Question

A 50.00-mL solution of 0.0229 M trimethylamine (Kb = 6.5 × 10–5) is titrated with a 0.0178 M solution of hydrochloric acid as the titrant. What is the pH of the base solution after 13.35 mL of titrant have been added? (Kw = 1.00 × 10–14)

Answer 1

Because 13.35 mL of acid have been added, and we started out with 50 mL of analyte, there are a total of 63.35 mL of analyte solution at this point. Hence, the molarity of TEA is the following:

0.0229 mol TEA / 0.06335 L solution in flask = 0.36148 M of TEA

The molarity of TEA.HCl is = 0.0178 M of TEA.HCl / 0.06335 L solution in flask = 0.28097 M of TEA.HCl

Now we can use the Henderson-Hasselbalch approximation:

pOH = pK_b + log TEA.HCl / TEA

pOH = 6.5 X10^-5 + log 0.28097 / 0.36148

= 6.39 x 10^-5

pH =14−pOH

=14−0.000065

pH = 13.99