Â
NH3Â Â Â Â +Â Â H2O -------------->
NH4+ + OH-
  0.126
MÂ 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
0Â Â Â Â Â Â Â Â Â 0
             Â
0.126-x                                 Â
  x       Â
x
     Kb = [NH4+] [OH-] /[NH3]
1.8 x 10-5 = x.x / 0.126-x Â
Since x is very small, 0.126 -x = 0.126
  Then,
1.8 x 10-5 = x.x / 0.126
  x = 0.0015 M
[OH-] = 0.0015 M
pOH = - log [OH-] = - log (0.0015) = 2.82
Hence,
pH = 14 -pOH= 14- 2.82= 11.18
Therefore,
Initial pH = 11.18
--------------------------------------
Ans = a
