5
5
K w . 100 ´ 10 –14
b
() K – = = = . 141 ´ 10 –12
K , aH PO 4 . 711 ´ 10
2
, bH PO 4 –3
3
*The use of a p-function to express a concentration is covered in Chapter 2.