Read an authoritative source for a discussion of spectrophotometry. The two comp

ID: 531336 • Letter: R

Question

Read an authoritative source for a discussion of spectrophotometry. The two components of the solution you will be making in this experiment must be used with great care. List these two substances and describe the problems associated with each. 1, 10-Phenanthroline forms a bright red complex with iron(II). The complex has an analytical wavelength of 510 nm and has a high molar absorptivity. Hydroxylamine hydrochloride is added to keep the iron m a reduced +2 state, and the pH is controlled using an acetate buffer. A series of iron(II)-phenanthroline solutions were made by adding various amounts of 4.300 times 10^-4 M Fe^2+ to 5 mL each of hydroxylamine hydrochloride, acetate buffer, and 1, 10-phenanthroline. Distilled water was added to make the total volume of each solution 50.00 mL The percent transmittance of each solution was read, using a blank containing everything but the iron(II) solution. Data for the determination are Calculate the molar concentration of [Fe(phenanthroline)_3]^2+ in each solution.

Explanation / Answer

2) I need to know what experiment you are performing to answer this question. In particular I need to know which reagents you are using to tell you what problems are associated with the use of each.

3) The balanced chemical equation for the reaction is

Fe2+ + 3 o-phen -------> [Fe(o-phen)3]2+ (o-phen = 1,10-phenanthrolin)

As per the balanced stoichiometric equation,

1 mole Fe2+ = 1 mole iron(II)-phenanthrolin …..(1)

Take the first entry in the table as an example. We will work out this one completely.

Volume of Fe2+ taken = 1.00 mL.

Concentration of stock Fe2+ taken = 4.300*10-4 M.

Moles of Fe2+ taken = (volume of Fe2+)*(concentration of Fe2+ in mol/L) = (1.00 mL)*(4.300*10-4 mol/L) = 4.300*10-4 mmole.

As per the stoichiometric equation,

4.300*10-4 mmole Fe2+ = 4.300*10-4 mmole iron (II)-phenantholin

Volume of the final solution = 50.00 mL.

Concentration of iron(II)-phenanthrolin in the final solution = (moles of iron(II)-phenanthrolin)/(volume of iron(II)-phenanthrolin in the final solution) = (4.300*10-4 mmole)/(50.00 mL) = 8.600*10-6 mol/L = 8.600*10-4 M (ans).

Now, we can work out the following table:

Trial

Volume of 4.300*10-4 M Fe2+ (mL)

Millimoles of Fe2+ = millimoles of iron(II)-phenanthrolin

Concentration of iron (II)-phenanthrolin = (millimoles of iron(II)-phenanthrolin)/(volume of final solution)

1.00

4.300*10-4

8.600*10-6 M

2.00

8.600*10-4

1.720*10-5 M

5.00

2.150*10-3

4.300*10-5 M

7.00

3.010*10-3

6.020*10-5 M

9.00

3.870*10-3

7.740*10-5 M

10.00

4.300*10-3

8.600*10-5 M