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.
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 = ?
Thank you!! I got 15.6%!
I agree.
To calculate the mass percent of Al(OH)3 in the antacid sample, you need to determine the mass of Al(OH)3 present in the solution and then calculate the percentage of that mass in relation to the mass of the entire antacid sample.
Here's how you can continue from where you left off:
1. Calculate the molar mass of Al(OH)3:
Al = 26.98 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
Total molar mass = (26.98 g/mol) + 3*(1.01 g/mol) + 3*(16.00 g/mol) = 78.00 g/mol
2. Calculate the mass of Al(OH)3 that reacted with the HCl:
Moles of Al(OH)3 = 0.000376 mol (from your calculation)
Mass of Al(OH)3 = (0.000376 mol) * (78.00 g/mol) = 0.0293 g
3. Calculate the mass percent of Al(OH)3 in the antacid sample:
Mass of entire antacid sample = 0.188 g (as given)
Mass percent of Al(OH)3 = (0.0293 g / 0.188 g) * 100% ≈ 15.6%
Therefore, the mass percent of Al(OH)3 in the antacid sample is approximately 15.6%.
Remember to round your final answer to an appropriate number of significant figures based on the given information.