Can someone please help me answer the lab questions at the end. http://www2.ohlo
ID: 959213 • Letter: C
Question
Can someone please help me answer the lab questions at the end.
http://www2.ohlone.edu/people/jklent/labs/101B_labs/Vit%20C%20analysis.pdf
Data: Standardize NaOH Solution
Trial 1
Trial 2
1. Weight of KHP sample
0.804 g
0.806 g
2. Initial volume in buret
0.00 mL
0.00 mL
3. Final volume in buret
32.20 mL
32.30 mL
Data: Vitamin C Analysis
Brand Name: Up & Up
Tablet size from label: 500 mg
# of tabs used for each trial: 1
4. Weight of tablet(s) before grinding
0.653 g
0.653 g
5. Initial Volume in buret
0.00 mL
25.70 mL
6. Final volume in buret
23.70 mL
49.1 mL
Calculations: Molarity of NaOH
Trial 1
Trial 2
7. Moles of KHC8H4O4 used
3.94 x 10-3 mol
3.95 x 10-3 mol
8. Liters of NaOH used
0.0322 L
0.0323 L
9. Molarity of NaOH
0.122 M
0.122 M
10. Average Molarity
0.122 M
Calculations: Vitamin C Content
Trial 1
Trial 2
11. Liters of NaOH used
0.0232 L
0.0234 L
12. Moles of Vitamin C present
2.83 x 10-3 mol
2.85 x 10-3 mol
13. Miolligrams of Vitamin C in sample
408 mg
502 mg
14. Average milligrams per tablet
500 mg
15. % label correspondence
0 %
16. Binder in Tablets
30.6 %
Trial 1
Trial 2
1. Weight of KHP sample
0.804 g
0.806 g
2. Initial volume in buret
0.00 mL
0.00 mL
3. Final volume in buret
32.20 mL
32.30 mL
Explanation / Answer
Calculations for the molarity of NaOH are correct.
There are some errors in the calculation of vitamin - C content. The rectified data is given below
11. Liters of NaOH used:
trail-1: 0.0237 L
trail -2: 0.0234 L
12. Moles of Vitamin C present:
Trail-1: 0.0237L x 0.122 mol/L = 2.89x10-3 mol
Trail-1: 0.0234L x 0.122 mol/L = 2.85x10-3 mol
13. Miolligrams of Vitamin C in sample:
Trail-1: 2.89x10-3 mol x 176.12 g/mol = 0.509 g = 0.509 g x (1000 mg / g) = 509 mg
Trail-2: 2.85x10-3 mol x 176.12 g/mol = 0.502 g = 0.502g x (1000 mg / g) = 502 mg
14. Average milligrams per tablet:
= (509 mg + 502 mg) / 2 = 505 mg
16: Binder percentage in tablet:
[(0.653 - 0.505) / 0.653] x 100 = 22.7 %