Page 180 - Modern Analytical Chemistry

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Chapter 6 Equilibrium Chemistry 163

Both assumptions are acceptable (again, verify that this is true). The
+
concentration of H 3 O is calculated using K w
K w . 100 ´ 10 –14
[HO + ] = – = –4 = . 107 ´ 10 –11
3
[OH ] . 935 ´ 10
giving a pH of 10.97.

CH 3 CH 3 CH 3
pK = 2.348 pK = 9.867
a1
a2
+ + – –
H N C COOH H N C COO H N C COO
3
3
2
H H H Figure 6.11
Acid–base equilibria for the
H L + HL L – amino acid alanine.
2
6G.5 pH of a Polyprotic Acid or Base
A more challenging problem is to find the pH of a solution prepared from a
polyprotic acid or one of its conjugate species. As an example, we will use the
amino acid alanine whose structure and acid dissociation constants are shown in
Figure 6.11.
CH 3
pH of 0.10 M H 2 L + Alanine hydrochloride is a salt consisting of the diprotic weak
+
+
–
acid H 2 L and Cl . Because H 2 L has two acid dissociation reactions, a complete H N C COO –
2
systematic solution to this problem will be more complicated than that for a mono- H
protic weak acid. Using a ladder diagram (Figure 6.12) can help us simplify the L –
+
–
problem. Since the areas of predominance for H 2 L and L are widely separated, we 9.867
+
can assume that any solution containing an appreciable quantity of H 2 L will con-
+
–
tain essentially no L . In this case, HL is such a weak acid that H 2 L behaves as if it CH 3
were a monoprotic weak acid.
+
To find the pH of 0.10 M H 2 L , we assume that pH + H N C COO –
3
H
+
–
[H 3 O ] >> [OH ] HL
+
Because H 2 L is a relatively strong weak acid, we cannot simplify the problem fur- 2.348
ther, leaving us with
CH 3
+ 2
[ HO ]
3
K a = + H N C COOH
+
C HL + –[ HO ] 3 H
3
2
H L +
+
Solving the resulting quadratic equation gives the [H 3 O ] as 1.91 ´10 –2 M or a 2
–
+
pH of 1.72. Our assumption that [H 3 O ] is significantly greater than [OH ] is Figure 6.12
acceptable. Ladder diagram for the amino acid alanine.
pH of 0.10 M L – The alaninate ion is a diprotic weak base, but using the ladder dia-
gram as a guide shows us that we can treat it as if it were a monoprotic weak base.
Following the steps in Example 6.11 (which is left as an exercise), we find that the
pH of 0.10 M alaninate is 11.42.