Question

A 2.50-mL aliquot of a solution that contains 6.76 ppm iron(III) is treated with an appropriate...

A 2.50-mL aliquot of a solution that contains 6.76 ppm iron(III) is treated with an
appropriate excess of KSCN and diluted to 50.0 mL. What is the absorbance of the
resulting solution at 480 nm in a 2.50-cm cell? Assume a value of 7.00 x 103 for
absorptivity constant.

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Answer #1

Data.

Initial concentration of Fe3+ = 6.76 ppm = 6.76 x 10-3 g/L

Initial Molarity = [Fe3+] / molar mass Fe3+ = 6.76 x 10-3/55.845 = 1.21 x 10-4 M

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Reaction equation:

Fe3+ + SCN- ===> Fe(SCN)2+

Moles Fe(SCN)2+ = Vi x ni = 2.5ml/1000ml x 1.21 x 10-4 M

= 3.026 x 10-7 mol

Concentration Fe(SCN)2+ = moles/Vf = 3.026 x 10-7 mol/0.050 L

[Fe(SCN)2+] = 6.05 x 10-6

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Absorbance.

A = E x n x l

where,

E = absorptivity constant = 7 x 103

n = [Fe(SCN)2+]

l = cell length = 2.5

then,

A =  (7 x 103) x (6.05 x 10-6) x (2.5)

A = 0.1059

The absorbance for this reaction will be 0.11

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