Question

An analytical chemist is titrating 158.8 mL of a 0.5200 M solution of propylamine (C2H,NH, with a 0.4200 M solution of HIO3. The pKof propylamine is 3.46. Calculate the pH of the base solution after the chemist has added 223.4 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 =] ] x 5 ?

Answer 1

Mmol of base = 158.8×0.52

=> 82.576 mmol

Mmol of Acid = 223.4×0.42

=> 93.828 mmol

Excess acid = 11.252 mmol

[H+] = 11.252/382.2

= 0.0294 M

pH = 1.53