Vitamin C lab report sheet - Ciara Bell

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Kennesaw State University *

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Chemistry

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Dec 6, 2023

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docx

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Lab 10. Determination of Vitamin C by Redox Titration Name: Ciara Bell Due Date: 11.08.2023 1. Data and observations (46 points): Mass of KIO 3: (2 points) - 1.1075g Table 1: Data of the standardization of sodium thiosulfate. You can solve the average and standard deviation using Excel ( 22 points ): Trial 1 Trial 2 Trial 3 Trial 4 Initial volume (mL) 0.20 0.05 0.12 0.09 Final Volume (mL) 32.10 31.10 30.67 21.70 Delivered Volume (mL) 31.90 31.05 30.55 21.61 Average (mL) and standard deviation (mL) n=3 x=31.17 mL s x =0.6825 - G test : G calc =1.490 G table =1.463 1.490 > 1.463 = Reject Data Table 2: Data of the back titration of vitamin C You can solve the average and standard deviation using Excel ( 22 points ): Trial 1 Trial 2 Trial 3 Trial 4 Initial volume (mL) 0.02 0.45 0.30 0.23 Final Volume (mL) 16.05 13.87 13.25 14.94 Delivered Volume (mL) 16.03 13.42 12.95 14.71 Average (mL) and standard deviation(mL) n=4 x=14.28mL 1
s x =1.39 2. Data treatment and calculations Pre-lab questions (22 points): (1) In the lab you use potassium iodate (primary standard, KIO 3 ) for vitamin C determination. What are the molar masses of these two substances (2×2=4 points) ? MM KI O 3 = ( 39.098 ) + ( 126.90 ) + ( 15.999 x 3 ) = 214.001 g / mol MM C 6 H 8 O 6 = ( 12.011 x 6 ) + ( 1.008 x 8 ) + ( 15.999 x 6 ) = 176.124 g / mol (2) What is a back titration? How is the vitamin C concentration determined in this experiment by a back titration (3×2 = 6 points)? A back titration is a titration in which the molarity of an unknown compound is concluded by reacting it with a known amount and concentration of the excess reagent. Vitamin C’s concentration is determined by titrating it initially with KIO 3 and the doing a back titration with KIO 3 to find the molarity and amount of Vitamin C. (3) Equation 2-4 are all redox reactions, identify the reducing and oxidizing agents in each of the three equations (4×3=12 points)? ¿ + 3 H 2 O + ¿ 3 I 3 ¿ ¿ + 6 H ¿ ¿ + 8 I ¿ Eq. 2: I O 3 ¿ The element I in IO 3 - is reduced, so IO 3 - is the oxidizing agent. The element I in I - oxidized, so I - is the reducing agent. 2
+ ¿ ¿ + 2 H ¿ ¿ + H 2 O C 6 H 6 O 6 + 3 I ¿ Eq. 3: C 6 H 8 O 6 + I 3 ¿ The element I in I 3 - is reduced, so I 3 - is the oxidizing agent. The element C in C 6 H 8 O 6 oxidized, so C 6 H 8 O 6 is the reducing agent. ¿ 2 ¿ + 3 I ¿ 2 ¿ S 4 O 6 ¿ ¿ + 2 S 2 O 3 ¿ Eq . 4 : I 3 ¿ The element I in I 3 - is reduced, so I 3 - is the oxidizing agent. The element S in 2 ¿ S 2 O 3 ¿ oxidized, so 2 ¿ S 2 O 3 ¿ is the reducing agent 3
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