What is the ph of an aqueous 0.032 m pyridine, (c5h5n)? (kb for pyridine = 1.7 x 10–9)

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!