Question

Find the [OH−] in a 0.400 M solution of ethylamine (C2H5NH2). For ethylamine, Kb=5.6×10−4.

Answer 1

C2H5NH2 dissociates as:

C2H5NH2 +H2O     ----->     C2H5NH3+   +   OH-
0.4                   0         0
0.4-x                 x         x


Kb = [C2H5NH3+][OH-]/[C2H5NH2]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((5.6*10^-4)*0.4) = 1.497*10^-2

since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Kb = x*x/(c-x)
5.6*10^-4 = x^2/(0.4-x)
2.24*10^-4 - 5.6*10^-4 *x = x^2
x^2 + 5.6*10^-4 *x-2.24*10^-4 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 5.6*10^-4
c = -2.24*10^-4

Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 8.963*10^-4

roots are :
x = 1.469*10^-2 and x = -1.525*10^-2

since x can't be negative, the possible value of x is
x = 1.469*10^-2

so,
[OH-] = x = 1.47*10^-2 M

Answer: 1.47*10^-2 M