What is the pH of a 0.29 M aqueous solution of pyridine (C5H5N) given that it has a base dissociation constant (Kb) of 1.7 x 10^-9?

To find the pH of a 0.29 M solution of pyridine, we can start by determining the concentration of hydroxide ions (OH^–) produced in the solution. Pyridine is a weak base, and it partially ionizes in water according to the following equilibrium:

C₅H₅N + H₂O <=> C₅H₅NH⁺ + OH^–

Let’s denote the concentration of pyridine as [C₅H₅N] = 0.29 M. At equilibrium, if ‘x’ is the concentration of OH^– produced, then:

• [C₅H₅NH⁺] = x
• [OH^–] = x
• [C₅H₅N] = 0.29 – x

The base dissociation constant expression (Kb) for the reaction is given by:

Kb = &frac;[C₅H₅NH⁺][OH^–]}{[C₅H₅N]}

Substituting in the expressions for equilibrium concentrations, we have:

1.7 x 10^-9 = &frac;[x][x]}{[0.29 – x]}

Assuming ‘x’ is small compared to 0.29, we can simplify the equation to:

1.7 x 10^-9 = &frac;[x][x]}{[0.29]}

Solving for ‘x’:

x² = (1.7 x 10^-9)(0.29)

Calculating the right side:

x² = 4.93 x 10^-10

x = &sqrt;{4.93 x 10^-10} = 2.22 x 10^-5

Thus, [OH^–] = 2.22 x 10^-5 M. Now we can calculate the pOH:

pOH = -log[OH^–] = -log(2.22 x 10^-5) ≈ 4.65

Finally, we convert pOH to pH using the relationship:

pH + pOH = 14

pH = 14 – 4.65 ≈ 9.35

Therefore, the pH of a 0.29 M aqueous solution of pyridine is approximately 9.35.