Question

1. If the HCl used in this experiment was prepared at a concentration of 0.15M. Using your volume of HCl neutralized by your whole tablet, calculate the mass of active ingredient (CaCO3) in the tablet in milligrams.

2. Instead of using ratios for back titrations we can also use molarities, if our solutions are standardized. A 0.196g sample of antacid containing an unknown amount of triprotic base Al(OH)3 was reacted with 25.0mL of 0.111M HCl. The resulting solution was then titrated with 11.05mL of 0.132M NaOH solution. Calculate the mass percent of Al(OH)3 in the antacid sample.

Answer 1

A 0.196g sample of antacid containing an unknown amount of triprotic base Al(OH)3 was reacted with 25.0mL of 0.111M HCl. The resulting solution was then titrated with 11.05mL of 0.132M NaOH solution. Calculate the mass percent of Al(OH)3 in the antacid sample.

The moles of HCl used = Molarity X volume = 0.111 X 25mL = 2.775 mmoles

The moles of NaOH used = Molarity X volume = 0.132 X 11.05 = 1.459 mmoles

So moles of HCl used against antacid = 2.775-1.459 = 1.316 mmoles

The reaction of aluminium hydroxide with HCl is

3HCl + Al(OH)3 ----> 3H2O + AlCl3

So here with each mole of aluminium trihydroxide three moles of HCl will react

So moles of Al(OH)3 present in antacid = Moles of HCl / 3 = 1.316 / 3 = 0.439 mmoles

mass of Al(OH)3 present = Moles X molar mass = 0.439 millimoles X 78 = 34.24 mg = 0.03424 grams

Mass % of Al(OH)3 = 0.03424 X 100 / 0.196 = 17.47%