Since Ka*Kb = 10-14
Thus, Kb = 10-14/(5.7*10-10) =
1.75*10-5
The reaction taking place is :
                    Â
NH3 + H2O ----> NH4+
+ OH-
Initial              Â
1.3Â 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 0
Eqb              Â
1.3-x                Â
x          x
Thus, Kb =
[NH4+]*[OH-]/[NH3] =
x*x/(1.3-x) = 1.75*10-5
Solving we get :
x = [OH] = 4.76*10-3 M
Thus, pOH = -log([OH-]) = 2.32
Thus, pH = 14- pOH = 11.68
