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?

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

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?

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

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?

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

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 K_c for this solution?

556

Ex.

K_c = [ FeSCN²⁺ ] / [ Fe³⁺ ] [ SCN^- ]

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