What is the pH of a 0.0100 M sodium benzoate solution? Kb for the benzoate anion (C7H5O2–) = 1.5 × 10–10?

2 Answers

• Ph Of Sodium Benzoate

  Source(s): https://owly.im/a0UM1

• Let B^-1 = C7H5O2^-1

  B^-1 + H2O <---> BH + OH^-1

  Kb = [BH][OH-] / [B^-1]

  [BH] = [OH^-1]

  1.5x10^-10 = [OH-]^2 / ([B^-1] - [OH^-1])

  Assume [OH-] << [B^-]

  1.5x10^-10 = [OH-]^2 /[B-]

  [OH-]^2 = 1.5x10^-10 x 0.0100

  [OH-] = 1.22x10^-6

  pOH = 5.91

  pH = 14.00 - 5.91 = 8.09