What is the [OH⁻]

Question

What is the [OH⁻] of a 1.90 M solution of pyridine (C₅H₅N, Kb = 1.70 × 10⁻⁹)?

Answer 1

Looking at the hydrolysis of pyridine, we have ...

C₅H₅N + H2O ==> C₅H₅N⁺ + OH^- Kb = 1.70x10-9

Kb = [C₅H₅N⁺][OH^-] / [C₅H₅N]

1.70x10^-9 = (x)(x) / 1.90-x (assume x is small relative to 1.90 and ignore it in the denominator)

1.70x10^-9 = x² / 1.90

x² = 3.23x10^-9

x = 5.68x10^-5 (this is small relative to 1.90 so above assumption was valid)

[OH-] = 5.68x10^-5

pOH = -log 5.68x10^-5 = 4.25

pH = 14 - 4.25

pH = 9.75