Lab 10 Report

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1 Lab 10. Determination of Vitamin C by Redox Titration Name: X Lab Date: November 2, 2023 Student Data Used: X 1. Data and observations (46 points): • Mass of KIO 3: (2 points) 1.1075 g • 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 mL 0.05 mL 0.12 mL 0.09 mL Final Volume (mL) 32.10 mL 31.10 mL 30.67 mL 21.70 mL Delivered Volume (mL) 31.90 mL 31.05 mL 30.55 mL 21.61 mL Average (mL) and standard deviation (mL) 31.17 mL ± 0.683 mL Grubbs Test for Outliers: Trial 4 had a suspicious value (21.61 mL delivered): Average mL delivered = 28.78 mL → |21.61 − 28.78| 4.811 = 1.490 > 1.463; Trial 4 can be discarded with 95% confidence. • 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 mL 0.45 mL 0.30 mL 0.23 mL Final Volume (mL) 16.05 mL 13.87 mL 13.25 mL 14.94 mL Delivered Volume (mL) 16.03 mL 13.42 mL 12.95 mL 14.71 mL Average (mL) and standard deviation(mL) 14.28 mL ± 1.385 mL

2 Grubbs Test for Outliers: Trial 1 had a suspicious value (16.03 mL delivered): Average mL delivered = 14.28 mL → |16.03 − 14.28| 1.385 = 1.264 < 1.463; Trial 1 can be kept with 95% confidence. 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) ? Ascorbic Acid C 6 H 8 O 6 : C * 6 = 12.011 g/mol * 6 = 72.066 g/mol H * 8 = 1.0078 g/mol * 8 = 8.0624 g/mol O * 6 = 15.999 g/mol * 6 = 95.994 g/mol C 6 H 8 O 6 = 72.066 + 8.0624 + 95.994 = 176.1224 g/mol Potassium Iodate KIO 3 : K = 39.098 g/mol I = 126.90 g/mol O * 3 = 15.999 g/mol * 3 = 47.997 g/mol KIO 3 = 39.098 + 126.90 + 47.997 = 213.995 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)? Back titration involves the addition of excess standard reagent to an analyte. Excess reagent is then titrated with a 2 nd reagent or with a standard solution of the analyte. It is used when molar concentration of the excess reagent is known, but the strength or concentration of the analyte needs to be determined. This titration method is commonly associated with insoluble salts. For this reaction involving vitamin C, back titration is useful because we will make an excess of Iodide and pass the endpoint for Ascorbic acid, then back titrate with the known reagent, Thiosulfate, to determine the Ascorbic acid concentration. Harris, D. C., & Lucy, C. A. (2020). Quantitative chemical analysis (p. 288). Macmillan Learning.

3 (3) Equation 2-4 are all redox reactions, identify the reducing and oxidizing agents in each of the three equations (4×3=12 points)? I tried my best to show and explain these. To preface, here is the reaction for triiodide: ? 2 (??) + ? − ↔ ? 3 − ; In this equation, Iodide is neutral and Iodine ion has a -1 charge. Iodide is insoluble because it is nonpolar, so treatment with Iodine ion produces polar, soluble triiodide. Equation 2: ?? 3 − + 8? − + 6? + ↔ 3? 3 − + 3? 2 ? ?? 3 − + 8? − + 6? + ↔ (3? 2 + 3? − ) + 3? 2 ? ? 𝑥 ? 3 −2 + 8? − + 6? + ↔ (3? 2 0 + 3? −1 ) + 3? 2 ? For I in IO 3 - : ? + ((−2) ∗ 3) = −1 → ? + −6 = −1 → ? = 5 . So, in Iodate, Iodine has a +5 charge. There is an overall 5e- change. I - in ?? 3 − is reduced from +5 to 0. Then, 5 I - from 8I - are oxidized from -1 to 0. Reduced and Oxidant: I - in 𝑰𝑶 𝟑 − Oxidized and Reductant: 5 I - from 8I - Electrons Transferred: 5e- Equation 3: (Charges shown below element symbol) ? 6 𝑥 ? 8 +1 ? 6 −2 + ? 3 − + ? 2 ? ↔ ? 6 𝑥 ? 6 +1 ? 6 −2 + 3? − + 2? + ? 6 𝑥 ? 8 +1 ? 6 −2 + (? 2 0 + ? − ) + ? 2 ? ↔ ? 6 +1 ? 6 +1 ? 6 −2 + 3? − + 2? + For C in C 6 H 8 O 6 : 6? + 8(1) + 6(−2) = 0 → 6? + 8 − 12 = 0 → 6? − 4 = 0 → 6? = 4 → ? = 2 3 . For C in C 6 H 6 O 6 : 6? + 6(1) + 6(−2) = 0 → 6? + 6 − 12 = 0 → 6? − 6 = 0 → 6? = 6 → ? = 1. Carbon in C 6 H 8 O 6 is increased by 1/3 when converted to C 6 H 6 O 6 . 1/3 * 6 = 2 e- transferred. So, C is oxidized from 2/3 to 1 when converted from C 6 H 8 O 6 to C 6 H 6 O 6 . I 2 , from I 3 - , is reduced to one of the I - in 3 I - , and 2 e- are transferred.

