Question

Calculate the pH at the equivalence point of a solution when 25.00mL of a 0.10M monoprotic weak acid, HA, is titrated with 25.00mL of 0.10M of NaOH.

The weak acid's Ka value is 1.4*10^-5.

Answer 1

no of moles of HA   = molarity* volume in L

                                 = 0.1*0.025   = 0.0025moles

no of moles of NaOH   = molarity* volume in L

                                    = 0.1*0.025 = 0.0025 moles

                HA    +     NaOH ---------------> NaA + H2O

I              0.0025      0.0025                        0

C           -0.0025     -0.0025                       0.0025

E              0              0                              0.0025

         NaA ---------------> Na^+ + A^-

0.0025 M                                       0.0025M

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

I          0.0025                              0         0

C          -x                                     +x       +x

E        0.0025-x                            +x          +x

             Kb   = Kw/ka

                    = 1*10^-14/1.4*10^-5   = 7.14*10^-10

              Kb   = [HA][OH-]/[A^-]

                7.4*10^-10 = x*x/0.0025-x

              7.4*10^-10*(0.0025-x) = x^2

                  x   = 1.36*10^-6

                  [OH-]   = x = 1.36*10^-6M

                  POH   = -log[OH-]

                             = -log1.36*10^-6

                              = -log0.00000136

                               = 5.8664

                  PH         = 14-POH

                               = 14-5.8664    = 8.1336 >>>>answer