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
