Acid-Base Equilibria

 

54) Ka = 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

 

      Ka = [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