Question

Part A: How many mL of a 100 mg/dL Ferrozine solution are needed to complex   5.5×10⁻⁸ moles   of iron?

Part B: How many moles of Fe³⁺ can be reduced by   1.3 mL    of   0.291 M   hydroxylamine solution?

Part C: As part of this laboratory, you may have to add iron to a solution. If you have   16 mL    of a solution that contains   37   μg dL^-1,  how many mL of 500 μg dL^-1 iron must be added to reach a concentration of 150 μg dL^-1?

Thank you in advance! I'm really struggling with this one!

Answer 1

Ans. A. Balanced reaction: Fe²⁺ + 3 FZ^2- -------> Fe(FZ₃)^4-

Stoichiometry: 1 mol Fe²⁺ requires 3 mol Ferrozine (FZ) for complexion.

Using stoichiometry, the number of moles of FZ required to complex 5.5x 10^-8 mol Fe²⁺ is equal to-

            Moles of FZ required = 3 x (5.5x 10^-8 mol) = 1.55 x 10^-7 mol

            Mass of 1.55 x 10^-7 mol FZ = Moles x Molar mass

                                                = 1.55 x 10^-7 mol x (492.46 g/ mol)                                  

                                                = 7.63313 x 10^-5 g

                                                = 7.63313 x 10^-2 mg                         ; [1 g = 10³ mg]

                                                = 0.0763313 mg

So, to complex the given moles of Fe, we need 0.0763313 mg of FZ.

Now,

Volume of FZ solution required = Mass of FZ required / Concentration of FZ solution

                                                = 0.0763313 mg / (100 mg/ dL)

                                                = 0.000763312 dL

                                                = 0.763312 mL                                ; [1 dL = 100.0 mL]

                                                = 763.31 uL

Thus, required volume of FZ solution = 0.763312 mL = 763.31 uL

Ans. B. Balanced reaction: 2Fe³⁺ + 2NH₂OH ----> 2Fe²⁺ + N2 + 2H⁺ + 2H2O

Stoichiometry: 2 mol Fe³⁺ is reduced by 2 mol hydroxylamine (NH₂OH).

Now,

Moles of NH₂OH in 1.3 mL, 0.291 M sample = Molarity x Volume (in Liters)

                                                = 0.291 M x 0.0013 L                       ; [1 L = 1000 mL]

                                                = (0.291 mol/ L) x 0.0013 L             ; [1 M = 1 mol/ L]

                                                = 0.0003783 mol

Using stoichiometry-

            Number of moles of Fe³⁺ reduced = Number of moles of NH₂OH present

                                                            = 0.0003783 mol

Ans. C.

Let the required volume = Y dL.

So, after adding Y dL (from stock solution) to solution 1 (16.0 mL, 37 ug/dL), final volume of this solution becomes (16.0 + Y) dL. And the resultant concentration will be 150 ug/dL

So, solution 1 is our desired resultant solution, with volume V1 = (16.0 + Y) dL ; and concertation, C1 = 150 ug/dL

Consider stock solution as solution 2. Since we require to take Y dL from this solution for addition, the required volume, V2 = Y dL. The concertation of this solution is given, C2 = 500 ug/dL

Now,

Using C1V1 = C2V2

C1= Concentration of initial solution 1, V1= volume of initial solution 1 [= solution 1]     

C2= Concentration of final solution 2, V2= volume of final solution 2 [= solution 2]

            (150 ug/ dL) x (16.0 +Y) dL = (500 ug/dL) x Y dL

            Or, (16.0 +Y) dL =[(500 ug/dL) x Y dL] / (150 ug/ dL)

            Or, (16.0 +Y) dL = 3.33333Y dL

            Or, 16.0 +Y = 3.33333Y

            Or, Y = 16.0 / 2.33333

            Hence, Y = 6.857 dL

Thus, required volume of solution 2 to be mixed with solution 1 = 6.857 dL