Question

Enter your answer in the provided box. What is the original molarity of an aqueous solution of ammonia (NH3) whose pH is 11.10 at 25°C? (Kb for NH3 = 1.8 × 10−5) M

Answer 1

Let the concentration of NH3 c molar

use:
pH = -log [H+]
11.1 = -log [H+]
[H+] = 7.943*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)/(7.943*10^-12)
[OH-] = 1.259*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 = 1.259*10^-3*1.259*10^-3/(c-1.259*10^-3)
c-1.259*10^-3 = 8.805*10^-2
c=8.931*10^-2 M
Answer: 8.93*10^-2 M