In the laboratory, a general chemistry student measured the pH of a 0.314 M aqueous solution of phenol (a weak acid), C6H5OH to be 5.269. Use the information she obtained to determine the Ka for this acid.

Ka = 9.0974 E-12

Explanation:

C6H5OH + H2O ↔ C6H5O - + H3O+

∴ Ka = ([H3O+]*[C6H5O-]) / [C6H5OH]

∴ C C6H5OH = 0.314 M

pH = - Log [H3O+]

∴ pH = 5.269 = - Log [H3O+]

⇒ [H3O+] = 5.3827 E-6 M

mass balanced:

⇒ C C6H5OH = 0.314 M = [C6H5O-] + [C6H5OH] ... (1)

charge balanced:

⇒ [H3O+] = [C6H5O-] + [OH-] ... [OH-] is neglected, it comes from the water

⇒ [H3O+] = [C6H5O-] ... (2)

(2) in (1):

⇒ 0.314 M - [H3O+] = [C6H5OH]

⇒ [C6H5OH] = 0.314 M - 5.3827 E-6 M

⇒ [C6H5OH] = 0.31399 M

replacing in Ka:

⇒ Ka = ((5.3827 E-6) * (5.3827 E-6)) / (0.31399) = 9.0974 E-12