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-path length cell. For comparison, a 10.0-mL reference sample of 6.74 times 10^-4 M Fe^3+ was treated with HNO_3 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 path length of the runoff cell was 2.41 cm. What was the concentration of iron in Uncle Wilbur's runoff? M

Explanation / Answer

Beer–Lambert law

A = l c

A = absorbance
= a constant ratio of (A) / (l) (c)
l = length of cell
c = concentration



The runoff sample exhibted the same absorbance as the reference when the pathlength of the runoff cell was 2.48 cm because of "The reference was placed in a cell with a 1.00 cm light path

" l c".... of the standard when l = 1.00cm
is equal to
" l c".... of the unknown when l = 2.41cm


so
() (1.00) (c of std) = () (2.41) (c of unk)
which simplifies by cancelling out the absorptivity constant:
(1.00) (c of std = (2.41) (c of unk)

so
c of unk = (c of std) / (2.41)

======================================...

Now using the dilution formula, find the diluted, analyzed concentration of the standard
C1V1 = C2V2
(6.74 x 10^-4 M Fe 3+) (10.0 mL) = C2 (diluted to 50.0 mL)
C2 = 1.34 X 10^-4 Molar Fe+3 ... is the concentration of standard as analyzed

now using the ratio of std to unk found earlier,
find the concentration of the unknown as analyzed:
c of unk = (c of std) / (2.41)

c of unk = (1.34 X 10^-4 Molar Fe+3) / (2.41)

c of unk = 5.56 X 10^-5 Molar Fe+3..... is the concentration of unknown as analyzed


============

using the dilution formula, we find the original concentration of the unknown before it was diluted to be analyzed:
C1V1 = C2V2

C1 (25.0 ml) = (5.56 X 10^-5 Molar Fe+3 as analyzed) (diluted to 100.0 mL)

C1 = 2.22 X 10^-4 Molar

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.