Question

A student performing our lab last semester determined that 27.68mL of base reacts with 25.0 mL of stomach acid. They then dissolved 1.666g of antacid into 200 mL of stomach acid and then removed a 20.0mL of the resulting solution. The titration of this 20 mL sample took 11.12mL of base. Calculate the volume of acid neutralized by the base in this 20 mL sample. Show your work to receive credit.

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
Part III:	Trial 1	Trial 2
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:
	Trial 1	Trial 2
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

Accroding to the titration,

25 mL of stomach acid needs 27.68 mL of NaOH to be neutralized. So, 1 mL of base or NaOH needs ( 25 / 27.68 ) mL = 0.9032 mL of stomach acid to be neutralized.

Now, during the titration of stomach acid after adding antacid 11.12 mL of base was needed to titrate 20 mL of stomach acid + antacid solution. From the volume ratio we know, 11.12 mL of base neutralizes ( 11.12 x 0.9032 ) mL = 10.04 mL of stomach acid. So, this the volume neutralized by the base. So, the remaining 9.96 mL of stomach acid was neutralized by antacid from 20 mL of solution.