Please read the following experiment reagents / procedure and answer the subsequ

Question

Please read the following experiment reagents / procedure and answer the subsequent question.

Reagent

Hydroquinone: (20mL) Freshly prepared solution of 10g/ L in DI water (provided).

o-Phenanthroline: (25mL) Dissolve 2.5 g in 100 mL of ethanol and add 900 mL of DI water (provided).

Trisodium citrate: (20mL) 25g/L. Dissolve 2.5 g of sodium citrate in 90 mL DI water in a 100 mL volumetric flask fill to the mark. Make this yourself. 100 mL could be enough for 5-6 people.

6 M HCl: (25mL) Dilute 124 mL of concentrated HCl up to 250 mL with DI water (provided).

Standard Fe (0.04 mg Fe/mL = 40ppm Fe): Dissolve 0.1405 g of reagent grade Fe(NH4)2(SO4)26H2O (FM 392.14) in DI water in a 500 mL volumetric flask containing 0.5 mL of concentrated sulfuric acid (H2SO4).

Spectrophotometer: Spectronic Genesys 20

Procedure

1. Pipet 10.00 mL of standard Fe solution into a beaker and measure the pH with pH paper.

2. Add sodium citrate solution one drop at a time until a pH of around 3.5 is reached. Count the drops needed (could be between 20-40)

3. Pipet a fresh 10.00 mL aliquot of Fe standard into a 100 mL volumetric flask and add the same number of drops of citrate solution as used in step #2.

4. Add 2.00 mL of hydroquinone solution and 3.00 mL of o-phenanthroline solution, then dilute to the mark with DI water, and mix well.

5. Prepare three more solutions from 5.00, 2.00, and 1.00 mL of your Fe standard and prepare a blank containing no Fe. Use sodium citrate solution in proportion to the volume of Fe solution. (if 10 mL of Fe required 30 drops, then 5 mL of Fe standard should require 15 drops, etc.).

6. Allow the solutions to stand for at least 10 minutes. Then measure the absorbance of each solution at 510 nm in a 1cm cell. (The color is stable, so all solutions may be prepared and all the absorbances measured at once.)

7. Make a graph of absorbance versus concentration in ppm of Fe in the standards. Find the slope and intercept by the method of least squares. Include your equation of the line and your R2 value on your graph. Calcualte the molarity of Fe(o-phenanthroline)3 2+ in each solution and find the average molar absorptivity.

8. Place one table of the iron containing vitamin in a 125 mL flask or 100 mL beaker and boil gently (in the hood) with 25 mL of 6M HCl for 15 minutes. Most of the tablet will dissolve leaving some insoluble binders behind. Filter the solution directly into a 100 mL volumetric flask. Wash the beaker/flask several times with small portions of water to complete quantitative transfer. Allow the solution to cool to room temperature and dilute to the mark with DI water. This is solution A

9. Dilute 5.00 mL of solution A to 100.00 mL in a 100.00 mL volumetric flask.                                 This is solution B.

10. Determine how many drops of citrate solution are needed to bring 10.00 mL of iron tablet solution B to pH 3.5.

11. Transfer 10.00 mL of solution B to a 100.00 mL volumetric flask. Add the required amount of citrate from step #10. Then add 2.00 mL of hydroquinone and 3.00 mL of o-phenanthroline. Dilute to the mark with DI water and mix well. This is solution C.

12. Measure the absorbance of solution C. Make sure you measure all your standards and your unknown solution C in the same instrument on the same day at roughly the same time.

13. Using the calibration curve, find the total amount of Fe (in mg) in the tablet.

What would be the concentration in mg of the unknown tablet if the slope calculated from the calibration curve was 0.20 and the absorbance for the unknown was 0.249? Please show all intermeditate calculations and be unit conscious here. [I believe the slope units should be 1/(ppm*cm-1)*(1cm); since the plot of Absorbance vs ppm Fe]. Keep in mind that this was solution C that was measured in the spectrophotometer.

Explanation / Answer

Slope of the calibration curve gives us Molar absorptivity = 0.20 1/(ppm.cm)

Absorbance of the unknown = 0.249

From, Beer Lambert's law,

A = e c l

c = A / e .l

= 0.249 / (0.20 ppm-1.cm-1)(1.00 cm)

= 1.245 ppm

Solution A: tablet in 100 mL DI water

5 mL of solution A is diluted to 100 mL ----Solution B

10 mL of solution B is diluted to 100 mL ----Solution C

The concentration measured was that of Fe in solution C

Using dilution law,

CBVB = CC VC

CB = 1.245 ppm ( 100 mL) / 10.00 mL

CB = 12.45 ppm

Similarly,

CAVA = CBVB

CA = 12.45 ppm (100 mL) / 5 mL

= 249 ppm = 249 mg/L or 0.249 mg /mL

We have 100 mL solution.

So Mass of the Fe in the tablet = 0.249 mg/mL * 100 mL = 24.9 mg Fe

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