Question

11. Consider the titration of 25.00 mL of 0.200 M methyl amine (CH3NH2). The titrant is 0.120 M HCI. Calculate each of the following: a. the initial pH of the base. b. the pH at 5.00 mL added acid c. the pH at the HCI needed to reach the equivalence point d. the volume of added acid required to reach the equivalence point. e. the pH at the equivalence point.

Answer 1

weak base vs strong acid

a) before addition of HCl(initial pH of base)

pkb of CH3NH2 = 3.38

pOH = 1/2(pkb-logC)

      = 1/2(3.38-log0.2)

      = 2.04

pH = 14-2.04 = 11.96

b) after addition of 5.00 ml HCl

no of mol of CH3NH2 = 25*0.2 = 5.00 mmol

no of mol of HCl = CH3NH3+ = 5*0.12 = 0.6 mmol

   pOH = pkb+log(CH3NH3+/CH3NH2)

       = 3.38+log(0.6/(5.00-0.6))

       = 2.51

pH = 14-2.51 = 11.49

c) pH at 1/2 the HCl needed to reach the equivalencepoint.

   pOH = pkb = 3.38

pH = 14-3.38 = 10.62

d)
no of mol of CH3NH2 = 25*0.2 = 5.00 mmol

no of mol of HCl = 5 mmol

volume of HCl required = n/M

                          = 5/0.12

                          = 41.67 ml

e) concentration of salt(CH3NH3+) formed = 5/(41.67+25)

                                         = 0.075 M

pH of salt = 7-1/2(pka+logC)

               = 7-1/2(3.38+log0.075)

              = 5.90