Question

What is the pH of an aqueous 0.032 M pyridine, (C5H5N)? (Kb for pyridine = 1.7 x 10–9)

Answer 1

pH + H₂O ⇋ [pOH]⁺ + OH⁻

Kb = [pOH⁺][OH⁻] / [ pH] = 1.7×10⁻⁹

Let x mol dissociate:

[pOH⁺] = [OH⁻] = x [pH] = 0.032 - x ≈ 0.032

[x] [x] / [0.032] = 1.7×10⁻⁹

x₂ = 0.032 * 1.7 x 10⁻⁹

x = 5.44 x 10⁻¹¹ = [OH⁻]

pOH = -log [OH⁻] = -log[ 5.44 x 10⁻¹¹] = 10.26

pH = 14.00 - pOH

pH = 14.00 - 10.26

PH = 3.74

hope this helps!