The ph of a 1.4m solution of acetic acid (hch3co2) is measured to be 2.30. calculate the acid dissociation constant of acetic acid. round your answer to significant digits.

Ka = 1.8 × 10⁻⁵

Explanation:

Acetic acid is a weak acid that dissociates according to the following equation.

HCH₃CO₂(aq) ⇄ CH₃CO₂⁻(aq) + H⁺(aq)

The concentration of the acid is Ca = 1.4 M.

The pH is 2.30. The concentration of H⁺ is:

pH = -log [H⁺]

[H⁺] = anti log (-pH) = anti log (-2.30) = 5.01 × 10⁻³ M

We can find the acid dissociation constant (Ka) using the following expression.

[H⁺] = √(Ca × Ka)

Ka = [H⁺]²/Ca

Ka = (5.01 × 10⁻³)²/1.4

Ka = 1.8 × 10⁻⁵

you need to give more information

the answer is c i believe