Question

An analytical chemist is titrating 209.0 ml of a 0.7700 M solution of isopropylamine ((CH), CHNH,) with a 0.5300 M solution of HIO;. The pK, of isopropylamine is 3.33. Calculate the pH of the base solution after the chemist has added 337.8 ml of the HIO, solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HIO; solution added. Round your answer to 2 decimal places. pH = 0 xn ?

Answer 1

Titration

initial moles of isopropylamine present = 0.7700 M x 209 ml = 160.93 mmol

moles of HIO3 added = 0.5300 M x 337.8 ml = 179.034 mmol

all of isoproylamine is neutralized

excess [HIO3] left = 18.104 mmol/(209 + 337.8) ml = 0.0331 M

HIO3 is a strong acid dissociates completely

[H+] = 0.0331 M

pH = -log[H+]

      = -log(0.0331) = 1.48