Calculate [H3O+]
In order to calculate hydronium ion concentration we use
following expression
pH= - log [H3O+]
Therefore
[H3O+]= antilog (-pH)
10^(-9.85)=1.41E-10
Using hydronium ion concentration, calculate hydroxide ion
concentration
[OH-]=1.0E-14/hydronium ion
concentration=1.0E-14/1.41E-10=7.1E-5
Now show dissociation reaction of given base.
Lets B is the base
B(aq)Â Â Â Â +
H2O(l)Â Â ----> BH+(aq)Â Â Â Â Â
+Â Â OH-(aq)
IÂ Â Â Â Â Â Â Â
0.400Â 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 Â Â Â Â Â Â Â
0
CÂ Â Â Â Â Â Â Â Â Â
-x                           Â
+x                      Â
+x
EÂ Â Â Â
(0.400-x)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
x                  Â
x
We know x = [OH-] = 7.1E-5 M
Equilibrium concentration of B(aq) = 0.400-7.1E-5=0.3999 M
[BH+]=x=7.1E-5 M
Use kb expression
Kb = [BH+][OH-]/[B]
(7.1E-5)^2/0.3999=1.25 E-8
So the kb of this base is 1.25 x 10^-8
