Question

Four tablets of an iron dietary supplement were destroyed by digestion, leaving the iron(III) in aqueous solution freed of organics. This mixture was diluted to 2.00L. Two 5.00 mL aliquots were treated as below to determine the iron by the method of standard addition.

a) One 5 mL aliquot was treated with 25.00 mL of ligand solution, diluted to 50.0 mL, and read 0.489 As in a 1.0cm cell at 606 nm.

b) To a second 5 mL aliquot, a 15.0 mL portion of 1.00 ppm iron(III) was added. The combined solution was treated with 25.00 mL of ligand solution and diluted to 50.0 mL. The resultant solution had an absorbance of 0.625 under the same conditions as solution a. How many mg of iron are present in each tablet of the supplement?

Please answer with an explanation (step by step) as how to solve.

Answer 1

4 tablets diluted to 2 L

a) 5 ml aliquot + 25 ml ligand ---> diluted to 50 ml

Absorbance = 0.489 = ebc = ec x 1 = ec

b) Another 5 ml aliquot + 15 ml of 1.00 ppm Fe(III) std. + 25 ml ligand ---> diluted to 50 ml

concentration of standard in dilute solution = 1 ppm x 15/50 = 0.3 ppm

Total absorbance = 0.625

Absorbance due to 0.3 ppm Fe(III) std. = 0.625 - 0.489 = 0.136

concentration of Fe(III) in 50 ml aliquot = 0.3 x 0.489/0.136 = 1.08 ppm

and in 5 ml aliquot = 10.8 ppm

Concentration of Fe(III) in 2 L tablet solution = 10.8 x 2000/5 = 4320 ppm

concentration of Fe per tablet = 4320/4 = 1080 mg