Question

Calculate the pH of a 0.166 M aqueous solution of pyridine (C5H5N, Kb = 1.5×10-9) and the equilibrium concentrations of the weak base and its conjugate acid.

pH =

[C5H5N]equilibrium = M

[C5H5NH+]equilibrium = M

Answer 1

Pyridine dissociate as

C₅H₅N + H₂O C₅H₅NH⁺ + OH^-

K_b = [C₅H₅NH⁺] [OH^-] / [C₅H₅N]

pyridine is monobesic base therefore

[C₅H₅NH⁺] = [OH^-] = x

K_b = [x][x] /  [C₅H₅N]

K_b =[x]² /  [C₅H₅N]

[x]² = K_b X  [C₅H₅N]

[x]² = 1.5 X 10^-9 X0.166 = 2.49 X 10^-10

[x] = 1.578 X 10^-5

[C₅H₅NH⁺] = [OH^-] = x = 1.578 X 10^-5

[C₅H₅NH⁺] equilibrium = 1.578 X 10^-5

[C₅H₅NH] equilibrium = intial [C₅H₅NH] - [C₅H₅NH⁺] equilibrium = 0.166 - 1.578 X 10^-5 = 0.16598 M

[C₅H₅NH] equilibrium = 0.16598 M

pOH = -log[OH^-] = -log(1.578 X 10^-5) = 4.8

pH = 14 - pOH = 14 - 4.8 = 9.2

pH of pyridine = 9.2