What is the kb of a 0.0200 m ( at equilibrium) solution of methyl amine, ch3nh2, that has a ph of 11.40?

Answer 1

I wonder if you mean 0.0200 m or 0.002 M? Do you know there is a difference? I will assume you mean M.
..CH3NH2 + HOH --> CH3NH3^+ + OH^-
I..0.02.............0..........0
C...-x..............x..........x
E..0.02-x...........x..........x

Kb = (CH3NH3^+)(OH^-)/(CH3NH2)
Substitute the E line into the Kb expression and solve for Kb. You will need to convert pH to pOH and use pOH = -log(OH^-). That gives you x.