Question

Determine the pH of each of the following solutions.

Determine the pH of each of the following solutions. (a) 0.211 M hypobromous acid (weak acid with Ka = 2.5e-09) Ка 3 Зе-08). (b) 0.235 M hypochlorous acid (weak acid with Ka (c) 0.586 M pyridine (weak base with K 1.7e-09)

Answer 1

a)
HBrO dissociates as:

HBrO          ----->     H+   + BrO-
0.211                 0         0
0.211-x               x         x


Ka = [H+][BrO-]/[HBrO]
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)*0.211) = 2.297*10^-5

since c is much greater than x, our assumption is correct
so, x = 2.297*10^-5 M



So, [H+] = x = 2.297*10^-5 M


use:
pH = -log [H+]
= -log (2.297*10^-5)
= 4.6389
Answer: 4.64

b)
HClO dissociates as:

HClO          ----->     H+   + ClO-
0.235                 0         0
0.235-x               x         x


Ka = [H+][ClO-]/[HClO]
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 ((3*10^-8)*0.235) = 8.396*10^-5

since c is much greater than x, our assumption is correct
so, x = 8.396*10^-5 M



So, [H+] = x = 8.396*10^-5 M


use:
pH = -log [H+]
= -log (8.396*10^-5)
= 4.0759
Answer: 4.08

c)
C5H5N dissociates as:

C5H5N +H2O     ----->     C5H5NH+   +   OH-
0.586                   0         0
0.586-x                 x         x


Kb = [C5H5NH+][OH-]/[C5H5N]
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)*0.586) = 3.156*10^-5

since c is much greater than x, our assumption is correct
so, x = 3.156*10^-5 M



So, [OH-] = x = 3.156*10^-5 M


use:
pOH = -log [OH-]
= -log (3.156*10^-5)
= 4.5008


use:
PH = 14 - pOH
= 14 - 4.5008
= 9.4992
Answer: 9.50