Procedure:
1. Calculate the mass of iron(II) ammonium sulfate hexahydrate, Fe(NH©þ)©ü(SO©þ)©ü*6H©üO (MW = 392.14) required to make 100 mL of solution that is 0.2M in Fe©÷⁺ ions. You will be asked to show your calculation in the Assignment.
2. Take a 100 mL volumetric flask from the Glassware shelf and place it on the workbench.
3. Add the required mass of iron(II) ammonium sulfate hexahydrate to the flask, and add 30 mL H©üO to dissolve the compound and release the water of hydration. Then fill with water to the mark to make a 100 mL solution.
4. Take an Erlenmeyer flask from the Glassware shelf and place it on the workbench.
5. Pour 20 mL of iron(II) ammonium sulfate solution from the volumetric flask into the Erlenmeyer flask.
6. Acidify the iron(II) ammonium sulfate solution by adding 5 mL H©üSO©þ.
7. Take a burette from the Glassware shelf and place it on the workbench.
8. Fill the burette with 20 mL KMnO©þ solution from the Chemicals shelf.
9. Titrate the iron(II) ammonium sulfate solution with KMnO©þ until the endpoint is reached indicated by the first appearance of a pink color that indicates excess MnO©þ⁻ ion.
10. NOTE: With this redox reaction, five Fe⁺©÷ ions react with just one MnO©þ- ion, so it will only take a few mL of KMnO©þ to reach the endpoint. GO SLOWLY.
11. Perform at least one quick titration to determine the general range of the endpoint. Refill the burette to 20 mL KMnO©þ, and perform a second titration to an exact endpoint of 1-2 drops of KMnO©þ⁻.
The solution turned a very light pink after 3.70mL of KMno4- was added.
Questions:
1. Show your calculation for the mass of iron(II) ammonium sulfate hexahydrate used to prepare 100 mL of solution that has a concentration of 0.2M of Fe©÷⁺ ions.
0.2 mol/L x 0.1 L = 0.02 mol
0.02 mol x 392.14 g/mol = 7.8 g
2. Record and calculate the following for the final, exact titration:
(a) Volume of Fe©÷⁺ Solution (mL): ??
(b) Concentration of Fe©÷⁺ ions (mol/L): ??
(c) Volume of KMnO©þ (mL): 3.70mL??
(d) In order to calculate the concentration of KMnO©þ (mol/L), you must remember that the stoichiometry of the redox reaction indicates that 5 Fe©÷⁺ ions are needed for every 1 Mn©þ⁻ ion. At the equivalence point, then:
(moles of Fe©÷⁺) = (moles of KMnO©þ⁻)
which is written using molar concentrations as:
(C1 * V1) Fe = 5*(C2 * V2)KMnO©þ
Now solve for C2, the molar concentration of the KMnO©þ solution.
I could solve part D if I had help on solving parts A,B, and C!!
Procedure:
1. Calculate the mass of iron(II) ammonium sulfate hexahydrate, Fe(NH©þ)©ü(SO©þ)©ü*6H©üO (MW = 392.14) required to make 100 mL of solution that is 0.2M in Fe©÷⁺ ions. You will be asked to show your calculation in the Assignment.
You have some funny symbols and I can't read most of them. However, I think I can guess what some are.
You want how many mols of the iron compd? That is M x L = ?
Then g = mols/molar mass. You know mols and molar mass solve for grams.
2. Take a 100 mL volumetric flask from the Glassware shelf and place it on the workbench.
3. Add the required mass of iron(II) ammonium sulfate hexahydrate to the flask, and add 30 mL H©üO to dissolve the compound and release the water of hydration. Then fill with water to the mark to make a 100 mL solution.
4. Take an Erlenmeyer flask from the Glassware shelf and place it on the workbench.
5. Pour 20 mL of iron(II) ammonium sulfate solution from the volumetric flask into the Erlenmeyer flask.
20 mL will be how many mols of the iron compd? mols = M x L = ?
6. Acidify the iron(II) ammonium sulfate solution by adding 5 mL H©üSO©þ.
7. Take a burette from the Glassware shelf and place it on the workbench.
8. Fill the burette with 20 mL KMnO©þ solution from the Chemicals shelf.
9. Titrate the iron(II) ammonium sulfate solution with KMnO©þ until the endpoint is reached indicated by the first appearance of a pink color that indicates excess MnO©þ⁻ ion.
10. NOTE: With this redox reaction, five Fe⁺©÷ ions react with just one MnO©þ- ion, so it will only take a few mL of KMnO©þ to reach the endpoint. GO SLOWLY.
mols KMnO4 = 1/5 mol Fe
11. Perform at least one quick titration to determine the general range of the endpoint. Refill the burette to 20 mL KMnO©þ, and perform a second titration to an exact endpoint of 1-2 drops of KMnO©þ⁻.
The solution turned a very light pink after 3.70mL of KMno4- was added.
Questions:
1. Show your calculation for the mass of iron(II) ammonium sulfate hexahydrate used to prepare 100 mL of solution that has a concentration of 0.2M of Fe©÷⁺ ions.
0.2 mol/L x 0.1 L = 0.02 mol
0.02 mol x 392.14 g/mol = 7.8 g
This looks ok.
