Question

Sodium benzoate (C6H5CO2Na) is used as a food preservative.

Part A Calculate the pH in 0.060 M sodium benzoate; Ka for benzoic acid (C6H5CO2H) is 6.5×10−5. Express your answer using two decimal places. pH = nothing SubmitPrevious AnswersRequest Answer Incorrect; Try Again; 4 attempts remaining Part B Calculate the concentrations of all species present (Na+, C6H5CO−2, C6H5CO2H, H3O+ and OH−) in 0.060 M sodium benzoate. Express your answers using two significant figures. Enter your answers numerically separated by commas.

Answer 1

C6H5CO2Na(aq) -------------> C6H5CO2^- (aq) + Na^+ (aq)

0.06M                                      0.06M                     0.06M

[Na^+]   = 0.06M

[C6H5CO2^- ]   = 0.06M

           C6H5Co2^- (aq) + H2O ---------------> C6H5CO2H (aq) + OH^- (aq)

I            0.06                                                          0                       0

C            -x                                                             +x                     +x

E          0.06-x                                                        +x                      +x

       Kb   = Kw/Ka

                 = 1*10^-14/6.5*10^-5   = 1.54*10^-10

              Kb    =   [C6H5CO2H][OH^-]/[C6H5CO2^-]

             1.54*10^-10   = x*x/0.06-x

             1.54*10^-10 *(0.06-x)   = x^2

               x   = 3.04*10^-6

          [OH^-]    = x   = 3.04*10^-6 M

         POH   = -log[OH^-]

                   = -log3.04*10^-6

                    = 5.5171

        PH     = 14-POH

                  = 14-5.5171   = 8.4829

part-B

C6H5CO2Na(aq) -------------> C6H5CO2^- (aq) + Na^+ (aq)

0.06M                                      0.06M                     0.06M

[Na^+]   = 0.06M

[C6H5CO2^- ]   = 0.06M

[C6H5CO2H]   = x   = 3.04*10^-6M

[H3O^+]    = Kw/[OH^-]

                 = 1*10^-14/3.04*10^-6   = 3.28*10^-9 M