Question

What is the pH of a 0.360 M solution of Methylamine?

Methylamine, CH3NH2, has a Kb = 4.40 x 10-4.

Thank you!

Answer 1

Lets write the dissociation equation of CH3NH2

CH3NH2 +H2O -----> CH3NH3+ + OH-

0.36 0 0

0.36-x x x

Kb = [CH3NH3+][OH-]/[CH3NH2]

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((4.4*10^-4)*0.36) = 1.259*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)

4.4*10^-4 = x^2/(0.36-x)

1.584*10^-4 - 4.4*10^-4 *x = x^2

x^2 + 4.4*10^-4 *x-1.584*10^-4 = 0

Let's solve this quadratic equation

Comparing it with general form: (ax^2+bx+c=0)

a = 1

b = 4.4*10^-4

c = -1.584*10^-4

solution of quadratic equation is found by below formula

x = {-b + √(b^2-4*a*c)}/2a

x = {-b - √(b^2-4*a*c)}/2a

b^2-4*a*c = 6.338*10^-4

putting value of d, solution can be written as:

x = {-4.4*10^-4 + √(6.338*10^-4)}/2

x = {-4.4*10^-4 - √(6.338*10^-4)}/2

solutions are :

x = 1.237*10^-2 and x = -1.281*10^-2

since x can't be negative, the possible value of x is

x = 1.237*10^-2

so.[OH-] = x = 1.237*10^-2 M

we have below equation to be used:

pOH = -log [OH-]

= -log (1.237*10^-2)

= 1.91

we have below equation to be used:

PH = 14 - pOH

= 14 - 1.91

= 12.09

Answer: 12.09