# Chemical Equilibrium Lab

Pages: 3 (534 words) Published: October 12, 2011
Stephany Matos
Chem Lab: Chemical Equilibrium
Lab 52
Synopsis
Iron (III) ions react with thiocyanate ions (SCN-) to form iron (III) thiocyanate, FeSCN2+. It is represented in the equation below:

Fe3+ (aq) + SCN- (aq) FeSCN2+ (aq)
Therefor the equilibrium constant for this reaction is:

KC = [FeSCN2+]/([Fe3+]•[SCN-]
For this experiment we were able to determine the equilibrium constant KC for this reaction. First we prepared five different mixtures with known initial concentrations of iron (III) and thiocyanate ion. By using the colorimeter we determined the absorbance of each mixture once it reached equilibrium. Looking at the equation above we see that the mole to mole ration between SCN- and FeSCN2+ is 1:1 so we can simply set up a ratio using the absorbancy to find the consentration of FeSCN2+ : (where Absorbance 1, Absorbance 2 and [FeSCN2+ ] 2 are already known) Absorbance 1 = [FeSCN2+ ] 1

Absorbance 2 = [FeSCN2+ ] 2
We then pluged the determined value of [FeSCN2+ ] 1 to find the concentration of [Fe3+ ] . [Fe3+ ] = [Fe3+ ](initial) - [FeSCN2+ ]

Although we prepared five different equilibrium solutions, each having different concentrations, the equilibrium constant KC for each of the reactions had similar values. By calculating the average and comparing each of the values to the average there was a percent error of _____.

Post Lab Questions:

Write the KC expression for the reaction 1.1?

KC = [FeSCN2+]/([Fe3+]•[SCN-]

Calculate the initial concentration of [Fe3+ ] in mixtures 1-4 (total volume was 4mL)? [Fe3+]i = 0.002M (0.002L) = 1.0e-3M
0.004L

Calculate the initial concentration of SCN- in mixtures 1-4 (total volume was 4mL)? M1V1=M2V2 -> 0.2M(x) = 0.002M(50mL) -> x = 0.5mL (amount of SCN- (mL) added) [SCN- ]= 0.002M (0.0005L) = 2.5e-4 M

0.004L

Calculate [FeSCN2+] (control) in cuvette #5?
[FeSCN2+] = [SCN- ] because in the balanced equation we see that for every 1M of [SCN- ] used 1M of [FeSCN2+] is produced so...

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