What is the pH of a 1.4 M pyridine solution that has Kb = 1.7 × 10-9? The equation for the dissociation of pyridine is C5H5N(aq) + H2O(l) ⇌ C5H5NH+(aq) + OH-(aq). What is the pH of a 1.4 M pyridine solution that has Kb = 1.7 × 10-9? The equation for the dissociation of pyridine is C5H5N(aq) + H2O(l) ⇌ C5H5NH+(aq) + OH-(aq). 4.31 9.69 8.72 10.69

pH = 9.69

Explanation:

When pyridine (C₅H₅N) is added to water, the equilibrium that occurs is:

C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq) Kb = 1.7x10⁻⁹

Where Kb is defined as:

Kb = 1.7x10⁻⁹ = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]

If you have a solution of 1.4M C₅H₅N, the equilibrium concentration of each specie is:

[C₅H₅N] = 1.4 - X

[C₅H₅NH⁺] = X

[OH⁻] = X

Where X represents the reaction coordinate

Replacing in Kb expression:

1.7x10⁻⁹ = [X] [X] / [1.4 - X]

2.38x10⁻⁹ - 1.7x10⁻⁹X =  X²

0 = X² + 1.7x10⁻⁹X - 2.38x10⁻⁹

Solving for X:

X = -0.0000488M → False answer, there is no negative concentrations

X = 0.0000488M → Right answer

Thus, [OH⁻] = 0.0000488M. As pOH = -log [OH⁻]

pOH = 4.31

Knowing pH = 14 - pOH

pH = 9.69

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