Question

Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 3.00 ✕ 102 mL of solution and then titrate the solution with 0.128 M NaOH.

C6H5CO2H(aq) + OH-(aq) C6H5CO2-(aq) + H2O(ℓ)

What are the concentrations of the following ions at the equivalence point?

Na+, H3O+, OH-

C6H5CO2-

___M Na+

___M H3O+

___M OH-

___M C6H5CO2-

What is the pH of the solution?

Answer 1

no of moles of benzoic acid = W/G.M.Wt

                                               = 0.235/122 = 0.001924 moles

   no of moles of NaOH   = molarity * volume in L

          0.001924              = 0.128* volume in L
volume in L                   = 0.001924/0.128 = 0.015L

total volume                    = 0.015 + 0.3 = 0.315L
molarity of C6H5COO-   = no of moles/volume

                                       = 0.001924/0.315   = 0.0061 M
       C6H5COO^- + H2O ---------> C6H5COOH + OH^-
                  kb   = x*x/0.0061-x

                  1.63*10^-10   = x²/0.0061-x

                  1.63*10^-10*(0.0061-x)   = x²

                               x = 9.97*10^-7

               [OH-] = x = 9.97*10^-7 M
               [C6H5COO-] = 0.0061-0.000000997 = 0.00609M
                [H3O+] = Kw/[OH-]

                               = 1*10^-14/9.97*10^{-7   =} 1*10^{-7 M}

                   [Na+]   = 0.0061M