Question

Determine the pH of each of the following solutions. (a) 0.607 M hydrosulfuric acid (weak acid with Ka = 9.5e-08). (b) 0.155 M hypoiodous acid (weak acid with Ka = 2.3e-11). (C) 0.549 M pyridine (weak base with Kb = 1.7e-09).

Answer 1

a)
H2S dissociates as:

H2S          ----->     H+   + HS-
0.607                 0         0
0.607-x               x         x


Ka = [H+][HS-]/[H2S]
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 ((9.5*10^-8)*0.607) = 2.401*10^-4

since c is much greater than x, our assumption is correct
so, x = 2.401*10^-4 M



So, [H+] = x = 2.401*10^-4 M


use:
pH = -log [H+]
= -log (2.401*10^-4)
= 3.6195
Answer: 3.62

b)
HIO dissociates as:

HIO          ----->     H+   + IO-
0.155                 0         0
0.155-x               x         x


Ka = [H+][IO-]/[HIO]
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 ((2.3*10^-11)*0.155) = 1.888*10^-6

since c is much greater than x, our assumption is correct
so, x = 1.888*10^-6 M



So, [H+] = x = 1.888*10^-6 M


use:
pH = -log [H+]
= -log (1.888*10^-6)
= 5.724
Answer: 5.72

c)
C5H5N dissociates as:

C5H5N +H2O     ----->     C5H5NH+   +   OH-
0.549                   0         0
0.549-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)*0.549) = 3.055*10^-5

since c is much greater than x, our assumption is correct
so, x = 3.055*10^-5 M



So, [OH-] = x = 3.055*10^-5 M


use:
pOH = -log [OH-]
= -log (3.055*10^-5)
= 4.515


use:
PH = 14 - pOH
= 14 - 4.515
= 9.485
Answer: 9.48