Methylamine, CH3NH2, has a Kb = 4.40 x 10-4.
What is the pH of a 0.250 M solution of
methylamine?
CH3NH2 dissociates as:
CH3NH2 +H2O -----> CH3NH3+ + OH-
0.25 0 0
0.25-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 ((4.4*10^-4)*0.25) = 1.049*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/(0.25-x)
1.1*10^-4 - 4.4*10^-4 *x = x^2
x^2 + 4.4*10^-4 *x-1.1*10^-4 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 4.4*10^-4
c = -1.1*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 = 4.402*10^-4
roots are :
x = 1.027*10^-2 and x = -1.071*10^-2
since x can't be negative, the possible value of x is
x = 1.027*10^-2
So, [OH-] = x = 1.027*10^-2 M
use:
pOH = -log [OH-]
= -log (1.027*10^-2)
= 1.9884
use:
PH = 14 - pOH
= 14 - 1.9884
= 12.0116
Answer: 12.01
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