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- (It is not this one)

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

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

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

Answer 1

1) Your HCl is a strong acid and so is going to separate into ions.

The H+ ions are then going to react with your weak base,

(CH3CH2)3N, giving you the equation below:

(CH3CH2)3N(aq) + H+(aq) + Cl-(aq) <-> (CH3CH2)3NH+(aq) + Cl-(aq)

Removing spectator ions, your net ionic equation is:

(CH3CH2)3N(aq) + H+(aq) <-> (CH3CH2)3NH+(aq)

At equivalence point, the amount of acid added will be equal to the amount of base you started with (in moles). So

first figure out how many moles you start with:

(1.55 mol/L (CH3CH2)3N)*(1L/1000mL)*(135.00mL)=0.20925 mol

To get the same number of moles of HCL:

0.25*V = 0.20925

V = 837 ml

At the equivalence point, because 1 mol of acid (or 1 mol of H+) has been added for every mole of base initially

present, we can assume all base has been converted to it's conjugate acid.

[(CH3CH)2NH+ ] = 0.20925 / (837+135) = 0.215 M

(CH3CH2)3NH+(aq) <-> H+(aq) + (CH3CH2)3N(aq)

0.215 0 0

0.215-x x x

Ka = [H+] [(CH3CH2)3N ] / [(CH3CH2)3NH+]

Ka = Kw / Kb = x^2 / [0.215-x] = 2.5*10^-11

x^2 + 2.5*10^-11 x - 5.375*10^-12 = 0

x = 2.3*10^-6 M

pH = -log x = 5.63

2]

(CH3CH2)3N + HCl ----> (CH3CH2)3NH+

BUffer system weak base and its salt

pOH = pKb + log [Salt] / [Base]

pH + pOH = 14

pH = 3.4

pOH = 3.397 + log [(CH3CH2)3NH+] / [(CH3CH2)3N]

[(CH3CH2)3NH+] / [(CH3CH2)3N] = 1

concentration must be equal to get pH = 10.96

Moles of (CH3CH2)3N = 2* moles of HCl then after titration we will be having equal moles of base and salt

Moles of HCl = 1.55*135 / 1000*2 = 0.1046

volume of HCl needed = 0.1046 / 0.25 = 418.5 ml

Answer is 418.5 ml of HCl needed

3]

When 527.31 ml of HCl is added moles of HCl will be greater than half of the moles of (CH3CH2)3N

So .....it leads to more amount of (CH3CH2)3NH+ and Cl- presenting in the species

answer is B