Please help with this question, I have no clue how to go about it.

Question

1.3188g of antacid is weighed and mixed with 75.00mL of excess 0.1746 M HCl. The excess acid required 27.20 mL of 0.09767 M NaOH for back titration. Calculate the neutralizing power of the antacid in terms of mmol H+ per gram of antacid.

Answer 1

mmoles HCl added = 0.1746 x 75.00 mL = ??
mmoles NaOH to titrate the excess acid = 27.20 mL x 0.09767 M NaOH = ?? moles NaOH.
The difference is the mmoles of the antacid in the 1.3188 g. That difference divided by the 1.3188 will be what you are looking for.

Answer 2

mmoles HCl added= 0.1746 x 75.00mL= 13.095 moles

mmles NaOH titrate the excess acid= 27.20 mL x 0.09767 M NaOH= 2.6566 moles

But I don't understand how to find the mmoles in 1.3188g.

Thanks.

Answer 3

Hi,

I also don't understand what you meant by the difference is the mmoles of the antacid in the 1.3188 g.

Please help, when possible.