Acid-Base Equilibria

54) K_a = 3.3 x 10^-4          T = 25°C          V = 250 mL

       m = 2 x 500 mg = 1000 mg HC₉H₇O₄             

      [HC₉H₇O₄] = n/V = 1000 mg HC₉H₇O₄ x 1 g/10³ mg x 1 mol HC₉H₇O₄/180.17 g HC₉H₇O₄/(250 mL x 1 L/10³ mL)

      [HC₉H₇O₄] = 0.0222 M

                 HC₉H₇O₄(aq) H⁺(aq) + C₉H₇O₄^-(aq)

      [ ]_i          0.0222                       0                0

      [ ]_c                 -x                         +x              +x

      [ ]_e             0.0222-x                   x                x

      K_a = [H⁺] x [C₉H₇O₄^-]/[HC₉H₇O₄]

      3.3 x 10^-4 = x · x/(0.0222-x) ≈ x²/0.0222

      [H⁺] = 2.7 x 10^-3 M

      %ion = [H⁺]/[HC₉H₇O₄] x 100%

      %ion = 2.7 x 10^-3 M/0.0222 M x 100% = 12%

      %ion > ≈ 5%, therefore the assumption 0.0222-x ≈ 0.0222 is not valid and the quadratic formula must be used.

      3.3 x 10^-4 = x²/(0.0222-x)

      x² + 3.3 x 10^-4x -7.3 x 10^-6 = 0

      x = (-b ± (b² - 4ac)^1/2)/2a

      [H⁺] = (-3.3 x 10^-4 ± ((3.3 x 10^-4)² - 4 x (-7.3 x 10^-6))^1/2)/2 = 2.5 x 10^-3 M

      pH = -log[H⁺] = -log(2.5 x 10^-3) = 2.60