Question

A student prepared the following two mixtures and recorded their absorbances in a cuvette with a 1.00 cm path length:

Mix #	.20 M Fe3	.020 M Fe3	3.25*10^-4M SCN	Absorbance
1	10.0 mL	--	10.0 mL	.708
2	--	10.0 mL	10.0 mL	.423

1. Write the balanced equilibrium reaction of Fe3 (aq) with SCN (aq) to form [FeSCN]2 (aq).

2. Write the equilibrium formation constant equation for the reaction (Kf expression).

Determination of moral absoptivity

3. Assuming that all of the SCN added to mixture #1 exists as [FeSCN]2 at equilibrium, determine [FeSCN2]eq for mixture #1.

4. Use the absorbance and [FeSCN2] eq for mixture #1 to calculate the molar absorptivity of [FeSCN]2.

Determination of Kf from mixture #2

5. Use Beer's Law and the molar absoptivity that you calculated from mixture #1 to calculate [FeSCN2]eq for mixture #2.

6. Calculate [Fe3]i and [SCN]i for mixture #2.

7. Use your answers to the previous two questions to calculate both [Fe3]eq and [SCN]eq for mixture #2.

8. Use the data for [FeSCN2]eq, [Fe3]eq, and [SCN]eq to calculate Kf for mixture #2.

Answer 1

1. Fe³⁺ (aq) + SCN^- (aq) <---> [FeSCN]²⁺(aq)

2. K_f = [FeSCN²⁺]/[ Fe³⁺][ SCN^-]

3. [FeSCN²⁺]_eq = [ SCN^-]_initial = 10mL * (3.25*10^-4)M/20 mL = 1.625*10^-4 M

4. According to Beer's Law, A = εcl

or, ε = A/cl = 0.708 /(1.625*10^-4 M)(1 cm)

= 4357 M^-1 cm^-1

5. According to Beer's Law, A = εcl

or, c = A/εl = [FeSCN²⁺]_eq for mixture #2

= 0.423/(4357 M^-1 cm^-1)( 1 cm)

= 9.7*10^­5 M

6. [Fe³⁺]_initial = 10mL * (0.02)M/20 mL = 0.01 M

[SCN^-]_initial = 10mL * (3.25*10^-4)M/20 mL

= 1.625*10^-4 M

7. [Fe³⁺ ]_eq = [Fe³⁺]_initial - [FeSCN²⁺]_eq for mixture #2

= 0.01 M - 9.7*10^­5 M

= 9.903*10^-3 M

[SCN^-]_eq = [SCN^-]_initial - [FeSCN²⁺]_eq for mixture #2

= 1.625*10^-4 M - 9.7*10^­5 M

= 6.55*10^-5 M

8. K_f = [FeSCN²⁺]/[ Fe³⁺][ SCN^-]

= 9.7*10^­5 / (9.903*10^-3)( 6.55*10^-5)

= 149.54