A student prepared the following two mixtures and recorded their absorbances in

Question

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

Mix #

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.

Mix #

Explanation / Answer

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

2. Kf = [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 = cl

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

= 4357 M-1 cm-1

5. According to Beer's Law, A = cl

or, c = A/l = [FeSCN2+]eq for mixture #2

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

= 9.7*10­5 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*10­5 M

= 9.903*10-3 M

[SCN-]eq = [SCN-]initial - [FeSCN2+]eq for mixture #2

= 1.625*10-4 M - 9.7*10­5 M

= 6.55*10-5 M

8. Kf = [FeSCN2+]/[ Fe3+][ SCN-]

= 9.7*10­5 / (9.903*10-3)( 6.55*10-5)

= 149.54

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