1. Calculate the mass of iron(II) ammonium sulfate hexahydrate, Fe(NH©ž)©ü(SO©ž)©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©ž.
7. Take a burette from the Glassware shelf and place it on the workbench.
8. Fill the burette with 20 mL KMnO©ž solution from the Chemicals shelf.
9. Titrate the iron(II) ammonium sulfate solution with KMnO©ž until the endpoint is reached indicated by the first appearance of a pink color that indicates excess MnO©ž⁻ ion.
10. NOTE: With this redox reaction, five Fe⁺©÷ ions react with just one MnO©ž- ion, so it will only take a few mL of KMnO©ž 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©ž, and perform a second titration to an exact endpoint of 1-2 drops of KMnO©ž⁻.
The solution turned a very light pink after 3.70mL of KMno4- was added.
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©ž (mL): 3.70mL??
(d) In order to calculate the concentration of KMnO©ž (mol/L), you must remember that the stoichiometry of the redox reaction indicates that 5 Fe©÷⁺ ions are needed for every 1 Mn©ž⁻ ion. At the equivalence point, then:
(moles of Fe©÷⁺) = (moles of KMnO©ž⁻)
which is written using molar concentrations as:
(C1 * V1) Fe = 5*(C2 * V2)KMnO©ž
The prof has given you a formula. Just fill in the blanks from above and solve for concn KMnO4.
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