Let the concentration of B c molar
use:
pH = -log [H+]
10.06 = -log [H+]
[H+] = 8.71*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)/(8.71*10^-11)
[OH-] = 1.148*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)
9*10^-7 = 1.148*10^-4*1.148*10^-4/(c-1.148*10^-4)
c-1.148*10^-4 = 1.465*10^-2
c=1.48*10^-2 M
Answer: 0.0148 M
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