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!