2. Record and calculate the following for the final, exact titration:
(a) Volume of Fe©÷⁺ Solution (mL): ??
Didn't you take 20 mL of the Fe soln?
(b) Concentration of Fe©÷⁺ ions (mol/L): ??
Isn't this 0.2M
(c) Volume of KMnO©þ (mL): 3.70mL??
(d) In order to calculate the concentration of KMnO©þ (mol/L), you must remember that the stoichiometry of the redox reaction indicates that 5 Fe©÷⁺ ions are needed for every 1 Mn©þ⁻ ion. At the equivalence point, then:
(moles of Fe©÷⁺) = (moles of KMnO©þ⁻)
which is written using molar concentrations as:
(C1 * V1) Fe = 5*(C2 * V2)KMnO©þ
The prof has given you a formula. Just fill in the blanks from above and solve for concn KMnO4.
Sure, I'll be happy to help you with parts (A), (B), and (C)!
(a) Volume of Fe²⁺ Solution (mL): From the given information, it states that 20 mL of Fe²⁺ solution was poured into the Erlenmeyer flask.
(b) Concentration of Fe²⁺ ions (mol/L): Given that the Fe²⁺ solution was prepared by dissolving the required mass of iron(II) ammonium sulfate hexahydrate in 100 mL of water to make a 0.2 M solution. So the concentration of Fe²⁺ ions is 0.2 M.
(c) Volume of KMnO₄ (mL): It states that the solution turned a very light pink after 3.70 mL of KMnO₄ was added.
Now, you can use the given volume and concentration of Fe²⁺ ions (from part B) and the volume of KMnO₄ (from part C) to calculate the concentration of KMnO₄ (C2) in part (D).
To solve parts A, B, and C, we need to use the information provided in the procedure.
(a) Volume of Fe²⁺ Solution (mL): In step 5, it is stated that 20 mL of the iron(II) ammonium sulfate solution is poured into the Erlenmeyer flask. Therefore, the volume of the Fe²⁺ solution used is 20 mL.
(b) Concentration of Fe²⁺ ions (mol/L): In the calculation for the mass of iron(II) ammonium sulfate hexahydrate used in step 1, we determined that 7.8 g is required to make 100 mL of a 0.2 M Fe²⁺ solution. Therefore, the concentration of Fe²⁺ ions is 0.2 M.
(c) Volume of KMnO₄ (mL): In the question, it is mentioned that the solution turned a very light pink after 3.70 mL of KMnO₄ was added.
Now, let's solve part D:
(d) In order to calculate the concentration of KMnO₄ (mol/L), we can use the stoichiometry of the redox reaction. It states that 5 Fe²⁺ ions react with 1 MnO₄⁻ ion.
At the equivalence point, the number of moles of Fe²⁺ will be equal to the number of moles of KMnO₄ used.
Let's assume:
C1 = Concentration of Fe²⁺ (mol/L)
V1 = Volume of Fe²⁺ Solution (L)
C2 = Concentration of KMnO₄ (mol/L)
V2 = Volume of KMnO₄ used (L)
Using the information from part D:
(C1 * V1) Fe = 5*(C2 * V2) KMnO₄
Substituting the known values:
(0.2 * 0.02) Fe = 5*(C2 * 0.00370) KMnO₄
Simplifying the equation:
0.004 Fe = 0.0185 KMnO₄
Now, we can solve for the concentration of KMnO₄ (C2):
C2 = (0.004 Fe) / (0.0185 KMnO₄)
Since we know the concentration of Fe²⁺ is 0.2 M, we can substitute this value:
C2 = (0.004 * 0.2) / (0.0185 KMnO₄)
Now, you can calculate the exact concentration of KMnO₄ by substituting the values into the equation.
Sure, I can help you solve parts A, B, and C of the question.
Part A: Volume of Fe²⁺ Solution (mL)
The volume of Fe²⁺ solution used in the titration is not given in the problem statement, so we cannot calculate it. You need to provide the volume of Fe²⁺ solution used in the titration in order to answer this question.
Part B: Concentration of Fe²⁺ ions (mol/L)
The concentration of Fe²⁺ ions can be calculated using the formula:
Concentration (mol/L) = moles / volume (L)
Since we have already calculated the mass of iron(II) ammonium sulfate hexahydrate used (7.8 g), we can convert it to moles using the molar mass of the compound (392.14 g/mol):
moles = mass / molar mass
moles = 7.8 g / 392.14 g/mol
moles = 0.0199 mol (rounded to four decimal places)
Now, we can calculate the concentration of Fe²⁺ ions:
Concentration = moles / volume
Concentration = 0.0199 mol / 0.100 L (since we are preparing 100 mL of solution)
Concentration = 0.199 mol/L (rounded to three decimal places)
Part C: Volume of KMnO₄ (mL)
As given in the problem statement, the volume of KMnO₄ added is 3.70 mL.
So, to summarize:
(a) Volume of Fe²⁺ Solution (mL): Please provide the volume used in the titration.
(b) Concentration of Fe²⁺ ions (mol/L): 0.199 mol/L (rounded to three decimal places).
(c) Volume of KMnO₄ (mL): 3.70 mL.