Calculate the pH at 25 °C

Question

Calculate the pH at 25 °C of a 0.99 M solution of trimethylammonium chloride ((CH₃)₃ NHCI): Note that trimethylamine ((CH₃)₄N) is a weak base with a

pK_b of 4.19.

Round your answer to 1 decimal place.

pH =

Answer 1

(CH₃)₃NH⁺ + H₂O ==> (CH₃)₃N + H₃O⁺ ((CH₃)₃NH⁺ acts as an acid in this hydrolysis reaction)

pKa for (CH₃)₃NH⁺ = 14 - 4.19 = 9.81

Ka for (CH₃)₃NH⁺ = 1.55x10^-10

Ka = [(CH₃)₃N] [H₃O⁺] / [(CH₃)₃NH⁺]

1.55x10^-10 = (x)(x) / 0.99 - x (asssume x is small relative to 0.99 and ignore it in the denominator)

1.55x10^-10 = x² / 0.99

x² = 1.53x10^-10

x = [H₃O⁺] = 1.24x10^-5 M

pH = -log [H₃O⁺] = -log 1.24x10^-5

pH = 4.9