Acid-Base Equilibria

 

66) [BrO-] = 1.15 M          Kb = 4.0 x 10-6

      

                  BrO-(aq) + H2O(l) HBrO(aq) + OH-(aq)

      [ ]i           1.15                                   0                0

      [ ]c                 -x                                  +x              +x

      [ ]e              1.15-x                               x                x

 

      Kb= [HBrO] x [OH-]/[BrO-]

      4.0 x 10-6 = x · x/(1.15-x) ≈ x2/1.15

      [OH-] = 2.1 x 10-3 M

 

      %ion = [HBrO]/[BrO-] x 100%

      %ion = 2.1 x 10-3 M/1.15 M x 100% = 0.18%

      %ion < ≈ 5%, therefore the assumption 1.15-x ≈ 1.15 is valid.

 

      pOH = -log[OH-] = -log(2.1 x 10-3) = 2.68

      pH + pOH = 14.00

      pH = 14.00 - pH = 14.00 - 2.68 = 11.32