Question

A buffer that contains 0.16 M of a base, B and 0.27 M of its conjugate acid BH+, has a pH of 8.56. What is the pH after 0.02 mol of Ba(OH)2 are added to 0.52 L of the solution?

Answer 1

no of moles of base B = molarity * volume in L

                                    = 0.16*0.52 = 0.0832 moles

no of moles of conjugate base = molarity * volume in L

                                                  = 0.27*0.52 = 0.1404 moles

      POH = 14-PH

              = 14-8.56 = 5.44

       POH = PKb + log[B]/[BH⁺]

        5.44     = Pkb + log0.1404/0.0832

       5.44      = Pkb +0.2272

      Pkb      = 5.44-0.2272

                   = 5.2128

After addition of 0.02moles of Ba(OH)2

no of moles of base B   = 0.0832+0.02 = 0.1032moles

no of moles of conjugate acid BH⁺ = 0.1404-0.02 = 0.1204 moles

   PoH   = Pkb + log[BH⁺]/[B]

             = 5.2128+ log0.1204/0.1032

              = 5.2128+0.06694   = 5.2797 >>>>answer