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Prelab for Determination of the Equilibrium Constant for a Chemical Reaction SD

front 1

In a similar experiment, 5.00 mL of a 4.00 x 10-4 M (4.00 times 10 to the minus 4th power M) solution of F e3+ was used. How many moles of F e3+ were present in the solution?

back 1

2.00 x 10-6 moles (2.00 times 10 to the minus 6th power moles)

Ex.

5.00 mL -> 0.005 L

0.005 L x ( 4.0 x 10-4 mol / 1L ) = 2.00 x 10-6 moles

front 2

It was determined by spectrophotometry that the moles of FeSCN2+ present at equilibrium was 2.00 x 10-4 moles (2.00 times 10 to the minus 4th power moles). If the sample was prepared using 4.00 x 10-3 moles (4.00 times 10 to the minus 3rd power moles) of Fe3+, how many moles of Fe3+ were present at equilibrium?

back 2

3.8 x 10-3 moles (3.8 times 10 to the minus 3rd power moles)

Ex.

Eq'm moles Fe3+ = Initial moles Fe3+ - Eq'm moles FeSCN2+

(4.00 x 10-3) - (2.00 x 10-4) = 3.8 x 10-3 moles

front 3

Please refer to the graph in the lab document. If the absorbance of a solution containing FeSCN2+ is found to be 0.7500, what is the concentration of this solution?

back 3

1.6 x 10-4 M (1.6 times 10 to the minus 4th power M)

Ex.

c = ( A / E )

c = 0.7500 / 447 nm = 1.6 x 10-4 M

front 4

It was found that an equilibrium solution contained 3.00 x 10-4 M Fe3+ (3.00 times 10 to the minus 4th power M F e 3+), 3.00 x 10-4 M SCN- (3.00 times 10 to the minus 4th power M SCN minus) and 5.00 x 10-5 M FeSCN2+ (5.00 times 10 to the minus 5th power M F e SCN 2+ ). What is Kc for this solution?

back 4

556

Ex.

Kc = [ FeSCN2+ ] / [ Fe3+ ] [ SCN- ]

5.00 x 10-5 M FeSCN2+ / (3.00 x 10-4 M Fe3+ ) (3.00 x 10-4 M SCN-)