Question

Calculate the values of the hydroxide ion concentration, pOH, and pH for a 0.20 M solution of ammonia

Answer 1

Kb for NH3 is 1.8*10^-5

NH3 dissociates as:

NH3 +H2O     ----->     NH4+   +   OH-
0.2                   0         0
0.2-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)*0.2) = 1.897*10^-3

since c is much greater than x, our assumption is correct
so, x = 1.897*10^-3 M



so,
[OH-] = x = 1.897*10^-3 M


use:
pOH = -log [OH-]
= -log (1.897*10^-3)
= 2.72


use:
PH = 14 - pOH
= 14 - 2.72
= 11.28

Answer:
[OH-] = 1.897*10^-3 M
pOH = 2.72

pH = 11.28