Question

A 0.500 m solution of a weak base has a ph of 9.75. what is the base hydrolysis constant, kb, for the weak base?

Answer 1

when we have PH = 9.75
So we can get POH = 14 - 9.75 = 4.25
and when POH = - ㏒[OH-]
by substitution:
4.25 = -㏒[OH-]
∴[OH] = 5.6x10^-5
from this reaction equation:
BOH ↔ B+ + OH-
∴[OH-] = [B+]= 5.6x10^-5 M
and Equ [BOH] = 0.5 m - X
= 0.5 - (5.6x10^-5)
= 0.4999
∴ Kb = [OH-][B+]/[BOH]
= (5.6x10^-5)^2 / 0.4999
= 6.27x10^-9