What is the pKb of a weak base KA if the pH of a 0.079-M solution of KA is 9.40?
use:
pH = -log [H+]
9.4 = -log [H+]
[H+] = 3.981*10^-10 M
use:
[OH-] = Kw/[H+]
Kw is dissociation constant of water whose value is 1.01*10^-14 at
25 oC
[OH-] = (1.01*10^-14)/[H+]
[OH-] = (1.01*10^-14)/(3.981*10^-10)
[OH-] = 2.537*10^-5 M
A- dissociates as:
A- +H2O ----->
HA + OH-
7.9*10^-2
0 0
7.9*10^-2-x
x x
Kb = [HA][OH-]/[A-]
Kb = x*x/(c-x)
Kb = 2.537*10^-5*2.537*10^-5/(0.079-2.537*10^-5)
Kb = 8.15*10^-9
use:
pKb = -log Kb
= -log (8.15*10^-9)
= 8.0888
Answer: 8.09
Answer: 8.09
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