Question

A 25.0-mL sample of a 0.310 M solution of aqueous trimethylamine is titrated with a 0.388 M solution of HCI. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pK_b of (CH_3)_3N = 4.19at 25degree C. pH after 10.0 mL of add have been added: pH after 20.0 mL of add have been added: pH after 30.0 mL of acid have been added:

Answer 1

pDuring the titration, chloride salt of trimethyl amine is formed. The pH of the titration solution would depend on the amount of trimethyl amine left and the chloride salt formed during titration. In case of a mixture of a weak base and its salt, the p(OH)is given by the well known relation.

p(OH) = pKb+ log (salt)/(base)

With the addition of 10ml of HCl solution, the amount of HCl added to the titration solution = 10x0.388= 3.88 mmole

Amount of chloride salt of trimethylamine formed = 3.88 mmole

Amount of trimethylamine left behind = 3.88 mmole

Therefore p(OH) of the titration solution = pKb + log (salt)/(base)

Since the volume of the solution is same for both acid and the salt. The ratio of their amounts in millimoles can be taken as the ratio of their molar concentrations.

Thus, p(OH)= 4.19 + log 3.88/3.88

= 4.19 + log 1 = 4.19 + 0

= 4.19

pH + p(OH) = 14

pH = 14 - p(OH)

= 14- 4.19

= 9.81

Therefore pH after addition of 10ml of HCl = 9.81

Similarly the pH after the addition of 20 ml of HCl is 9.50 and

The pH after the addition of 30 ml of HCl is 9.30