Page 182 - Modern Analytical Chemistry

1400-CH06 9/9/99 7:41 AM Page 165

Chapter 6 Equilibrium Chemistry 165

We can simplify this equation further if K a1 K w << K a1 K a2 C HL , and if K a1 << C HL ,
giving

[HO + ] = KK a2 6.42
3
a1

+
For a solution of 0.10 M alanine, the [H 3 O ] is
4
1
[HO + ] = ( .487 ´ 10 –3 )( .358 ´ 10 –10 ) = . 7 807 10 –7 M
´
3
or a pH of 6.11. Verifying that the assumptions are acceptable is left as an exercise.
Triprotic Acids and Bases, and Beyond The treatment of a diprotic acid or base is
easily extended to acids and bases having three or more acid–base sites. For a tripro-
tic weak acid such as H 3 PO 4 , for example, we can treat H 3 PO 4 as if it was a mono-
–
2–
protic weak acid, H 2PO 4 and HPO 4 as if they were intermediate forms of diprotic
3–
weak acids, and PO 4 as if it was a monoprotic weak base.
EXAMPLE 6.12

Calculate the pH of 0.10 M Na 2 HPO 4 .
SOLUTION

2–
We treat HPO 4 as the intermediate form of a diprotic weak acid
2–
–
3–
H 2 PO 4 (aq) t HPO 4 (aq) t PO 4 (aq)
where the equilibrium constants are K a2 = 6.32 ´10 –8 and K a3 = 4.5 ´10 –13 .
Since the value of K a3 is so small, we use equation 6.41 instead of equation 6.42.
4
6
10
10
1
0
6
( .32 ´ 10 –8 )( .5 ´ 10 –13 )( . ) + ( .32 ´ –8 )( .00 10 –14 )
´
[HO + ] =
3
. 0 10 + . 6 32 ´ 10 –8
= . 186 ´ 10 –10
or a pH of 9.73.
6G.6 Effect of Complexation on Solubility
The solubility of a precipitate can be improved by adding a ligand capable of
forming a soluble complex with one of the precipitate’s ions. For example, the
solubility of AgI increases in the presence of NH 3 due to the formation of the
+
soluble Ag(NH 3 ) 2 complex. As a final illustration of the systematic approach
to solving equilibrium problems, let us find the solubility of AgI in 0.10 M
NH 3 .
We begin by writing the equilibria that we need to consider
–
+
AgI(s) t Ag (aq)+I (aq)
+
+
Ag (aq) + 2NH 3 (aq) t Ag(NH 3 ) 2 (aq)
–
+
NH 3(aq)+H 2O(l) t OH (aq)+NH 4 (aq)
+
–
2H 2O(l) t H 3O (aq)+OH (aq)