Using concentration and Kb to find pH

Question

What is the pH of a 0.050 M triethylamine, (C₂H₅)₃N, solution?

K_b for triethylamine is 5.3 × 10^–4

Answer 1

(C₂H₅)₃N + H₂O <==> (C₂H₅)₃NH⁺ + OH^- .. hydrolysis of (C₂H₅)₃N

(C₂H₅)₃N + H₂O <==> (C₂H₅)₃NH⁺ + OH^-

...0.05.................................0.................0.........Initial

...-x...................................+x.................+x........Change

..0.05-x...............................x...................x........Equilibrium

Kb = [(C₂H₅)₃NH⁺][OH^-] / [(C₂H₅)₃N]

5.3x10^-4 = (x)(x) / 0.05 - x and assume x is small relative to 0.05 and ignore it

5.3x10^-4 = x² / 0.05

x² = 2.65x10^-5

x = 0.00515 M (note this is ~ 10% of 0.05 so our above assumption was not valid)

Return to the Kb expression and solve using the quadratic formula...

5.3x10^-4 = (x)(x) / 0.05 - x

x² = 2.65x10^-5 -5.3x10^-4x

x² + 5.3x10^-4x - 2.65x10^-5 = 0

x = [OH^-] = 0.0049 M

pOH = -log [OH^-] = -log 0.0049

pOH = 2.31

pH = 14 - pOH

pH = 11.69