Final answer:
The pH of a 0.50 M solution of ethylamine, with a Kb of 5.6 x 10⁻⁴, is approximately 3.23. This is calculated by finding the concentration of hydroxide ions produced by the base, converting that to pOH, and then to pH using the relation pH + pOH = 14.
Step-by-step explanation:
To calculate the pH of a 0.50 M solution of ethylamine (C₂H₅NH₂), we need to use its base dissociation constant (Kb = 5.6 x 10⁻⁴). Ethylamine is a weak base that partially ionizes in water to form ethylammonium ions (C₂H₅NH₃⁺) and hydroxide ions (OH⁻).
The base ionization equation for ethylamine is given as:
C₂H₅NH₂ + H₂O ⇌ C₂H₅NH₃⁺ + OH⁻
We can set up an equilibrium expression using Kb:
Kb = [C₂H₅NH₃⁺][OH⁻] / [C₂H₅NH₂]
Assuming x is the concentration of OH⁻ produced, the equilibrium equation becomes:
5.6 x 10⁻⁴ = x² / (0.50 - x)
For simplification, assuming (0.50 - x) ≈ 0.50, we can solve for x. Thus, x² = (0.50)(5.6 x 10⁻⁴) and x ≈ 1.68 x 10⁻⁴ M.
The pOH is calculated as -log[OH⁻], so pOH = -log(1.68 x 10⁻⁴) ≈ 10.77. To find the pH, we use the relationship pH + pOH = 14, thus pH ≈ 14 - 10.77 = 3.23.