Question

The pH of an aqueous solution of 0.313 M trimethylamine (a weak base with the formula (CH3)3N) is

Answer 1

(CH3)3N dissociates as:

(CH3)3N +H2O -----> (CH3)3NH+ + OH-

0.313 0 0

0.313-x x x

Kb = [(CH3)3NH+][OH-]/[(CH3)3N]

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 ((6.4*10^-5)*0.313) = 4.476*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)

6.4*10^-5 = x^2/(0.313-x)

2.003*10^-5 - 6.4*10^-5 *x = x^2

x^2 + 6.4*10^-5 *x-2.003*10^-5 = 0

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

a = 1

b = 6.4*10^-5

c = -2.003*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 = 8.013*10^-5

roots are :

x = 4.444*10^-3 and x = -4.508*10^-3

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

x = 4.444*10^-3

So, [OH-] = x = 4.444*10^-3 M

use:

pOH = -log [OH-]

= -log (4.444*10^-3)

= 2.3522

use:

PH = 14 - pOH

= 14 - 2.3522

= 11.65

Answer: 11.65