Question

Calculate the pH of a 0.0198 M aqueous solution of dimethylamine ((CH),NH, Kn=5.9x104) and the equilibrium concentrations of the weak base and its conjugate acid. PH M [(CH3)2NH)equilibrium [(CH3)2NH2 Lequilibrium = C =

Answer 1

(CH3)2NH dissociates as:

(CH3)2NH +H2O -----> (CH3)2NH2+ + OH-

1.98*10^-2 0 0

1.98*10^-2-x x x

Kb = [(CH3)2NH2+][OH-]/[(CH3)2NH]

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 ((5.9*10^-4)*1.98*10^-2) = 3.418*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)

5.9*10^-4 = x^2/(1.98*10^-2-x)

1.168*10^-5 - 5.9*10^-4 *x = x^2

x^2 + 5.9*10^-4 *x-1.168*10^-5 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 5.9*10^-4

c = -1.168*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 = 4.708*10^-5

roots are :

x = 3.136*10^-3 and x = -3.726*10^-3

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

x = 3.136*10^-3

So, [OH-] = x = 3.136*10^-3 M

use:

pOH = -log [OH-]

= -log (3.136*10^-3)

= 2.50

use:

PH = 14 - pOH

= 14 - 2.50

= 11.50

[(CH3)2NH] = 0.0198 - x

= 0.0198 - 3.136*10^-3 M

= 0.0167 M

[(CH3)2NH] = x

= 3.136*10^-3 M

Answer:

PH = 11.50

[(CH3)2NH] = 0.0167 M

[(CH3)2NH] = 3.14*10^-3 M