360
Acid-Base Equilibria
easily calculated from values of /£,. For example, if we take the product of the
expressions for K, and A: h for acetic acid and the acetate ion, we get
_ [H
+ ][C 2 H 3 0 2 -]
[HC 2 H 3 0 2 ][OH-] _ f
-
3 o 2j x
[C2 H 3 0 2 -]
-
[
from which we may solve for AT h :
This equation is a general expression, in which K, is the ionization constant
for the weak acid or weak base that is formed on hydrolysis. For the hydrolysis
of the C 2 H 3 O 2 " ion, we get
_ 1.00 x 1Q-" _
_ [HC 2 H 3 0 2 ][OH-]
Ah
1.74xlO-»
3 -
/3Xlu
-
[C 2 H 3 0 2 -]
For the hydrolysis of the NHJ ion, we get
_
" ~ 1.74 x 10-' ~ '
X 10
-
If the salt has ions of both a weak acid and a weak base,
1.00 x 1Q-"
_ [NH 3 ][H+]
-
We may use the hydrolysis constant to compute the pH of a salt solution, as
illustrated in the following problems.
PROBLEM:
Calculate the pH of a 0.0500 M NaC 2 H 3 O 2 solution.
SOLUTION:
Write the chemical equation, showing the equilibrium concentrations above the
symbols.
0.0500 - x
x
x
C 2 H 3 O 2 - + H 2 O *± HC 2 H 3 O 2 + OHAs in previous problems, we represent the small unknown concentration of OH'
by x. Because HC 2 H 3 O 2 is formed simultaneously with OH" and in equimolar
amounts, its concentration also is.r. The wrchydrolyzed concentration of C 2 H 3 02"
is 0.0500 - x. Substitution of these values into the K h expression gives