Question

Calculate the pH and concentrations of CH3NH2 and CH3NH3 in a 0.0395 M methylamine (CH3NH2) solution. The Kb of CH3NH2 = 4.47 × 10^-4.

Answer 1

CH3NH2 dissociates as:

CH3NH2 +H2O     ----->     CH3NH3+   +   OH-
3.95*10^-2                   0         0
3.95*10^-2-x                 x         x


Kb = [CH3NH3+][OH-]/[CH3NH2]
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 ((4.47*10^-4)*3.95*10^-2) = 4.202*10^-3

since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Kb = x*x/(c-x)
4.47*10^-4 = x^2/(3.95*10^-2-x)
1.766*10^-5 - 4.47*10^-4 *x = x^2
x^2 + 4.47*10^-4 *x-1.766*10^-5 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 4.47*10^-4
c = -1.766*10^-5

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 = 7.083*10^-5

roots are :
x = 3.984*10^-3 and x = -4.431*10^-3

since x can't be negative, the possible value of x is
x = 3.984*10^-3

so,
[CH3NH3+] = x = 3.98*10^-3 M
[CH3NH2] = 0.0395 - x = 0.0395 - 3.984*10^-3 = 0.0355 M