Question

What is the pH at this point? The KA for HOAc is 1.8x10^-5.

A titration of 25.0 mL of 0.200 M HOAc requires 50.0 mL of 0.100 M NaOH to reach equivalence.

a. 5.22

b. 11.05

c. 2.95

d. 8.78

e. 7.00

Answer 1

no of moles of HOAc = molarity *volume in L

                                    = 0.2*0.025 = 0.005moles

no of moles of NaOH   = molarity *volume in L

                                     = 0.1*0.05   = 0.005moles

               HOAc + NaOH --------------> NaOAc + H2O

            0.005mole      0.005moles        0.005moles

        NaOAc (aq) --------------> Na^+ (aq) + OAc^- (aq)

total volume   = 25 +50 = 75ml   = 0.075L

molarity of NaOAc   = no of moles/volume in L

                                = 0.005/0.075   = 0.067M

               OAc^- (aq) + H2O (l) --------------> HOAc(aq) + OH^- (aq)

I              0.067                                              0                    0

C              -x                                                   +x                   +x

E           0.067-x                                             +x                    +x

                    Kb   = Kw/Ka

                            = 1*10^-14/(1.8*10^-5)

                             = 5.6*10^-10

                 Kb   = [HOAc][OH^-]/[OAc^-]

             5.6*10^-10 = x*x/(0.067-x)

             5.6*10^-10*(0.067-x) = x^2

                 x = 6.125*10^-6

           [OH^-]   = x   = 6.125*10^-6M

            POH   = -log[OH^-]

                        = -log6.125*10^-6

                        = 5.2

         PH   = 14-POH

                  =14- 5.2

                 = 8.78

d. 8.78 >>>>answer