32)
pH = 10.63
[H+] = 10^-pH
        = 10^-10.63
         = 2.34 x
10^-11 M
[H+] = 2.34 x 10^-11 M
33)
[H+] = 2.34 x 10^-11 M
[OH-] = Kw / [H+]
        = 1.0 x 10^-14
/ 2.34 x 10^-11
       = 4.27 x 10^-4 M
[OH-]= 4.27 x 10^-4 M = x
NH4OH (aq) ----------------> NH4+(aq) + OH-(aq)
CÂ 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 ---------> initial (I)
-x                                      Â
+x              Â
+x ----------> change (C)
C-x                                    Â
x                Â
x ------------> equilibrium (E)
Kb = [NH4+][OH-]/[NH4OH]
Kb = x^2 / C-x
here x is know from above problem but initial concentration of
NH4OH is not given so it is needed to calculate Kb
   Â
