Question

Fe(III) forms a complex with thiocyanate, Fe(SCN)²⁺, that has a molar absorptivity of 7.00 x 10³ at 580 nm. A 5.00 ml aliquot of a solution that contains 47.4 ppm Fe³⁺ is reacted with excess thiocyanate and diluted to 25 ml.

a) What fraction of 580 nm light is absorbed in a 1.00 cm pathlength cell by this solution?

Consider 5.00 ml of this solution diluted to a volume of 50ml

b) What fraction of 580 nm light is absorbed in a 1.00 cm pathlength cell by the diluted solution?

Answer 1

In simple terms, we are asked to find the absorbances of the two solutions.

a) Molar mass of Fe(III) = 55.845 g/mol.

We start with 47.4 ppm Fe³⁺. Therefore, the solution contains 47.4 mg/L Fe³⁺.

Molar concentration of Fe³⁺ = (47.4 mg/L)*(1 g/10³ mg)*(1 mole/55.845 g) = 8.4878*10^-4 mol/L = 8.4878*10^-4 M ≈ 8.488*10^-4 M

Write down the balanced chemical equation as

Fe³⁺ (aq) + SCN^-(aq) ------> Fe(SCN)²⁺ (aq)

[Fe³⁺] = [Fe(SCN)²⁺] = 8.488*10^-4 M

Find the concentration of [Fe(SCN)]²⁺ in the dilute solution using the dilution equation as

V₁*S₁ = V₂*S₂

====> (5.00 mL)*(8.488*10^-4 M) = (25.00 mL)*S₂

====> S₂ = (5.00)*(8.488*10^-4 M)/(25.00) = 1.6976*10^-4 M

Find the absorbance of the dilute solution using Beer’s law as

A = ε.c.l where ε = molar absorptivity = 7.00*10^-3 M^-1cm^-1, c = 1.6976*10^-4 M and l = 1.00 cm. Plug in values to obtain

A = (7.00*10³ M^-1cm^-1).(1.6976*10^-4 M).(1.00 cm) = 1.18832 ≈ 1.188 (ans).

b) In this part, we dilute the solution prepared in the last step further. Find out the concentration in the new solution by using the dilution law as

(5.00 mL)*(1.6976*10^-4 M) = (50.00 mL)*S

===> S = (5.00)*(1.6976*10^-4 M)/(50.00) = 1.6976*10^-5 M

Find the absorbance as

A’ = ε.c.l = (7.00*10³ M^-1cm^-1).(1.6976*10^-5 M).(1.00 cm) = 0.118832 ≈ 0.1188 (ans).