Question

Find the [OH−] in a 0.340 M solution of ethylamine (C2H5NH2). For ethylamine, Kb=5.6⋅10−4.

Answer 1

for simplicity lets write weak base as BOH

C2H5NH2 +H2O   ----->C2H5NH3+   +   OH-
0.3400                   0         0
0.3400-x                 x         x

Kb = [C2H5NH3+][OH-]/[C2H5NH2]
Kb = x*x/(c-x)
since kb is small, x will be small and it can be ignored as compared to c
So, above expression becomes

Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((5.60E-4)*0.3400) = 1.38E-2

[OH-] = x = 1.38E-2 M

Answer: 1.38*10^-2 M