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A 0.40 M solution of a salt NaA has pH=8.95. What is the value of Ka...

A 0.40 M solution of a salt NaA has pH=8.95. What is the value of Ka for the parent acid HA?

Solutions

Expert Solution

                    NaA(aq) -----------> Na^+(aq) + A^-(aq)

PH   = 8.95

POH = 14-PH

         = 14-8.95   = 5.05

POH = 5.05

-log[OH-]   = 5.05

   [OH-]     = 10^-5.05   = 8.9*10^-6 M

                A^- + H2O --------------> HA + OH^-

          [HA]   = [OH-] at equilibrium

               Kb   = [OH^-][HA]/[A^-]

                       = 8.9*10^-6*8.9*10^-6/0.4

                        = 1.98*10^-10

              Ka     = Kw/Kb

                        = 1*10^-14/1.98*10^-10

                          = 5.05*10^-5   >>>>answer


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