Question

Enter your answer in the provided box.

What is the original molarity of an aqueous solution of ammonia

(

NH₃

)

whose pH is 11.16 at 25

°

C?

(

K_b for NH₃ = 1.8

×

10⁻⁵

)

M

Answer 1

Let the concentration of NH3 c molar

use:

pH = -log [H+]

11.16 = -log [H+]

[H+] = 6.918*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)/(6.918*10^-12)

[OH-] = 1.445*10^-3 M

NH3 dissociates as:

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

c 0 0

c-x x x

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

Kb = x*x/(c-x)

1.8*10^-5 = 1.445*10^-3*1.445*10^-3/(c-1.445*10^-3)

c-1.445*10^-3 = 0.1161

c=0.1175

Answer: 0.117 M