A 2.50-mL aliquot of a solution that contains 7.9 ppm iron(III) (55.85 g/mol) is

Question

A 2.50-mL aliquot of a solution that contains 7.9 ppm iron(III) (55.85 g/mol) is treated with an appropriate excess of KSCN to form a complex Fe(SCN)^2+ and diluted to 50.0 mL. The complex Fe(SCN)^2+ has a red color with a molar absorptivity () of 7.00 times 10^3 L cm^-1 mol^1 at 580 nm. (a) What is the absorbance of the resulting solution at 580 nm in a 2.50-cm cell? (b) Sketch a photometric titration curve in the box at the right for Fe(III) with thiocyanate ion, both species are colorless, when a photometer with a green filter is used to collect data. (c) Why is a green filter used?

Explanation / Answer

3a) We start with 7.9 ppm Fe (III). We know that 1 ppm = 1 mg/L; therefore, the concentration of Fe present is 7.9 mg/L.

Molar mass of Fe = 55.85 g/mol.

Molar concentration of Fe (III) = (7.9 mg/L)*(1 g/1000 mg)*(1 mole/55.85 g) = 1.4145*10-4 mole/L = 1.4145*10-4 M

Moles of Fe (III) in the sample = (2.50 mL)*(1.4145*10-4 mol/L) = 3.53625*10-4 mmole.

The balanced chemical equation is

Fe3+ (aq) + SCN- (aq) -------> [Fe(SCN)]2+ (aq)

As per the stoichiometric reaction,

1 mole Fe3+ = 1 mole Fe(SCN)2+

Therefore, moles of Fe(SCN)2+ formed = 3.53625*10-4 mmole.

Molar concentration of Fe(SCN)2+ formed = (3.53625*10-4 mmole)/(50.0 mL) = 7.0725*10-6 mol/L.

Use Beer’s law:

A = *c*l where = molar absorptivity; c = concentration of Fe(SCN)2+ and l = path length of the solution.

Plug in values.

A = (7.00*103 L cm-1 mol-1)*(7.0725*10-6 mol/L)*(2.50 cm) = 0.12377 0.124 (ans).

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