Question

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?

Answer 1

C5H5N(aq)+H2)⇄C5H5NH+(aq)+OH-(aq) 

Answer 2

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