The following three steps axe carried out in the experiment: 0.1212g of Fe(NH_4)

Question

The following three steps axe carried out in the experiment: 0.1212g of Fe(NH_4)_2(SO_4)_2 middot 6H_2O is diluted to 100 ml using a volumetric flask (i.e., 100.00mL) 2.00 mL of this diluted solution is pipetted into a 50-ml. volumetric flask. To the solution in this flask. 15.00 mL of deionized water is added About 50 mg of ascorbic acid also is added and dissolved in the solution. Using a graduated cylinder. 15.0 mL ol 0.0020 M bipyridyl complexing agent is added This turns the solution to a red color. Finally, a pH 4.7 buffer solution is added to the volumetric flask until the total volume of the solution is 50.00 mL. 2.00 mL is transferred from the 50-ml. volumetric flask to a clean test tube, and sufficient deionized water is added to the test tube to make a total solution volume of 10.00 mL. Calculate the molar concentration of Fe(II) at the end of Step 1. at the end of step 2, and at the end of step 3.

Explanation / Answer

Ans. Step 1: Moles of Fe(NH4)2(SO4)2.6H2O = Mass/ molar mass

                                                            = 0.1213 g / (392.14 g mol-1)

                                                            = 3.093 x 10-4 moles

1 mol Fe(NH4)2(SO4)2.6H2O contains 1 mol Fe(II). So the number of moles of Fe(II) is equal to the number of moles of Fe(NH4)2(SO4)2.6H2O.

Therefore,

            Number of moles of Fe(II) = 3.093 x 10-4 moles

Final volume made upto = 100.0 mL                         ; (Volume of the volumetric flask)

                                                = 0.100 L                    ; [1 L = 1000.0 mL]

Now,

            Concentration of Fe(II) = Moles of Fe(II) / Volume of solution made upto

                                                = 3.093 x 10-4 moles / 0.100 L

                                                = 3.093 x 10-3 mol/ L

                                                = 3.093 x 10-3 M

Thus, molarity of this solution = 3.093 x 10-3 M

Step 2: Complexation does not change the identity of Fe(II). Thus, amount of Fe(II) in the solution remains the same.

2.0 mL of 3.093 x 10-3 M Fe(II) is diluted to a final volume of 50.0 mL to yield the final solution of step 2.

Using formula: M1V1 = M2V2

M1= molarity of solution 1, V1= volume of solution 1      ;[Solution, Step 1]

M2= molarity of solution 2, V2= volume of solution 2      , [Solution, step 2]

            Or, 3.093 x 10-3 M x 2.0 mL = M2 x 50.0 mL

            Or, M2 = (3.093 x 10-3 M x 2.0 mL) / 50.0 mL = 1.237 x 10-4 M

Thus, [Fe(II)] at the end of step 2 = 1.237 x 10-4 M

Step 3: 2.0 mL of 1.237 x 10-4 M Fe(II) of solution 2 is diluted to a final volume of 10.0 mL to yield the final solution of step 3.

Now,

            1.237 x 10-4 M x 2.0 mL = M2 x 10.0 mL

            Or, M2 = (1.237 x 10-4 M x 2.0 mL) / 10.0 mL = 2.474 x 10-5 M

Thus, [Fe(II)] at the end of step 3 = 2.474 x 10-5 M

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.