What is the value of Kb for a base that is 0.021 % ionized in a 0.037 M solution?

Answer 1

First of all we have to assume for simplicity that we are talking about a monovalent base BOH
.. .. .. .. .. BOH <=> B+ +OH-
Initial .. .. .. C
Dissociate .x
Produce .. .. .. .. .. .. x .. .. ..x
At Equil .. .C-x .. .. .. x .. .. ..x

Kb= [B+][OH-] /[BOH] = x^2/(C-x)
The degree of dissociation is a=x/C => x=aC
Substitute x in the Kb equation and you get

Kb= (aC)^2/(C-aC) =a^2C/(1-a) =
= (0.00021^2)*0.037 / (1-0.00021) =1.63*10^-9