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4 Reduced and Oxidant: I 2 in 𝑰 𝟑 − Oxidized and Reductant: C in C 6 H 8 O 6 Electrons Transferred: 4e- Equation 4: (Charges shown below element symbol) ? 3 − + 2? 2 ? 3 2− ↔ ? 4 ? 6 2− + 3? − (? 2 0 + ? − ) + 2? 2 𝑥 ? 3 2− ↔ ? 4 𝑥 ? 6 2− + 3? − For S 2 in 2S 2 O 3 2- : [2(?) + 3(−2)] = −2 → 2? − 6 = −2 → 2? = 4 → ? = 2 For S 4 in S 4 O 6 2- : 4? + 6(−2) = −2 → 4? − 12 = −2 → 4? = 10 → ? = 2.5 4 Sulfur atoms total are exchanging electrons. S 2 in 2S 2 O 3 2- is oxidized from 2 to 2.5, which is .5e- change. .5 * 4 = 2e- transferred. 2 I - from I 3 - are reduced from -1 to 0. 1 * 2 = 2e- transferred. Reduced and Oxidant: I - in 𝑰 𝟑 − Oxidized and Reductant: S in S 2 O 3 2- Electrons Transferred: 4e-

5 Post-lab Questions. Data treatment (32 points): (4) Calculate the concentration (molarity) of KIO 3 (3 points) and the amount of IO 3 - in 25.00 mL of the solution (3 points). 1.1075 ? ??? 3 × 1 ??? 213.9950 ? ??? × 1 ? 0.500 ? = 0.01035 ? ??? 3 0.01035 ? ??? 3 × (0.025 ?) = 2.588 × 10 −4 ??? ??? 3 → ?: ?? 3 𝑖? 1: 1, ?? ?ℎ??? ??? 2.588 × 10 −4 ??? ?? 3 50.00 ?? = 2 × 2.588 × 10 −4 ??? ?? 3 = 5.175 × 10 −4 ??? ?? 3 (5) Use equation 2 (stoichiometry!) to find out the moles of I 3 - involved in each titration of using 25.00 mL of KIO 3 solution ( 3 points ). ?? 3 − + 8? − + 6? + ↔ 3? 3 − + 3? 2 ? 2.588 × 10 −4 ??? ?? 3 × 3 ??? ? 3 − 1 ??? ?? 3 − = 7.763 × 10 −4 ??? ? 3 − 50.00 ??: 2 × 7.763 × 10 −4 ??? ? 3 − = 1.553 × 10 −4 ??? ? 3 − (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 ). ? 3 − + 2? 2 ? 3 2− ↔ ? 4 ? 6 2− + 3? − 7.763 × 10 −4 ??? ? 3 − × 2 ??? ? 2 ? 3 2− 1 ??? ? 3 − = 1.553 × 10 −4 ??? ? 2 ? 3 2− ?????𝑖?? ?? ? 2 ? 3 2− : 1.553 × 10 −4 ??? ? 2 ? 3 2− × 1 ? 0.03117 ? = 0.04981 ? ? 2 ? 3 2− ???𝑖????? (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 ). 0.04981 ? ? 2 ? 3 2− × 0.01428 ? ? 2 ? 3 2− ???𝑖????? 𝑖? ???? ?𝑖????𝑖?? = 7.113 × 10 −4 moles S 2 O 3 2−

6 b) Use equation 4 to figure out the moles of I 3 - reacted with the delivered S 2 O 3 2- ( 3 points ). ? 3 − + 2? 2 ? 3 2− ↔ ? 4 ? 6 2− + 3? − 7.113 × 10 −4 moles S 2 O 3 2− × 1 mol I 3 − 2 mol S 2 O 3 2− = 3.556 × 10 −4 ??? ? 3 − ???? ???? ?𝑖????𝑖?? c) With the results in step (5) and (7b), find out the moles of I 3 - that has reacted with ascorbic acid ( 3 points ). Moles of I 3 - from a 50.00 mL titration of KIO 3 - minus moles of I 3 - in back titration 1.553 × 10 −3 ??? ? 3 − − 3.556 × 10 −4 ??? ? 3 − = 1.197 × 10 −3 ??? ? 3 − ??????? ?𝑖?ℎ 𝐴. 𝐴. d) Use equation 3 to figure out the moles of reacted ascorbic acid solution and molarity ( 3 points ). ? 6 ? 8 ? 6 + ? 3 − + ? 2 ? ↔ ? 6 ? 6 ? 6 + 3? − + 2? + 1: 1 ???𝑖? ?? ? 3 − ?? ? 6 ? 8 ? 6 → 1.197 × 10 −3 ??? ? 6 ? 8 ? 6 50.00 ?? ? 6 ? 8 ? 6 ??? ?? ?? ??𝑖?𝑖??? 250.0 ?? ?????𝑖?? → 1.197 × 10 −3 ??? ? 6 ? 8 ? 6 × 250.0 ?? 50.00 ?? = 5.985 × 10 −3 ??? 𝐴. 𝐴. 5.985 × 10 −3 ??? 𝐴. 𝐴. 0.2500 ? = 0.02394 ? 𝐴. 𝐴. (8) Find out the mass of vitamin C in one tablet ( 3 points ) and calculate the relative error of your result ( 2 points ). 5.985 × 10 −3 ??? 𝐴. 𝐴. → ???? ?? ??? ?????? 𝑖? 1000 ?? → 5.985 × 10 −3 ??? 𝐴. 𝐴. × 176.1224 ? 𝐴. 𝐴. 1 ??? 𝐴. 𝐴. = 1.054 ? × 1000 ?? = 1054 ?? ?????𝑖?? ????? = |1054 ?? − 1000 ??| 1000 ?? = 0.054 ??????? ????? = |1054 ?? − 1000 ??| 1000 ?? × 100% = 5.4%

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7 (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 indicator is added, the solution will instantly begin to react with iodine and turn a deep blue color. The iodide ions will be oxidized by the indicator and once completely reacted, the titration will be complete, and the solution will become colorless as all the iodide is consumed. Harris, D. C., & Lucy, C. A. (2020). Quantitative chemical analysis (p. 381, 385-387). Macmillan Learning.