A 250 mL sample of a 0100 M solution of aqueous trimethylami

A 25.0 mL sample of a 0.100 M solution of aqueous trimethylamine is titrated with a 0.125 M solution of HCl. What is the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added:

Solution

This is a strong acid and strong base titration.

At 10.00 ml H₃O⁺ conc. is very small so [OH^-] conc.:-

[OH^-] = original no. of mol of amine - no. of mol of HCl addef / total volume

[OH^-] = 25.00 (0.100) - 10.00 (0.125) / 25 + 10

         = 2.5 - 1.25 / 35

         = 1.25 /35

        = 0.03571 = 3.57 (10^-2) M

[H₃O⁺] = K_w / 3.57 (10^-2)

           = 1.00 (10^-14) /3.57 (10^-2)

           = 2.85 (10^-13)

pH = -log [2.85(10^-13)]

     = - 0.455 -(-13)

    pH = 12.54

At 20.00 ml ; this is equivalence point because [H₃O⁺]=[OH^-]

[H₃O⁺] = under the root K_w

           = under the root 1.00 (10^-14)

          = 1.00 (10^-7)

pH = - log[1.00(10^-7)]

pH = 7.00

At 30.00 ml;[H₃O⁺] = 30.00 (0.125) - 25.00 (0.100) / 55

                               = 3.75 - 2.5 / 55

                               = 1.25 / 55

                              = 0.0227 = 2.27 (10^-2)

                pH = -log[H₃O⁺]

                    = - log [2.27 (10^-2)]

                   = -0.356 -(-2)

                 pH = 1.644