Calculating Kb of a weak base?

Question

Could someone please help me step-by-step through this sort of problem? I'm having trouble wraping my head around it.

Here's an example:
A 0.35 M solution of an unknown weak base has a pOH of 5.75. Calculate the Kb.

Answer 1

XOH <----> X+ + OH-
pOH = 5.75 => [OH-] = 10^-5.75 = 1.78 x 10^-6 M = [X+]
[XOH] = 0.35 - 1.78 x 10^-6 = 0.35 M
Kb = ( 1.78 x 10^-6)^2 / 0.35 = 9.05 x 10^-10

Answer 2

Recognize first that the reaction you are dealing with is:

BOH(aq) --> B+(aq) + OH-(aq)

Since you are dealing with a weak base, the reaction only goes a very little bit to the right (the equilibrium lies very far to the left).

The equilibrium constant expression for this reaction will be:

Kb = [B+][OH-]/[BOH]

So, convert your pOH into [OH-].
[B+] will be equal to [OH-] because every time you form an OH- ion, you also produce a B+ ion.
[BOH] will not change significantly from the 0.35 M that you began with.

Plug those values into your expression for Kb, and you're done.