Question

# 2.)a.)A 24.1 mL sample of 0.370 M methylamine, CH3NH2, is titrated with 0.352 M nitric acid....

2.)a.)A 24.1 mL sample of 0.370 M methylamine, CH3NH2, is titrated with 0.352 M nitric acid. At the equivalence point, the pH is

B.)A 27.8 mL sample of 0.243 M dimethylamine, (CH3)2NH, is titrated with 0.226 M nitric acid.
The pH before the addition of any nitric acid is

C.)A 23.2 mL sample of 0.389 M diethylamine, (C2H5)2NH, is titrated with 0.387 M hydrochloric acid.
After adding 33.3 mL of hydrochloric acid, the pH is

a) At equivalence M1V1 = M2V2

Thus 24.1 x 0.370 = V x 0.352

Thus V = 25.33 mL

Thus the concentration of salt formed at equivalence = 24.1x0.370/(24.1+25.33)

= 0.1803 M

It is a salt of weak base and strong acid whose pH is given by

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

= 1/2 [ 14 - 3.36 - log 0.1803]

= 5.692

B) Before addition of acid , it the weak base solution . The pH of weak base is given by

pH = 14 - pOH

= 14 -1/2[pkb -logC]

= 14 - 1/2 [ 3.25 - log 0.243]

= 12.06

C)

B + HCl --------------> BH+ + Cl-

23.2x0.389 33.3x0.387 0 0 initial mmoles

=9.0248 =12.8871

0 3.8623 9.0248 - after reaction

Thus the solution contains a strong acid and a weak acidiic salt .

The pH of this solution is given by {H=] from strong acid only.

[H+] = 3.8623/(56.5) = 0.06835 M

Thus pH = - log [H+]

= - log(0.06825)

= 1.1652