Question

A 25.0-mL sample of a 0.250 M solution of aqueous trimethylamine is titrated with a 0.313 M solution of HCl. Calculate the pH of the solution after 10.0, 20.0. and 30.0 ml of acid have been added; pKb of (CH3)3N = 4.19 at 25degreeC. pH after 10.0 mL of acid have been added; 4.19 Number Did you find the pH or the pOH of this solution? pH after 20.0 mL of acid have been added: Number 3.65 pH after 30.0 mL of acid have been added: Number 1.24

Answer 1

a) answer   pH= 9.80

HCl            + (CH3)3N-----------------> (CH3)3NH+Cl-

10 x0.313          25 x0.250                      0             

3.13                  6.25                         0   ------------initial

0                       3.12                        3.13 ------------ after reaction

solution contain salt and base it forms buffer

pOH= pKb + log [salt/base]

       = 4.19 + log[3.13/3.12]

        = 4.191

pH= 14-pOH

pH= 9.8

b) answer : pH= 3.653

HCl            + (CH3)3N-----------------> (CH3)3NH+Cl-

20 x0.313          25 x0.250                      0             

6.26                  6.25                         0   ------------initial

0.01                      0                        6.25 ------------ after reaction

solution contain strong acid

so [H+] = 0.01/ toatl volume

     [H+] = 0.01/(25+20)

      [H+] = 2.2 x10^-4

pH= -log[H+]

pH= 3.653