Question

Determine the pH during the titration of 39.9 mL of 0.349 M triethylamine ((C2H5)3N , Kb = 5.2×10-4) by 0.349 M HBr at the following points. (a) Before the addition of any HBr (b) After the addition of 17.7 mL of HBr (c) At the titration midpoint (d) At the equivalence point (e) After adding 61.0 mL of HBr

Answer 1

Kb of base = 5.2 x10^-4

So pKb = -log 5.2 x10^-4 = 3.28 4

The reaction is

(C2H5)3N + HBr ----------> (C2H5)3NH+Br-

39.9x0.349 0 0 initial mmoles

Q1) Before addition of HBr

(C2H5)3N + HBr ----------> (C2H5)3NH+Br-

39.9x0.349 0 0 initial mmoles

The solution is of weak base whose pOH = 1/2[ pKb-logC]

= 1/2[3.284-log 0349] =1.87

thus pH = 14-1.87= 12.13

Q b) after addition of 17.7 mL HBr

(C2H5)3N + HBr ----------> (C2H5)3NH+Br-

39.9x0.349=13.9251 0 0 initial mmoles

- 17.7x0.349 = 6.1773 - change

7.7478 0 6.1773 equilibrium mmoles

Now the solution is a buffer consisting of a weak base and its conjugate acid.

Its pH is given by Hendersen equation as

pOH = pKb + log [conjugate acid]/[base]

= 3.284 + log [6.1773/7.7478]

=3.1856

pH = 14-pOh = 14-3.1856 =10.8144

c)

At the midpoint

At midpoint [conjugate acid] = [base ] and thus pOH = pKb= 3.284

Thus pH = 14-3.284=10.716

d) at the equivalence

(C2H5)3N + HBr ----------> (C2H5)3NH+Br-

39.9x0.349 0 0 initial mmoles

- 39.9x0.349 - change

0 0 13.9251

Thus the solution has only salt of weak base and strong acid which is acidic and its pHis given by

pH = 1/2[pKw-pKb-log C]

[salt = 13.9251/[39.9+39.9] =0.1745

and pH = 1/2[14-3.284 -log 0.1745]

=5.737

e)After 61 mL of Hbr

(C2H5)3N + HBr ----------> (C2H5)3NH+Br-

39.9x0.349 0 0 initial mmoles

- 61.0x0.349 =21.289 - change

0 7.3639 13.9251

thus the solution has excess of HBr and th eph is decided by the [H=] = [HBr]

so [Hbr] = mmoles/volume= 7.3639/(39.9+61)=0.07298 M

pH = -log[H+] = -log 0.07298 =1.1368