H2O → NH4+ + OH- KB  = [NH4+ ] [OH- ]    =  1.77 x 10-5           [NH3]Before neutralization,[NH4+ ]  = [OH-] = xThusKb  =   x2    = 1.77x 10-5        0.10Hence x = 1.33 x 10-5 = [OH-]  Now Kw = [H+][OH-]So [H+]  = Kw/ [OH-] = 10-14 / 1.33x 10-12 = 7.51 x 10-12       p H = - log(7.51 x 10-12) = 11.12On ADDITION of 25 ml of 0.1 M HCl solution to 50 ml of 0.1 M ammonia solution , 2.5 MOL of ammonia are neutralized. The resulting 75 ml solution CONTAINS the remaining unneutralised 2.5 mol of NH3 molecules and 2.5 mol of NH4+.                                         NH3    + HCl  → NH4+   + Cl-                                       2.5         2.5          0             0 At equilibrium                     0            0            2.5        2.5The resulting 75 ml of solution contains 2.5 mol of NH4+ ions and 2.5 mol of uneutralised NH3 molecules. This NH3 exists in the following equilibrium.NH4OH     ⇌      NH4+     + OH-0.033 M- y      y                  yWhere y = [OH-] = [NH4+]The FINAL 75ml solution after neutralization already contains 2.5 mol NH4+ ions, thus total concentration of NH4+ ions is given as:[NH4+]  = 0.033 + yAs y is small, [NH4OH] = 0.033 M and [NH4+] = 0.033MKb =  [NH4+]  [OH-]   = y (0.033)  = 1.77x 10-5 M            [NH4OH]          (0.033)Thus , y = 1.77 x 10-5 = [OH-][H+] = 10-14             = 0.56 x 10-9        1.77 x 10-5 Hence p H = 9.24