Question

Consider the titration of a 20.0 −mL sample of 0.100 M HC2H3O2 with 0.130 M NaOH. Determine each of the following. a. the volume of added base required to reach the equivalence point b.the pH after adding 6.00 mL of base beyond the equivalence point Express your answer using two decimal places.

Answer 1

   HC2H3O2 + NaOH ------------------> NaC2H3O2   + H2O

   1 mole         1 mole

CH3COOH                                                            NaOH

M1 = 0.1M                                                              M2 = 0.13M

V1   = 20ml                                                            V2 =

n 1 = 1                                                                    n2 = 1

          M1V1/n1    =     M2V2/n2

              V2          = M1V1n2/M2n1

                            = 0.1*20*1/0.13*1

                              = 15.38ml

Total volume at equalent point = 20+15.38    = 35.38ml

no of moles of NaOH = molarity * volume in L

                                    = 0.13*0.006   = 0.00078moles

molarity of NaOH = no of moles/total volume

                              = 0.00078/0.03538   = 0.022M

NaOH(aq) ------------------> Na^+ (aq) + OH^- (aq)

0.022M                                                     0.022M

[OH^-]   = [NaOH]

[OH^-]   = 0.022M

POH   = -log[OH^-]

            = -log0.022

           = 1.6575

PH    = 14-POH

         = 14-1.6575    = 12.3425 >>>>>answer