How many moles of HCl were neutralized by the antacid tablet?

Question

Aluminum hydroxide is a base that is the active ingredient in some over-the-counter antacids. Suppose you have 29.0mL of 0.180M HCl solution in a flask and you add an antacid tablet to the HCl. After allowing the antacid to react with the HCl solution, you titrate the solution with 0.20 M NaOH. It requires 8.45 mL of NaOH to reach the end point.

Answer 1

First, let's write the balanced equation for the reaction taking place:

3HCl + Al(OH)₃ ==> 3H₂O + AlCl₃ .. balanced equation for HCl reacting with aluminum hydroxide

In the 2nd reaction, we are adding NaOH because we used EXCESS HCl in the first reaction. So, this is referred to as a "back titration".

HCl + NaOH ==> H₂O + NaCl

If we know the moles of HCl added originally, and we know how much was left over after reacting with the Al(OH)₃, we can then determine the moles of HCl neutralized by the antacid.

moles HCl originally present = 29.0 ml x 1 L / 1000 ml x 0.180 mol / L = 5.22x10^-3 mols HCl

moles HCl left over, i.e. neutralized by NaOH =

8.45 ml NaOH x 1 L/1000 ml x 0.20 mol/L x 1 mol HCl/mol NaOH = 1.69x10^-3 mols HCl

moles HCl neutralized by the antacid = 5.22x10^-3 mol - 1.69x10^-3 mol = 3.53x10^-3 mols HCl