Let the concentration of B c molar
use:
pH = -log [H+]
10.31 = -log [H+]
[H+] = 4.898*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)/(4.898*10^-11)
[OH-] = 2.042*10^-4 M
B dissociates as:
B +H2O -----> BH+ + OH-
c 0 0
c-x x x
Kb = [BH+][OH-]/[B]
Kb = x*x/(c-x)
8.6*10^-7 = 2.042*10^-4*2.042*10^-4/(c-2.042*10^-4)
c-2.042*10^-4 = 4.847*10^-2
c=4.868*10^-2 M
Answer: 4.87*10^-2 M
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