#degree of ionization

A weak base BOH ( Kb = 2.0 × 10⁻⁵ ) has a 0.1 M solution with pH 10.5. What is the degree of ionization?

pOH = 14 - 10.5 = 3.5 , [OH-] = 10⁻³.⁵ ≈ 3.16 × 10⁻⁴ , α = ([OH-]/[BOH]) = (3.16 × 10⁻⁴/0.1) = 3.16 × 10⁻³ = 0.00316 .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases