Question

The pH of a 1.1M solution of hexanoic acid HC6H11O2 is measured to be 2.40. Calculate the acid dissociation constant Ka of hexanoic acid. Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

Ka = 1.4 x 10⁻⁴

Explanation:

HC₆H₁₁O₂ ⇄ H⁺ + C₆H₁₁O₂⁻

At equilibrium 1.1M HC₆H₁₁O₂ => [H⁺] = [C₆H₁₁O₂⁻] = 10^-pH = 10⁻²°⁴ = 4.0 x 10⁻³M

Ka = [H⁺][C₆H₁₁O₂⁻]/[HC₆H₁₁O₂] = (4.0 x 10⁻³)²/1.1 = 1.4 x 10⁻⁵    

Answer 2

Answer:

Ka = 1.4 x 10⁻⁴

Explanation:

HC₆H₁₁O₂ ⇄ H⁺ + C₆H₁₁O₂⁻

At equilibrium

1.1M HC₆H₁₁O₂ => [H⁺] = [C₆H₁₁O₂⁻] = 10^-pH = 10⁻²°⁴ = 4.0 x 10⁻³M

Ka = [H⁺][C₆H₁₁O₂⁻]/[HC₆H₁₁O₂] = (4.0 x 10⁻³)²/1.1 = 1.4 x 10⁻⁵