Question

In the lab a general chemistry student measures the ph of a 0.491 M aqueous solution of triethylamine to be 12.219 . Use this info to determine kb

Answer 1

pH = 12.219

C = 0.491 M

(CH2CH3)3N + H2O ↔ (CH2CH3)3NH+ + OH-

At equilibrium [OH-] = [(CH2CH3)3NH+]

pOH = 14 – 12.219 = 1.781

[OH-] = 10-1.781 = 0.01656

Kb = [OH-]·[(CH2CH3)3NH+] / [(CH2CH3)3N] = (0.01656)·(0.01656) / 0.491 = 5.6·10^-4