Question

24.

A)The pH of an aqueous solution of 0.475 M triethylamine (a weak base with the formula (C₂H₅)₃N) is_____?

B)The hydroxide ion concentration, [OH^-], of an aqueous solution of 0.475 M ammonia is ___________M.

Answer 1

A)

kb of triethylamine = 5.2*10^-4

(CH3)3N dissociates as:

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

0.475 0 0

0.475-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 ((5.2*10^-4)*0.475) = 1.572*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)

5.2*10^-4 = x^2/(0.475-x)

2.47*10^-4 - 5.2*10^-4 *x = x^2

x^2 + 5.2*10^-4 *x-2.47*10^-4 = 0

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

a = 1

b = 5.2*10^-4

c = -2.47*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 = 9.883*10^-4

roots are :

x = 1.546*10^-2 and x = -1.598*10^-2

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

x = 1.546*10^-2

So, [OH-] = x = 1.546*10^-2 M

use:

pOH = -log [OH-]

= -log (1.546*10^-2)

= 1.8108

use:

PH = 14 - pOH

= 14 - 1.8108

= 12.1892

Answer: 12.19

B)

Kb of NH3 = 1.8*10^-5

NH3 dissociates as:

NH3 +H2O -----> NH4+ + OH-

0.475 0 0

0.475-x x x

Kb = [NH4+][OH-]/[NH3]

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 ((1.8*10^-5)*0.475) = 2.924*10^-3

since c is much greater than x, our assumption is correct

so, x = 2.924*10^-3 M

So, [OH-] = x = 2.924*10^-3 M

Answer: 2.92*10^-3 M