Acid-Base Equilibria

66) [BrO^-] = 1.15 M          K_b = 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

      K_b= [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