Question

What are the equilibrium concentrations of NH3, NH4+, and OH-  in a 0.95 M solution of ammonia? Kb = 1.8x10^-5

What is the pH of the solution? pH =

Answer 1

NH3 dissociates as:

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

0.95 0 0

0.95-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)*0.95) = 4.135*10^-3

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

so, x = 4.135*10^-3 M

So, [OH-] = x = 4.135*10^-3 M

[NH4+] = x = 4.135*10^-3 M

[NH3] = 0.95 - x

= 0.95 - 4.135*10^-3 M

= 0.946 M

use:

pOH = -log [OH-]

= -log (4.135*10^-3)

= 2.3835

use:

PH = 14 - pOH

= 14 - 2.3835

= 11.6165

Answer:

[NH3] = 0.95 M

[NH4+] = 4.1*10^-3 M

[OH-] = 4.1*10^-3 M

PH = 11.62