Given :
[NH3] = 0.150 M
Ka = 5.55 E-10
Solution:
Kb is calculated as
Kb = 1.0 E-14 /ka = 1.0 E-14/5.55 E-10 = 1.8 E-5
ICE
              Â
NH3 Â Â Â Â +
         Â
H2OÂ Â ----- > NH4+Â Â + OH-
IÂ Â Â Â Â Â Â Â Â Â Â Â Â Â
0.150-x                                               Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â
0
CÂ Â Â Â Â Â Â Â Â Â Â Â
-x                                                          Â
+x         Â
+x
E
           Â
(0.150-x)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
x            Â
x
Ka = 1.8E-5 = x^2 /
0.150Â Â Â Â Â Â Â Â Â Â
since ka is very small we can neglect x in the denominator.
x = 1.644 E-3
[OH-] = 1.644E-3 M
pOH = - log (1.644 E-3 ) = 2.87
pH=14-2.87 = 11.2
pH = 11.2
