The pH of a 1.0M solution of butanoic acid HC4H7O2 is measured to be 2.41. Calculate the acid dissociation constant Ka of butanoic acid.

Ka = 1.52 E-5

Explanation:

CH3 - (CH2) 2-COOH ↔ CH3 (CH2) 2COO - + H3O+

⇒ Ka = [H3O+][CH3) CH2) 2COO-] / [CH3 (CH2) 2COOH]

mass balance:

⇒ C CH3 (CH2) 2COOH = [CH3 (CH2) 2COO-] + [CH3 (CH2) 2COOH] = 1.0 M

charge balance:

⇒ [H3O+] = [CH3 (CH2) 2COO-]

⇒ Ka = [H3O+]² / (1 - [H3O+])

∴ pH = 2.41 = - Log [H3O+]

⇒ [H3O+] = 3.89 E-3 M

⇒ Ka = (3.89 E-3) ² / (1 - 3.89 E-3)

⇒ Ka = 1.519 E-5