Question

A total volume of 33.15-mL of 0.1053-M NaOH is needed to titrate 24.5-mL of benzoic acid. (Benzoic acid is a monoprotic acid with a pKa = 4.20). Determine the pH when the following volumes of the 0.1053 M NaOH have been added

a) 30.00 mL (before the equivalence point)?

b) 33.15 mL (at the equivalence point)

c) 36.00 mL (after the equivalence point)

Answer 1

   C6H5COOH + NaOH ---------> C6h5COONa + H2O

1 mole             1 mole

no of moles of NaOH = molarity * volume in L

                                  = 0.1053*0.03315   = 0.00349moles

no of moles of C6H5COOH   = no of moles of NaOH

no of moles of C6H5COOH   = 0.00349 moles

no of moles of C6H5COOH   = molarity * volume in L

                0.00349               = molarity * 0.0245

        molarity                    = 0.00349/0.0245   = 0.142M

   a.                no of moles of NaOH   = 0.1053* 0.03 = 0.003159 moles

                         C6H5COOH + NaOH ---------> C6H5COONa + H2O

I                       0.142               0.003159             0

C                   -0.003159        -0.003159               0.003159

E                    0.1388               0                          0.003159

     PH   = PKa + log[C6H5COONa]/[C6H5COOH]

    Pka of 42.

    PH   = 4.2 + log0.003159/0.1388

            = 4.2 -1.6428   = 2.5572

b. [C6H5COO-] = no of moles/total volume

                             = 0.00349/0.05765 = 0.0605

                 C6H5COO- + H2O ----------------> C6H5COOH + OH-

I                0.00349                                             0                  0

C                -x                                                      +x                +x

E               0.00349-x                                              +x              +x

                  kb = Kw/ka

                         = 1*10^-14/6.3*10^-5   = 1.58*10^-10

                   Kb = x*x/0.00349-x

                   1.58*10^-10    = x²/0.00349-x

                  1.58*10^-10 (0.00349-x)   = x²

                         x = 7.42*10^-7

                 [OH-] = x = 7.42*10^-7 M

                     POH = -log 7.42*10^-7

                              = 6.1295

                       PH   = 14-POH

                              = 14-6.1295 = 7.8705

c. no of moles of NaOH = 0.1053*0.036 = 0.00379moles

                            C6H5COOH + NaOH ---------> C6H5COONa + H2O

             I              0.00349         0.00379                 0

           C            - 0.00349         -0.00349                 0.00349

          E                0                     0.0003              

               [OH-]     = 0.0003/0.05765   = 0.0052 M

             POH = -log0.0052   = 2.2839

           PH   = 14-POH

                   = 14-2.2839 = 11.7161