What is the pH of 0.166 M triethylammonium chloride, (C2H5)3NHCl. The Kb of triethylamine, (C2H5)3N, is 5.2 x 10-4.?

5.7

Explanation:

(C₂H₅)₃NHCl dissociates according to the following equation.

(C₂H₅)₃NHCl ⇒ (C₂H₅)₃NH⁺ + Cl⁻

The molar ratio of (C₂H₅)₃NHCl to (C₂H₅)₃NH⁺ is 1:1. Then, the concentration of (C₂H₅)₃NH⁺ is Ca = 0.166 M.

(C₂H₅)₃NH⁺ is the conjugate acid of (C₂H₅)₃N. Given the Kb of (C₂H₅)₃N, we can calculate Ka for (C₂H₅)₃NH⁺ using the following expression.

Ka × Kb = Kw

Ka = Kw / Kb

Ka = 1.0 × 10⁻¹⁴ / 5.2 × 10⁻⁴

Ka = 1.9 × 10⁻¹¹

(C₂H₅)₃NH⁺ dissociates according to the following equation.

(C₂H₅)₃NH⁺ ⇄ (C₂H₅)₃N + H⁺

We can calculate [H⁺] using the following expression.

[H⁺] = √(Ca × Ka) = √(0.166 × 1.9 × 10⁻¹¹) = 1.8 × 10⁻⁶

The pH is:

pH = -log [H⁺] = -log 1.8 × 10⁻⁶ = 5.7

1 is the answer to this problem