Practice question
Question
A weak base BOH has Kb = 4.0 × 10⁻⁴ . If its 0.02 M solution has a pH of 10.8, what is the percentage ionization?
Explanation
pH = 10.8 , pOH = 14 - 10.8 = 3.2 , [OH-] = 10⁻³.² ≈ 6.31 × 10⁻⁴ . [BOH] = 0.02 - 6.31 × 10⁻⁴ ≈ 0.0194 . Kb = ([OH-]²/[BOH]) = ((6.31 × 10⁻⁴)²/0.0194) ≈ 2.05 × 10⁻⁵ , but given Kb = 4.0 × 10⁻⁴ , ionization α = ([OH-]/[BOH]initial) = (6.31 × 10⁻⁴/0.02) = 0.03155 , % = 3.155%.
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