Question

What is the pH of 0.32 M triethylammonium chloride, (C2H5)3NHCl. The Kb of triethylamine, (C2H5)3N, is 5.2 x 10-4.

Answer 1

To get the pH of triethylammonium chloride, we first establish the dissociation in water:

(C2H5)3NHCl <-> (C2H5)3NCl^- + H₃O⁺

As it behaves as a weak acid, we need Ka, and we obtain it as follows:

Kw = Ka * Kb

Ka = Kw / Kb

Ka = 1x10^-14 / 5.2 x 10^-4 = 1.923 x 10^-11

We know use the definition of the constant:

Ka = [H3O⁺][A^-] / [HA]

1.923 x 10^-11 = x² / 0.32

We use x to define concentrations of hydronium ions and negative ion as we have a 1 on 1 stoichiometric relation, and we don't know their concentration.

We isolate x:

x = 2.48 x 10^-6

As this is the hydronium concentration, we use it to get pH:

pH = -log (2.48 x 10^-6) = 5.6