Question

Instead of using ratios for back titrations we can also use molarities if our solutions are standardized. A 0.188g sample of antacid containing an unknown amount of triprotic base Al(OH)3 was reacted with 25.0mL of 0.101M HCl. The resulting solution was then titrated with 10.55mL of 0.132M NaOH solution. Calculate the mass percent of Al(OH)3 in the antacid sample.

This is what I have done so far:
0.025L | 0.101mol/L = 0.00252 mol HCl 0.01055 L | 0.132mol/L = 0.001393 mol NaOH 0.00252-0.001393 = 0.001127 mol HCl neutralized by antacid 3:1 ratio 0.000376 mol Al(OH)3

I am confused as to where I go from finding mols of Al(OH)3 and don't understand how to find the total to take a percent from? This is probably really dumb but I can't figure it out.

Answer 1

You have done very well. I carried the numbers out a little further and obtained 0.000377 mols Al(OH)3. You did the hard part. The rest is really simple.
g Al(OH)3 = mols x molar mass
Then % Al(OH)3 = [g Al(OH)3/g sample]*100 = ?

Answer 2

Thank you!! I got 15.6%!

Answer 3

I agree.