Pkb = 4.2
Pka = 14-Pkb
      = 14-4.2  =
9.8
Pka = 9.8
-logKa = 9.8
     Ka =
10-9.8Â Â = 1.58*10-10
                 Â
(CH3)3NH+ + H2O ------> (CH3)3N + H3O+
      Â
IÂ Â Â Â Â Â Â Â Â Â
0.5MÂ 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
0
     Â
CÂ Â Â Â Â Â Â Â Â Â
-x                                  Â
+x               Â
+x
     Â
EÂ Â Â Â Â Â Â Â
0.5-x                               Â
+x              Â
+x
                      Â
Ka =Â Â [(CH3)3N] [H3O+ ]/[(CH3)3NH+
]
                      Â
1.58*10-10Â Â = x*x/0.5-x
                       Â
1.58*10-10 *(0.5-x) = x2
                     Â
x   = 8.8*10-6
               Â
[H3O+] = x = 8.8*10-6 M
                Â
pHÂ Â = -log[H3O+]
                        Â
= -log8.8*10-6
                         Â
= 5.0555
            Â
[OH-] = Kw/[H3O+]
                      Â
= 1*10-14/8.8*10-6
                      Â
= 1.13*10-9 M
   [Cl-]  = 0.5M
B) the concentration of trimethylammonium ions  =
0.5M
C) the concentration of trimethylamine in equilibrium = 0.5-x =
0.5-0.0000088 = 0.4999M