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For BOH B+ + OH- , Kb = (x²/0.01 - x) ≈ (x²/0.01) = 4.0 × 10⁻⁶ . Solving, x = [OH-] = sqrt4.0 × 10⁻⁸ = 2.0 × 10⁻⁴ . pOH = -log(2.0 × 10⁻⁴) ≈ 3.7 , pH = 14 - 3.7 = 10.3 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
α = 0.005 , [OH-] = 0.02 × 0.005 = 1.0 × 10⁻⁴ , pOH = 4 , pH = 14 - 4 = 10 . Check: Kb = ((1.0 × 10⁻⁴)²/0.02) = 5.0 × 10⁻⁷ (discrepancy, but closest).
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
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