If the kb of a weak base is 1.1 × 10-6 m, what is the ph of a 0.49 m solution of this base?

When Kb = 1.1 x 10^-6

∴ Ka = Kw / Kb

        = 1.1 x 10^-14 / (1.1 x 10 ^-6) 

       = 1 x 10^-8

when the reaction equation when we assume the base is B

B + H2O ↔ BH+   + OH-

∴ Ka = [BH+][OH-]/[B]

1 x 10^-8 = X^2 / 0.49

X^2 = 4.9 x 10^-9

∴X = 7 x 10^-5

∴[OH-] = 7 x 10^-5

∴POH = -㏒[OH]

           = -㏒7 x 10^-5

           = 4.15

∴ PH = 14 - POH

             14 - 4.15

         = 9.85