Consider a weak base "B". What is the pH of a 0.118 mol L-1 solution of B? The KB of B is 5.14 x 10-7?

_______________ B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq) …… Kb = 5.14 × 10⁻⁷
Initial: ________ 0.118 M ________ 0 M _____ 0 M
Change: ________ -y M _________ +y M ____ +y M
Equilibrium: __ (0.118 - y) M ______ y M ____ y M

As Kb is very small, the dissociation of B is to a very small extent.
We can assume that 0.118 ≫ y
and thus [B] at equilibrium = (0.118 - y) M ≈ 0.118 M

At equilibrium: Kb = [BH⁺] [OH⁻] / [B]
5.14 × 10⁻⁷ = y² / 0.118
y = √{(5.14 × 10⁻⁷) × 0.118} = 2.46 × 10⁻⁴

pOH = -log[OH⁻] = -log(2.46 × 10⁻⁴) = 3.61
pH = pKw - pOH = 14.00 - 3.61 = 10.39