Homework Answers
1)
use:
pH = -log [H+]
9.262 = -log [H+]
[H+] = 5.47*10^-10 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)/(5.47*10^-10)
[OH-] = 1.828*10^-5 M
B dissociates as:
B +H2O -----> BH+ + OH-
0.575 0 0
0.575-x x x
Kb = [BH+][OH-]/[B]
Kb = x*x/(c-x)
Kb = 1.828*10^-5*1.828*10^-5/(0.575-1.828*10^-5)
Kb = 5.812*10^-10
Answer: 5.812*10^-10
2)
use:
pH = -log [H+]
12.19 = -log [H+]
[H+] = 6.486*10^-13 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)/(6.486*10^-13)
[OH-] = 1.542*10^-2 M
B dissociates as:
B +H2O -----> BH+ + OH-
0.319 0 0
0.319-x x x
Kb = [BH+][OH-]/[B]
Kb = x*x/(c-x)
Kb = 1.542*10^-2*1.542*10^-2/(0.319-1.542*10^-2)
Kb = 7.829*10^-4
Answer: 7.829*10^-4
3)
use:
pH = -log [H+]
12.12 = -log [H+]
[H+] = 7.603*10^-13 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)/(7.603*10^-13)
[OH-] = 1.315*10^-2 M
B dissociates as:
B +H2O -----> BH+ + OH-
0.443 0 0
0.443-x x x
Kb = [BH+][OH-]/[B]
Kb = x*x/(c-x)
Kb = 1.315*10^-2*1.315*10^-2/(0.443-1.315*10^-2)
Kb = 4.024*10^-4
Answer: 4.024*10^-4
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