Chemical Equilibrium and Spectrophotometric Analysis

Question

We have a mixture of 100 mL of Fe³⁺ solution (1.00 × 10^–3 M) and 100 mL of A- ion solution (1.00 ×

10^–3 M). The mixture has a deep blue color, indicating the formation of a complex between Fe3+ and A-

. From spectrophotometric analysis, we find that 30.0% of the Fe³⁺ is in the form of a complex, where

there are two probable formulae: FeX²⁺ or FeX₃. 

a. Balance the following equations and calculate the concentrations of the remaining species.

Fe³⁺ + A^- ⇔ FeA²⁺,

Fe³⁺ + A^- ⇔ FeA₃

b. After reaching equilibrium states, write the equilibrium expressions and the equilibrium constants

for both cases in (a).

Answer 1

Fe³⁺ + A^- ==> FeA²⁺ .. balanced

Fe³⁺ + 3A^- ==> FeA₃ .. balanced

Initial moles Fe³⁺ = 100 ml x 1 L / 1000 ml x 1.00x10^-3 mol / L = 1x10^-4 moles Fe³⁺

Initial moles A^- = 100 ml x 1 L / 1000 ml x 1.00x10^-3 mol / L = 1x10^-4 moles A^-

Fe³⁺ + A^- ==> FeA²⁺

If 30% of the 1x10^-4 mols Fe³⁺ reacts, then at equilibrium we will have

7x10^-5 mols Fe³⁺ / 0.2 L = 3.5x10^-4 M

7x10^-5 mols A^- / 0.2 L = 3.5x10^-4 M

3x10^-5 mols FeA²⁺ / 0.2 L = 1.5x10^-4 M

Equilibrium expression: K = [FeA²⁺] / [Fe³⁺][A^-]

K = (1.5x10^-4) / (3.5x10^-4)(3.5x10^-4)

K = 1224

Fe³⁺ + 3A^- ==> FeA₃

If 30% of the 1x10-4 mols Fe³⁺ reacts, then at equilibrium we will have

7x10^-5 mols Fe³⁺ / 0.2 L = 3.5x10^-4 M

1x10^-5 mols A^- / 0.2 L = 5x10^-5 M

3x10^-5 mols FeA₃ / 0.2 L = 1.5x10^-4 M

Equilibrium expression: K = [FeA₃] / [Fe³⁺][A^-]³

K = (1.5x10^-4) / (3.5x10^-4)(5x10^-5)³

K = 3.43x10¹²