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 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
Chapter10: Potentiometry And Redox Titrations
Section: Chapter Questions
Problem 8P
Related questions
Question
This is a practice problem.
![Experiment 3. Assay of Cupric Sulfate
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 mlaay)*Nx meg
x 100 %
wt
Reaction (fillin the coefficients on the blank spaces):
- CUSO4 5H2O +_ KI →_ Cul +_ 12 +,
K2SO4 +_ H2O
and 2NazS2Oa + l2 + Na:S«Os + 2Nal
Reduction: Oxidizing agent
Cu*2 +
e- - Cu1
Oxidation: Reducing agent 21-1 → 12° +
* 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
Volume of titration/assay
Percentage
28.0 ml
15.1 ml
Calculation and Conclusion
On space below, solve the titer value and percentage purity of the sample then interpret.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F59b472b7-7944-4d7e-bd2a-377808f8a955%2Fb92f791b-caaf-4f4d-9ac7-93eda7fea32c%2Fwt4rxz_processed.png&w=3840&q=75)
Transcribed Image Text:Experiment 3. Assay of Cupric Sulfate
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 mlaay)*Nx meg
x 100 %
wt
Reaction (fillin the coefficients on the blank spaces):
- CUSO4 5H2O +_ KI →_ Cul +_ 12 +,
K2SO4 +_ H2O
and 2NazS2Oa + l2 + Na:S«Os + 2Nal
Reduction: Oxidizing agent
Cu*2 +
e- - Cu1
Oxidation: Reducing agent 21-1 → 12° +
* 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
Volume of titration/assay
Percentage
28.0 ml
15.1 ml
Calculation and Conclusion
On space below, solve the titer value and percentage purity of the sample then interpret.
Expert Solution
![](/static/compass_v2/shared-icons/check-mark.png)
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 4 steps with 1 images
![Blurred answer](/static/compass_v2/solution-images/blurred-answer.jpg)
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you