Answer : C) 1 & 2 only
Explanation
:
CH3NH2Â Â Â +Â Â H2OÂ Â Â
------------->Â Â CH3NH3+ +Â Â OH-
 Â
1Â 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 Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0
1 -
x                                                    Â
x               Â
x
Kb = x^2 / 1 - x
5 x 10^-4 = x^2 / 1 - x
x = 0.0223
[OH-] = [CH3NH3+] = 0.0223 M
[OH-] / [CH3NH3+] = 1
% ionization = 2.23 %
pH = 12.35
if concentration is = 1/2 M
CH3NH2Â Â Â +Â Â H2OÂ Â Â
------------->Â Â CH3NH3+ +Â Â OH-
  0.5
                                                     Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0
0.5-
x                                                    Â
x               Â
x
Kb = x^2 / 0.5 - x
5 x 10^-4 = x^2 / 0.5 - x
x = 0.0158
[OH-] = [CH3NH3+] = 0.0158 M
[OH-] / [CH3NH3+] = 1
% ionization = 3.16
pH = 12.20
here
[OH-] / [CH3NH3+] is constant
pH decreases.