Question

A 0.900 M aqueous solution of a weak base is 14% dissociated. What is the base dissociation constant? (A= 2.1e-2)

I started by making an ICE Chart.
......B + H2O (l) <-> BH+ (aq) + OH- (aq)
I...0.900...-...............0................0
C..-x........-...............+x...............+x
E..0.900-x................x.................x
and that Kb = [BH+][OH-]/[B]
I know that % dissociation means x/initial concentration * 100 but I don't know what to do next.

Answer 1

You're ALMOST there.
(Base) = 0.900 M . Dissociated 14% or 0.14
So BH^+ will be 0.900 x 0.14 = 0.126 M; therefore, BH^+ is 0.126 and (OH^-) = 0.126 M and B = 0.900 - 0.126 = ?
Plug all of that into the Kb expression and solve for Kb. Post your work if you get stuck. Thanks for sharing your thoughts and exactly the problem you're having. That really helps me help you.