Principle: lodometry The percentage purity of cupric sulfate set by the NF is 98.5 to 100.5% Procedure (reduce formulation to %, 1 trial only) 1. Weigh 650 mg of cupric sulfate (Cuso, · 5H;O) and dissolve in 50 ml of water. 2. Add 4 ml of acetic acid and 3 g of potassium iodide 3. Titrate the solution with 0.1 N sodium thiosulfate then adding 3 ml of starch TS as the end point is approached. 4. Perform a blank titration. (Blank titration is a solution containing all the reagents in the procedure except the analyte) 5. Calculate the percentage purity of the sample, using the formula: (mlgiana" mi) * N x meq 100 % % = Reaction (fillin the coefficients on the blank spaces): _CUSO4 5H2O + _ KI →_ Cul +_ 2 + _ K2SO4 + _ H2O and 2Nazs:Os + la → Na:S.Os + 2Nal Reduction: Oxidizing agent Cu*2 +. Cu* Oxidation: Reducing agent 211 - 12º + _ *- * Number of e- gain/loss of copper is ;which is the factor of cupric sulfate Data and Results Trial 1 Weight of cupriC sulfate Volume of blank 28.0 ml 15.1 ml Volume of titration/assay Percentage

Principle: lodometry The percentage purity of cupric sulfate set by the NF is 98.5 to 100.5% Procedure (reduce formulation to %, 1 trial only) 1. Weigh 650 mg of cupric sulfate (Cuso, · 5H;O) and dissolve in 50 ml of water. 2. Add 4 ml of acetic acid and 3 g of potassium iodide 3. Titrate the solution with 0.1 N sodium thiosulfate then adding 3 ml of starch TS as the end point is approached. 4. Perform a blank titration. (Blank titration is a solution containing all the reagents in the procedure except the analyte) 5. Calculate the percentage purity of the sample, using the formula: (mlgiana" mi) * N x meq 100 % % = Reaction (fillin the coefficients on the blank spaces): _CUSO4 5H2O + _ KI →_ Cul +_ 2 + _ K2SO4 + _ H2O and 2Nazs:Os + la → Na:S.Os + 2Nal Reduction: Oxidizing agent Cu2 +. Cu Oxidation: Reducing agent 211 - 12º + _ - Number of e- gain/loss of copper is ;which is the factor of cupric sulfate Data and Results Trial 1 Weight of cupriC sulfate Volume of blank 28.0 ml 15.1 ml Volume of titration/assay Percentage

Appl Of Ms Excel In Analytical Chemistry

2nd Edition

ISBN:9781285686691

Author:Crouch

Publisher:Crouch

Chapter10: Potentiometry And Redox Titrations

Section: Chapter Questions

Problem 8P

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M 23 (H) Inorganic Analytical Chemistry pard My courses CHEM 23 (H) FS-AY:20-21 Week 4-5: Quantitative Analysis by Gravimetry and Titration Quiz 4 Qu 10 The 500.0-mg sample of impure Na2 CO3 (FM: 105.96) required 22.00 mL of the HCI standard solution (the obtained concentration of HCI from the previous question, problem 5: Titrimetry short answer type) for complete neutralization. ut of 1. Give the balanced chemical equation between the sample, Na2CO3 and the standard, HCI. 2. What is the mass (g) of Na, CO3 in the sample? 3. Calculate the % (w/w) purity of a 500.0-mg sample of impure Na2 CO3 Note: Type your solution here or upload its pic in Jpeg or pdf format. Another option is to send it to me via email (subject: Fam name-Quiz 4) or messenger, privately. Do not forget to box the final answer and write your name. В I 1.
AaBbCcDdEe AaBbCcDc AaBbCcDdE AaBbCcDdF AaBb A v 三v Emphasis Heading 1 Normal Strong Sub PRE-LAB QUESTIONS 1. Calculate the molarity of a 6.15% (w/v%) sodium hypochlorite (NaOCI) solution. 2. Calculate the molarity of a solution of Erythrocin B (MW 879.9 g/mol) if a 0.5028-g sample is diluted to 100 mL in a volumetric flask and then two serial dilutions are carried out in which 5-mL aliquots are diluted to 100 mL using volumetric equipment. 3. Based on the table, what is the order of reaction for Red Dye #3 and bleach if the following graph gives a straight line: Order of m Straight Line Plot [Red Dye #3] vs time In[Red Dye #3] vs time 1/[Red Dye #3] vs time 1 ates) E Focus 目 MacBook Air
5. A 300.0 mg sample containing Na,CO3, NaHCO3 and NaOH and inert material either alone or in some combination was dissolved and titrated with 0.1000 M HCI the titration required 24.41 mL to reach the phenolphthalein endpoint. And an additional 8.67 mL to reach the methyl red endpoint. Determine the composition of the sample and calculate the percent of each titrated component.
DATA AND RESULTS: Titration Data (Standardization of EDTA) Example II Mass, g Caco, 0.0250 0.0251 0.0246 0.0254 Final reading, mL 25.00 24.91 24.87 25.31 Initial Reading, mL 0.00 0.02 0.02 0.02 Volume, mL EDTA 25.00 Molarity of EDTA, M 0.009992 B. Analysis Data Example II Volume of Sample, mL 25.00 24.90 25.10 25.0 Final reading, mL 27.46 25.08 25.06 24.99 Initial Reading, mL 0.00 0.02 0.02 0.02 Volume of EDTA, mL 27.46 Molarity of EDTA, M 0.009992 Ca" concentration, ppm 440 STATISTICAL EVALUATION OF RESULTS Standardization of EDTA Mean, M EDTA Median, M EDTA Standard Deviation Ci for u at 95% Prob. Level Reported value: M EDTA Ca" Concentration Mean, Ca" ppm Median, Ca" ppm Standard deviation Cl for u at 95% Prob. Level Reported value: Ca" ppm
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In this double titration experiment determining composition of soda ash, you are given the following data: M HCl titrant: 0.0501 M +/- 0.0507 Mass of soda ash sample, g: 0.2505 ± 2.0e-4 (Analytical Balance) Total volume of Sample Stock, mL: 75 ± 0.05 (100mL graduated cylinder) Volume of Sample Stock Aliquot, mL: 25 ± 0.3 Find the missing information. Show solutions
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In how much volume of diluted sulfuric acid should you dissolve a multivitamin tablet of approximately 1700 mg of Vitamin C to back analyze 10.00 mL of that solution, with 25.00 mL of 0.01 M KIO3 and 0.04 M thiosulfate to use approximately 8-10 mL of titrant ? You have 25, 50, 150, 250 and 500 mL flasks.
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