A Weak base is added to a 0.20M Concentration. The resulting solution has a [OH-] = 3.16x10^-6. What is the Kb and pH?

Denote the weak base as B. Then, the dissociation of the weak base can be represented by the following equation :
B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)

At equilibrium :
[OH⁻] = 3.16 × 10⁻⁶ M
Thus, [BH⁺] = [OH⁻] = 3.16 × 10⁻⁶ M
and [B] = 0.20 - (3.16 × 10⁻⁶) ≈ 0.20 M

Kb = [BH⁺] [OH⁻] / [B] = (3.16 × 10⁻⁶)² / 0.20 = 4.99 × 10⁻¹¹ (M)

pOH = -log[OH⁻] = -log(3.16 × 10⁻⁶) = 5.5
pH = 14.0 - 5.5 = 8.5