Vitamin C lab report sheet - Ciara Bell

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Dec 6, 2023

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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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Post-lab Questions. Data treatment (32 points): (4) Calculate the concentration (molarity) of KIO 3 (3 points) and the amount of IO 3 - in 50.00 mL of the solution (3 points). 1.1075 g KI O 3 1 x 1 mol KI O 3 214.001 g KI O 3 x 1 0.5000 L = 0.010350 M KI O 3 0.010350 M x 25.00 mL = 0.25875 mmol = 0.00025875 mol KI O 3 (5) Use equation 2 (stoichiometry!) to find out the moles of I 3 - involved in each titration of using 50.00 mL of KIO 3 solution ( 3 points ). − ¿ 3 mol I 3 − ¿ 1 mol KI O 3 = 0.00077625 mol I 3 ¿ 0.00025875 mol KI O 3 1 x ¿ (6) Use equation 4 to find the moles of reacted sodium thiosulfate ( 3 points ), and then use the average of delivered volume from table 1 to find out the molarity of sodium thiosulfate ( 3 points ). 4

2 − ¿ 1 molI 3 − ¿ = 0.0015525 mol S 2 O 3 ¿ 2 mol S 2 O 3 2 − ¿ ¿ 0.00077625 mol I 3 − ¿ 1 x ¿ ¿ 2 − ¿ 0.0015525 mol S 2 O 3 2 − ¿ 0.03117 L = 0.04981 M S 2 O 3 ¿ ¿ (7) Use data of back titration of vitamin C to find the molarity of vitamin C: a) Use the average volume of sodium thiosulfate in the back titration and its standardized concentration in step (6) to find out the moles of delivered S 2 O 3 2- . ( 3 points ). 2 − ¿ 0.04981 molS 2 O 3 2 − ¿ 1 L x 0.03117 L 1 = 0.00155258 mol S 2 O 3 ¿ ¿ b) Use equation 4 to figure find out the moles of I 3 - reacted with the delivered S 2 O 3 2- ( 3 points ). − ¿ 2 molS 2 O 3 2 − ¿ = 0.00077629 mol I 3 ¿ 1 mol I 3 − ¿ ¿ 0.00155258 mol S 2 O 3 2 − ¿ 1 x ¿ ¿ c) With the results in step (5) and (7b), find out the moles of I 3 - that has reacted with ascorbic acid ( 3 points ). 5

− ¿ 0.00077629 mol − 0.00077625 mol = 3.90 x 10 − 8 molI 3 ¿ d) Use equation 3 to figure out the moles of reacted ascorbic acid solution and molarity ( 3 points ). 1 molI 3 − ¿ x 1 0.05000 L = 7.77 x 10 − 7 M Ascorbic Acid 3.90 x 10 − 8 mol I 3 − ¿ 1 x 1 molC 6 H 8 O 6 ¿ ¿ (8) Find out the mass of vitamin C in one tablet ( 3 points ) and calculate the relative error of your result ( 2 points ). 7.77 x 10 − 7 mol Ascorbic Acid 1 L x 0.250 L 1 x 176.124 Ascorbic Acid 1 mol = 3.421 x 10 − 6 g (9) Read your textbook page 381, 385-387, explain the reactions when you add the starch indicator and when you observe the color change at the end point. When a starch indicator is added to a reaction, it reacts when I 2 is present, but it doesn’t perform a redox reaction. At the endpoint, the solution will show the color blue because of its complex with Iodine. 6

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