What is the ph of a 4.8 m pyridine solution that has kb = 1.9 * 10-9? the equation for the dissociation of pyridine is?

C5H5N (aq) + H2) ⇄C5H5NH + (aq) + OH - (aq)

Answer: 9.98

Explanation:

1) The equation for the dissociation of pyridine is:

C₅H₅N₅ (aq) + H₂O (l) ⇄ C₅H₅NH⁺ (aq) + OH⁻ (aq)

2) Kb equation:

Kb = [C₅H₅NH⁺ (aq) ] [OH⁻ (aq) ] / [C₅H₅N₅ (aq) ]

Where:

[C₅H₅NH⁺ (aq) ] = [OH⁻ (aq) ] ← from the equilibrium reaction

[C₅H₅N₅ (aq) ] = 4.8 M ← from the statement

⇒ 1.9 * 10 ⁻⁹ = x² / 4.8 ⇒ x² = 9.12 * 10⁻⁹

⇒ x = 9.55 * 10⁻⁵ = [OH⁻ (aq) ]

3) pOH

pOH = - log [OH⁻ (aq) ] = 4.02

4) pOH + pH = 14

⇒ pH = 14 - 4.02 = 9.98