Question

24. 50.00 mL of 0.10 M CH3COOH (Ka = 1.8 x 10-5) is titrated with a 0.10 M KOH solution. After 50.00 mL of the KOH solution is added, the pH in the titration flask will be A) 4.28 B) 8.72 C) 9.41 D) 11.24 E) 12.08 [Kb = 5.6 x 10-10 for acetate ion] 0.0025 - 0.25 .

Answer 1

no of moles of CH3COOH = molarity * volume in L

                                          = 0.1*0.05   = 0.005moles

no of moles of KOH    = molarity * volume in L

                                  = 0.1*0.05 = 0.005moles

total volume = 50 +50 = 100ml = 0.1L

   CH3COOH(aq) + KOH(aq) ----------> CH3COOK(aq) + H2O(l)

I     0.005                 0.005                            0

C    -0.005               -0.005                           0.005

E      0                      0                                   0.005

molarity of CHCOOK     = no of moles/volume in L

                                       = 0.005/0.1    = 0.05M

   CH3COO^- (aq) + H2O(l) ----------> CH3COOH(aq) + OH^- (aq)

I     0.05                                                       0                     0

C   -x                                                            +x                   +x

E     0.05-x                                                    +x                    +x

                 Kb   = Kw/ka

                          = 1*10^-14/(1.8*10^-5)   = 5.6*10^-10

     Kb   =   [CH3COOH][OH^-]/[CH3COO^-]

      5.6*10^-10   = x*x/(0.05-x)

      5.6*10^-10*(0.05-x)   = x^2

   x = 5.3*10^-6

    [OH^-]   = x   = 5.3*10^-6M

    POH   = -log[OH^-]

             = -log(5.3*10^-6)

           = 5.2757

PH     = 14-POH

          = 14-5.2757

         = 8.72

b. 8.72 >>>answer