Question

2. The pH of a 1.53 M solution of a weak base B is measured to be 10.74 Determine Kb of the base Determine K, of the weak bases's conjugate acid, HB* Determine (H+] in a 2.00 M solution of the chloride salt, HBCI Determine the pH of a 2.00 M solution of the chloride salt, HBCI

Answer 1

1)

use:
pH = -log [H+]
10.74 = -log [H+]
[H+] = 1.82*10^-11 M

use:
[OH-] = Kw/[H+]
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
[OH-] = (1.0*10^-14)/[H+]
[OH-] = (1.0*10^-14)/(1.82*10^-11)
[OH-] = 5.495*10^-4 M
B dissociates as:

B +H2O     ----->     HB+   +   OH-
1.53                   0         0
1.53-x                 x         x


Kb = [HB+][OH-]/[B]
Kb = x*x/(c-x)
Kb = 5.495*10^-4*5.495*10^-4/(1.53-5.495*10^-4)
Kb = 1.975*10^-7
Answer: 1.98*10^-7

2)
Given:
Kb = 1.975*10^-7

use:
Ka = Kw/Kb
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
Ka = (1.0*10^-14)/Kb
Ka = (1.0*10^-14)/1.975*10^-7
Ka = 5.063*10^-8
Answer: 5.06*10^-8

3)
HB+ dissociates as:

HB+          ----->     H+   + B
2                 0         0
2-x               x         x


Ka = [H+][B]/[HB+]
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 ((5.063*10^-8)*2) = 3.182*10^-4

since c is much greater than x, our assumption is correct
so, x = 3.182*10^-4 M



So, [H+] = x = 3.182*10^-4 M
Answer: 3.18*10^-4 M

4)
use:
pH = -log [H+]
= -log (3.182*10^-4)
= 3.4973
Answer: 3.50