Question

Determine the pH of each of the following solutions.

(a) 0.532 M phenol (weak acid with K_a = 1.3e-10).





(b) 0.277 M hypobromous acid (weak acid with K_a = 2.5e-09).





(c) 0.713 M pyridine (weak base with K_b = 1.7e-09).

Answer 1

a)
C6H5OH dissociates as:

C6H5OH          ----->     H+   + C6H5O-
0.531                 0         0
0.531-x               x         x


Ka = [H+][C6H5O-]/[C6H5OH]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((1.3*10^-10)*0.531) = 8.308*10^-6

since c is much greater than x, our assumption is correct
so, x = 8.308*10^-6 M



So, [H+] = x = 8.308*10^-6 M


use:
pH = -log [H+]
= -log (8.308*10^-6)
= 5.0805
Answer: 5.08

b)
HA dissociates as:

HA          ----->     H+   + A-
0.277                 0         0
0.277-x               x         x


Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((2.5*10^-9)*0.277) = 2.632*10^-5

since c is much greater than x, our assumption is correct
so, x = 2.632*10^-5 M



So, [H+] = x = 2.632*10^-5 M


use:
pH = -log [H+]
= -log (2.632*10^-5)
= 4.5798
Answer: 4.58

c)
B dissociates as:

B +H2O     ----->     BH+   +   OH-
0.713                   0         0
0.713-x                 x         x


Kb = [BH+][OH-]/[B]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((1.7*10^-9)*0.713) = 3.482*10^-5

since c is much greater than x, our assumption is correct
so, x = 3.482*10^-5 M



So, [OH-] = x = 3.482*10^-5 M


use:
pOH = -log [OH-]
= -log (3.482*10^-5)
= 4.4582


use:
PH = 14 - pOH
= 14 - 4.4582
= 9.5418
Answer: 9.54