Question

Instead of using ratios for back titrations we can also use molarities, if our solutions are standardized. A 0.196 g 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.

Part II:
1. Antacid Brand	Walgreens
2. Mass of Whole Tablet (g)	1.316
3. Mass of Crushed Tablet and Boat (g)	1.507
4. Mass of Boat after Tablet removed (g)	0.229
5. Mass of Tablet added to 200mL acid (g)	1.278
			If needed
Part III:	Trial 1	Trial 2	Trial 3
6. Stomach Acid (HCl) used (mL)	25.0	25.0
7. Initial Buret Reading (mL)	0.21	0.12
8. Final Buret Reading (mL)	33.95	33.61
9. Volume NaOH added (mL)	33.74	33.49
10. Average Volume NaOH Used (mL)	33.615
Part IV:			If needed
	Trial 1	Trial 2	Trial 3
11. Volume filtered acid added to flask (mL)	25.0	25.0
12. Initial Buret Reading (NaOH) (mL)	0.81	0.96
13. Final Buret Reading (NaOH) (mL)	26.31	27.12
14. Volume NaOH Added (mL)	25.50	26.16
15. Volume HCl Remaining in sample (mL)	18.96	19.46
16. Average HCl Remaining in 25 mL (mL)	19.21
17. Volume HCl Neutralized in 200mL Sample (mL)	46.32
18. Volume HCl Neutralized by Whole Tablet(mL)	47.70

Answer 1

The aluminum hydroxide is neutralized with the hydrochloric acid according to the following reaction:

ec.1

The excess acid is neutralized with a standardized solution of sodium hydroxide:

ec.2

Taking into account the amount of HCl added and its concentration we can calculate the total moles of HCl

From the second titration we can obtain the moles of excess HCl:

By difference between the total and excess moles we can know the moles that react with the aluminum hydroxide:

The ratio HCl:Al(OH)₃ is 3:1, that is, for each mole of aluminum hydroxide 3 mol of hydrochloric acid react; therefore, the grams of Al(OH)₃ (MW=78g/mol) are:

So, The percentage of Al(OH)₃ in the antacid is: