Question

Butyric acid is responsible for the odor in rancid butter. A solution of 0.25 M butyric acid has a pH of 2.71. What is the Ka for the acid? Multiple Choice 0.36 2.4 × 10–2 7.8 × 10–3 1.5 × 10–5

Answer 1

Answer:

Ka = 1.5 × 10⁻⁵

Explanation:

Butyric acid is a weak acid that ionizes according to the following equation:

CH₃-CH₂-CH₂-COOH(aq) ⇄ CH₃-CH₂-CH₂-COO⁻(aq) + H⁺(aq)

We can find the value of the acid dissociation constant (Ka) using the following expression:

$Ka=\frac{[H^{+}]^{2} }{Ca}$

where

[H⁺] is the molar concentration of H⁺

Ca is the initial molar concentration of the acid

We can find [H⁺] from the pH.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -2.71 = 1.95 × 10⁻³ M

Then,

$Ka=\frac{(1.95 \times 10^{-3})^{2} }{0.25} =1.5 \times 10^{-5}$