When I was a boy, Uncle Wilbur let me watch as he analyzed the iron content of r

Question

When I was a boy, Uncle Wilbur let me watch as he analyzed the iron content of runoff from his banana ranch. A 25.0-mL sample was acidified with nitric acid and treated with excess KSCN to form a red complex. (KSCN itself is colorless.) The solution was then diluted to 100.0 mL and put in a variable-pathlength cell. For comparison, a 10.0-mL reference sample of 6.80 104 M Fe3 was treated with HNO3 and KSCN and diluted to 50.0 mL. The reference was placed in a cell with a 1.00-cm light path. The runoff sample

exhibited the same absorbance as the reference when the pathlength of the runoff cell was 2.48 cm. What was the concentration of iron in Uncle Wilbur

Explanation / Answer

c1(reference) x V1(reference) = c2(reference) x V2(reference)

6.80 x 10^(-4) x 10.0 = c2(reference) x 50.0

c2(reference) = 1.36 x 10^(-4) M


Beer's law: A = ecl

Since A and e are constant => cl = constant


c2(reference) x l(reference) = c2(sample) x l(sample)

1.36 x 10^(-4) x 1.00 = c2(sample) x 2.48

c2(sample) = 5.484 x 10^(-5) M


c1(sample) x V1(sample) = c2(sample) x V2(sample)

c1(sample) x 25.0 = 5.484 x 10^(-5) x 100.0


Concentration of Fe = c1(sample) = 2.19 x 10^(-4) M

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