Answer:
1.51 × 10⁻⁵
Step-by-step explanation:
Step 1: Given data
pKa value: 4.82
Step 2: Write the reaction for the acid ionization of butanoic acid, HC₄H₇O₂
Butanoic acid is a weak acid that ionizes according to the following equation.
HC₄H₇O₂(aq) ⇄ C₄H₇O₂⁻(aq) + H⁺(aq)
The acid ionization constant for this reaction is Ka
Step 3: Calculate the value of the ionization constant for butanoic acid
We will use the following expression.
pKa = -log Ka
Ka = antilog -pKa = antilog -4.82 = 1.51 × 10⁻⁵