Question

Kb for NH3 is 1.8 X 10^-5. What is the pOH of a 0.50 M aquesous solution of NH4CL at 25 degree C?
a) 9.22
b) 11.23
c) 11.48
d) 4.78
e) 2.52

Please show work, I have gotten mulitple answers different answers so I'm confused on which steps I am doing or not doing. Thanks!!!

Answer 1

NH4Cl is a strong salt which dissociates as
NH4Cl ----> NH4+ + Cl-
NH4+ + H2O <-------> NH3 + H3O+

NH4+(aq)+H2O(l)⇌NH3(aq)+H3O+(aq)
I.......0.50.................0............0
C......(-x)................(+x)...........(+x)
E.....0.50-x...............x................x

The acid dissociation constant Ka is given by

Ka = Kw/Kb = 1.0 x 10^-14 / 1.8 x 10^-5 =5.6 x 10^-10

Ka = [NH3][H3O+]/[NH4+]

5.6 x 10^-10 = x^2 / 0.50-x
x = [H3O+]= 1.673 x 10^-5 M
pH = - log[H+] = 1.673 x 10^-5 = 4.78

pOH = 14-4.78 = 9.22

The answer is a) 9.22