Question

When I was a boy, Uncle WIlbur measured the iron content of runoff from his banana ranch. He acidified a 25.0-mL sample with HNO₃ and treated it with excess KSCN to form a red complex. He then diluted the solution to 100.0 mL and put it in a variable-pathlength cell. For comparison, he treated a 10.0-mL reference sample of 6.80 x 10^-4 M Fe³⁺ with HNO₃ and KSCN and diluted it to 50.0 mL. The reference was placed in a cell with a 1.00-cm pathlength. Runoff had the same absorbance as the ref. when the pathlength of the runoff cell was 2.48 cm. What was the conc. of iron in Uncle Wilbur's runoff?

Answer 1

Answer: Here we have to use the  Beer–Lambert law A =
A = absorbance
apsalon = a constant ratio of (A) / (l) (c)
l = length of cell
c = concentration

.... of the standard when l = 1.00cm is equal to  
".... of the unknown when l = 2.48cm

Hecen we get ,  (1.00) (c of std) = (2.48) (c of unk) and solving it we get ,

c of unk = (c of std) / (2.48) ................... (a)

using the dilution formula ,C1V1 = C2V2 we get ,

(6.80 x 10^-4 M Fe 3+) (10.0 mL) = C2 (diluted to 50.0 mL)

C2 = 1.36 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) / (2.48)

Hecne , c of unk = (1.36 X 10^-4 Molar Fe+3) / (2.48)

c of unk = 5.48 X 10^-5 Molar Fe+3..... is the concentration of unknown as analyzed

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

C1V1 = C2V2 , we get ,

C1 (25.0 ml) = (5.48 X 10^-5 Molar Fe+3 as analyzed) (diluted to 100.0 mL)

C1 = 2.19 X 10^-4 Molar

Hence it is all about the given question . Thank you .