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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.

Respuesta :

Answer:

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

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