Question

21.

A)In the laboratory, a general chemistry student measured the pH of a 0.475 M aqueous solution of triethylamine, (C₂H₅)₃N to be 12.210.
Use the information she obtained to determine the K_b for this base.
K_b(experiment) =

B)In the laboratory, a general chemistry student measured the pH of a 0.475 M aqueous solution of ammonia to be 11.453.
Use the information she obtained to determine the K_b for this base.
K_b(experiment) =

Answer 1

A)

use:

pH = -log [H+]

12.21 = -log [H+]

[H+] = 6.166*10^-13 M

use:

[OH-] = Kw/[H+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H+]

[OH-] = (1.0*10^-14)/(6.166*10^-13)

[OH-] = 1.622*10^-2 M

(C2H5)3N dissociates as:

(C2H5)3N +H2O -----> (C2H5)3NH+ + OH-

0.475 0 0

0.475-x x x

Kb = [(C2H5)3NH+][OH-]/[(C2H5)3N]

Kb = x*x/(c-x)

Kb = 1.622*10^-2*1.622*10^-2/(0.475-1.622*10^-2)

Kb = 5.733*10^-4

Answer: 5.73*10^-4

B)

use:

pH = -log [H+]

11.45 = -log [H+]

[H+] = 3.524*10^-12 M

use:

[OH-] = Kw/[H+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H+]

[OH-] = (1.0*10^-14)/(3.524*10^-12)

[OH-] = 2.838*10^-3 M

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)

Kb = 2.838*10^-3*2.838*10^-3/(0.475-2.838*10^-3)

Kb = 1.706*10^-5

Answer: 1.71*10^-5