CHEM 110L - Experiment 10 It's Back... Titration

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Chemistry

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Apr 3, 2024

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Experiment 10 It’s Back… Titration Total Points: 40 Student’s Name: Lab Section: NOTE: You must show your work for all calculations; no work, no credit. Mass of Na 2 S 2 O 3 ______2.503g_______ (2 pts) Molarity of Na 2 S 2 O 3 _____0.010073M_________ (Show your calculation) 2.503g x 1mol Na2S2O3/248.18g/1L = 0.010073 (1 pt) Density of Bleach _____0.10262_________ g/mL (Show your calculation) 31.625g-21.363g/100mL = 0.10262 Initial Samples 20 mL Diluted Bleach Treated with KI Volume of Na 2 S 2 O 3 Used During Sample Titration (3 pts) Titrant Added (mL) to: After Starch Added Titrant Added (mL) to: Total Volume Color: Brown Initial Buret Volume, mL Light Orange/ Dark Yellow Dark Blue End Point: Colorless (mL) Sample 1 0 10.15 18.86 29.01 Sample 2 0 8.820 8.080 16.90 1. Calculate the following for each trial. Make sure to show all your work to receive credit. a. (2 pts) Moles of Na 2 S 2 O 3 used to titrate the bleach solution. S1: 0.010073M x 0.2901L = 0.00292 moles S2: 0.010073M x 0.1690L = 0.001702 moles b. (2 pts) Moles of I 2 formed by reaction with bleach. S1: 0.00292mol/2 = 0.001461 moles S2: 0.001702mol/2 = 8.512E-4 moles c. (4 pts) Molarity of OCl - present in the bleach solution. (Remember to account for the dilution of the original sample.) S1: (0.001461mol)(1mol OCl-/1mol I2)/(0.02L)(10)= 0.73054M S2: (8.512E-4)(1mol OCl-/1mol I2)/(0.02L)(10)= 0.42558M d. (2 pts) Grams of NaOCl present in the bleach (in g/mL): S1: (0.73054M)(74.44g/mol) = 54.382g/L -> 0.054382g/mL S2: (0.42558M)(74.44g/mol)= 31.680g/L -> 0.031680g/mL e. (2 pts) Percent NaOCl present in the bleach: S1: (0.054382g/mL)/(0.10262g/mL) x 100 = 52.99% S2: (0.031680g/mL)/(0.10262g/mL) x 100 = 30.87%

Experiment 10 It’s Back… Titration Total Points: 40 2. (6 pts) Use the following equation to answer the following questions: I 2 -starch complex (aq) + 2 S 2 O 2- 3(aq) → starch + 2 I - (aq) + S 4 O 6 2- (aq) What is being reduced? ____Iodine I2_____________________________ What is the oxidizing agent? ______Iodine I2_______________________ How many electrons are being gamed by each iodine atom? _1 e-__ Post-lab Questions NOTE: You must show your work for all calculations; no work, no credit. 1. (6 pts) Balance the redox reaction first (it is in basic medium). OCl - (aq) + CrO 2 - (aq) → Cl - (aq) + CrO 4 2- (aq) 2OH- + 3ClO- + 2CrO 2 2- → 3Cl- + 2CrO 4 2- + H2O 2. (4 pts) A titration experiment is set up to use 0.777M bleach (NaOCl) to analyze CrO 2 - . If 50.0 mL of the CrO 2 - (aq) solution required 12.27 mL of bleach to react to completion, what would you calculation as the molarity of the CrO 2 - solution? 0.777M x 0.1227L = 0.93379 mol NaCl 2mol/3mol x 0.93379 mol NaCl = 0.635586 mol/0.50L = 0.127M 3. (6 pts) Excess I 2 can be used to determine vitamin C levels in solution. Vitamin C (ascorbic acid) reacts in the following manner: C 6 H 8 O 6(aq) + I 2 → C 6 H 6 O 6(aq) + 2 H + (aq) + 2 I - (aq) The I 2 that did not react with ascorbic acid is then determined in a thiosulfate reaction as you did in this experiment. If 22.07 mL of 0.8310 M I 2 was added to a sample containing vitamin C, the unused I 2 required 20.00 mL of 0.143 M S 2 O 3 2- to react. How many moles of vitamin C were originally present? I2 + 2 S2O3 -> 2I + S4O6 I2= 0.8310M x 0.2207L = 0.1834017 mol S203= 0.143M x 0.2000L = 0.0286mol /2mol = 0.0143mol 0.1834017mol – 0.0143mol = 0.1691017 moles of Vitamin C

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