Question

If the Kb of a weak base is 6.5 × 10-6, what is the pH of a 0.26 M solution of this base?

Answer 1

                       let weak base is MH3

            MH3 + H2O -----------> MH4^+ + OH^-

I         0.26                                     0           0

C         -x                                       +x           +x

E       0.26-x                                   +x           +x

                 Kb   =   [MH4^+][OH^-]/[MH3]

                 6.5*10^-6   = x*x/0.26-x

                 6.5*10^-6 *(0.26-x)   = x^2

                  x   = 0.0013

       [OH^-]    = x   = 0.0013M

         POH   = -log[OH^-]

                      = -log0.0013

                       = 2.886

         PH         = 14-POH

                       = 14-2.886    = 11.114