1644 A 250 mL sample of a 0100 M solution of aqueous trimeth

16.44.

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

K_b = 7.4 x 10^-5

Solution

millimoles of amine = 25 x 0.1 = 2.5

1) millimoles of HCl added = 10.0 x 0.125 = 1.25

2.5 - 1.25 = 1.25 millimoles base left

1.25 millimoles salt formed

pOH = pKb + log [salt ] / [base]

[salt] = [base]

so pOH = pKb

pKb = - log Kb = - log [7.4 x 10^-5]

pKb = 4.13

pOH = 4.13

pH = 14 - 4.13

pH = 9.87

2) millimoles of HCl added = 20 x 0.125 = 2.5

2.5 millimoles base present.

[salt] = 2.5 / 45 = 0.055 M

pOH = 1/2 [pKw + pKb log C]

pOH = 1/2 [14 + 4.13 + log 0.055]

pOH = 8.43

pH = 14 - 8.43

pH = 5.57

3) millimoles of HCl added = 30 x 0.125 = 3.75

3.75 - 2.5 = 1.25

1.25 millimoles acid left

[HCl] = 1.25 / 55 = 0.023 M

pH = - log [H⁺]

pH = - log [0.023]

pH = 1.64