What is the ph of a 0.21 m solution of pyridinium chloride (c5h5nh+cl-) ? (kb for pyridine is 1.5 * 10-9.) answer in units of ph?

Chemical reaction: C₅H₅NH⁺ + H₂O → C₅H₅N + H₃O⁺

c (C₅H₅NH⁺) = 0,21 M = 0,21 mol/dm³.

Kb (pyridine - C₅H₅N) = 1,5·10⁻⁹.

Ka (C₅H₅NH⁺) = 10⁻¹⁴ / 1,5·10⁻⁹.

Ka (C₅H₅NH⁺) = 6,67·10⁻⁶.

c (C₅H₅N) = c (H₃O⁺) = x.

Kb = c (C₅H₅N) · c (H₃O⁺) / c (C₅H₅NH⁺).

6,67·10⁻⁶ = x² / (0,21 M - x).

Solve quadratic equation: x = c (H₃O⁺) = 1,1·10⁻³ mol/dm³.

pH = - logc (H₃O⁺) = - log (1,1·10⁻³ mol/dm³).

pH = 2,9.