Hydrolysis
361
= 5.75 x 10-0 =
^
1.74 x lO"
5
'
(0.0500 -x)
Neglecting x compared to 0.0500, we have
=
5
-
75 x 1(r
'
0
x- = 2.88 x 10~
n
x = [OH-] = 5.36 x 106 M
pOH = -log (5.36 x 106
) = 5.27
pH = 14.00 - 5.27 = 8.73
PROBLEM:
If 25.0 ml of 0.200 M NaOH are added to 50.0 ml of 0.100 M HC 2 H 3 O 2 , what is
the pH of the resulting solution?
SOLUTION:
We must first find the number of moles of acid and base used, in order to determine whether there is an excess of either one.
Moles of NaOH added = (0.0250 liter) (o.200 j
= 0.00500 mole
Moles of HC 2 H 3 O 2 added = (0.0500 liter) ( 0.100
-H = 0.00500 mole
Because equal numbers of moles of acid and base are added, only the salt is
present; 0.00500 mole of NaC 2 H 3 O 2 in 0.0750 liter to give
„
0.00500 mole
[C2H3 °'
] = 0.0750 liter = °'
0667 M
Because only a salt is present in solution, the problem involves the hydrolysis
equilibrium
0.0667 - x
x
x
C 2 H 3 O 2 - + H 2 O *± HC 2 H 3 O 2 + OHjust as in the last problem. Solving in the same manner gives
K h = 5.75 x 10-'° = 0.0667 - x
0.0667
x * = (0.0667)(5.75 x IQ-'
0 ) = 3.84 x 10-"
x = [OH'] = 6.19 x 10B M
pOH = -log (6.19 x 1Q6 ) = 5.21
pH = 14.00 - 5.21 = 8.79