Question

The next three (3) problems deal with the titration of 135 mL of 1.55 M triethylamine, (CH3CH2)3N (Kb = 4.0 x 10-4) with 0.25 M HCl. Water will be a major species throughout the titration. The chemical species, in addition to water, that can be found in this reaction mixture during the titration are:

I. H+ II. OH- III. Cl- IV. (CH3CH2)3N V. (CH3CH2)3NH+

1) What is the pH of the solution at the equivalence point?

2) How many mL of HCl will need to be added to the triethylamine, (CH3CH2)3N solution to reach a pH of 10.60?

3) What will the major species in solution be when 527.31 mL of HCl has been added to the base solution?

a) (CH3CH2)3N

b) (CH3CH2)3NH+, Cl-

c) (CH3CH2)3N, (CH3CH2)3NH+, Cl-

d) (CH3CH2)3N, (CH3CH2)3NH+, Cl-, OH-

e) (CH3CH2)3N, (CH3CH2)3NH+, Cl-, H+

Answer 1

   (CH3CH2)3N(l) + HCl(aq) ---> (CH3CH2)3NH+(aq) + Cl^-(aq)

major species = iii.Cl^- , iv. (CH3CH2)3NH+.

at the equivalence point, gives formation of salt

pH = 7 - 1/2(pkb+logC)

   pkb = -logkb = -log(4*10^-4) = 3.4

   C = concentration of salt = ?

    no of mole of (CH3CH2)3N = 135*1.55/1000 = 0.21

    no of mole of HCl required to reach equivalence point = 0.21

   volume of HCl = 0.21/0.25 = 0.84 L

                             = 840 ml

   total volume of mixer = 840+135 = 975 ml

      concentration of salt = n/v = 0.21/0.975 = 0.2154 M

pH = 7 - 1/2( 3.4+log0.2154) = 5.63

2) before equivalence point mixture is buffer

   pH of buffer = 14 - (pkb+log(salt/base)

   pkb = 3.4

   salt = acid added = x mole

   base = triethylamine-acid added = 0.21-x mole

10.6 = 14 - (3.4+log(x/(0.21-x))

x = 0.105 mole

volume of HCl added = 0.105/0.25 = 0.42 L

                     = 420 ml

3. after addition of 527.31 ml of HCl , the mixture cross the half equivalence point.

so that , major specieses are b) (CH3CH2)3NH+, Cl-.