Introduction:
The equation of the reaction is Fe3+(aq) + SCN- (aq) Fe(NCS)2+ (aq).
The objective of this lab was to determine the equilibrium constant (K) for the formation of the ferric thiocyanate. This was done by creating the solutions and putting them into a spectrophotometer. This determines the absorbance at each wavelength. The maximum wavelength gives the absorbance at equilibrium.Equilibrium is the point where the amount of products equals the amount reactants. Equilibrium is needed to solve the K value. The equation for the K value is, K = [C]c [D]d / [A]a [B]b. A, B, C, and D are the concentrations of the reaction and the lower case a, b, c, and dare the molar ratios of the different reactants.
To start the lab a blank solution has to be made and then poured into a cuvette and put into the spectrophotometer make sure to wipe off the outside of the cuvette so the maximum amount of light can go through. This solution is created by adding both solution one and solution four. Next the maximum wavelength has to be determined by using the solution of the highest concentration of Fe(NCS)2+ .Also record the absorbance for this concentration every 10 nm beginning at 400 nm and ending at 500 nm. Next all the other solutions have to run through the spectrophotometer; however the maximum wavelength is the only thing that needs to be recorded here. Figures 1 arethe stock solutions needed. Figure 2 are the solutions for the standard curve. Figure three are the solutions for equilibrium. All of the following solutions were put into the spectrophotometer and the absorbances were recorded.
Using the varying molarities of ferric thyocianatewas used in a chart to find the standard curve. The standard curve is determined by plotting absorbance versus concentration of Fe(NCS)2+ which can be seen in Figure two. A trend line is then added to a graph this gives a y=mx+b equation. Y is equal to absorbance and x is equal to the concentration of Fe(NCS)2+.This equation can then be used to determine unknown concentrations. The concentrations are then plugged into the equation previously stated in paragraph one to solve for K. The K avg then needed to be determined which can be done by adding all the K values together and dividing by the number of K values.
Fig. 1
Solution # / Solution ContentsI / 0.20 M Fe(NO3)3
II / 2.0 x 10-3 M Fe(NO3)3
III / 3.0 x 10-3 M KSCN
IV / 0.5 M HNO3
Fig.2
[KSCN] / KSCN (sol III) / Fe(NO3)3 (sol I) / HNO3 (sol IV) / Total Volume3.0 x 10-4 M / 2.5 mL / 12.5 mL / 10 mL / 25 mL
6.0 x 10-4 M / 5.0 mL / 12.5 mL / 7.5 mL / 25 mL
9.0 x 10-4 M / 7.5 mL / 12.5 mL / 5.0 mL / 25 mL
1.2 x 10-3 M / 10.0 mL / 12.5 mL / 2.5 mL / 25 mL
1.5 x 10-3 M / 12.5 mL / 12.5 mL / 0 mL / 25 mL
Fig.3
Beaker # / 2.0 x 10-3 M Fe(NO3)3(Solution II) / 3.0 x 10-3 M KSCN
(Solution III) / 0.50 M HNO3
(Solution IV)
1 / 10 mL / 2.0 mL / 8.0 mL
2 / 10 mL / 4.0 mL / 6.0 mL
3 / 10 mL / 6.0 mL / 4.0 mL
4 / 10 mL / 8.0 mL / 2.0 mL
5 / 10 mL / 10.0 mL / 0 mL
Data Tables and graphs :
Wavelength (nm) / Absorbency400 / 1.102
410 / 0.999
420 / 0.960
430 / 0.966
440 / 0.968
450 / 0.941
460 / 0.891
470 / 0.822
480 / 0.742
490 / 0.645
500 / 0.560
Solution 5
Absorbance for the standard curve
Solution # / AbsorbencyI / 0.641
II / 0.358
III / 0.213
IV / 0.105
Equilibrium Solutions
Solution # / AbsorbencyI / 0.108
II / 0.205
III / 0.378
IV / 0.628
V / 1.295
Absorbance / Fe(NCS)
0.105 / 0.00005
0.213 / 0.00012
0.358 / 0.00027
0.641 / 0.00048
The Volume Used in Each Solution For Equilibrium
Fe(NO3)3 / I / II / III / IV / VInitial mL / 0 / 10 / 0 / 10 / 15
Final mL / 10 / 20 / 10 / 20 / 25
Change in Volume / 10 / 10 / 10 / 10 / 10
3.0 x10-3 KSCN / I / II / III / IV / V
Initial mL / 0 / 2 / 6 / 12 / 15
Final mL / 2 / 6 / 12 / 20 / 25
Change in Volume / 2 / 4 / 6 / 8 / 10
HNO3 / I / II / III / IV / V
Initial mL / 0 / 8 / 14 / 18 / 20
Final mL / 8 / 14 / 18 / 20 / 20
Change in Volume / 8 / 6 / 4 / 2 / 0
Calculations Separate Paper:
Conclusion:
The goal of this lab was to determine the K value of Fe(NCS). This goal was accomplished. By creating all the solutions from the chart and putting them in the spectrophotometer. This gives the absorbance of all the solutions. The trend line created from the data obtained gave the equation to solve for the concentrations. Once the concentrations were determined they could be put into the [C]c [D]d / [A]a [B]b to determine the K values. Then the average K value was found by adding all the K values together and dividing by the number of K values used. The K avg value determined was 6.025x102.
There were many errors that could of occurred in this lab. One error could have been that the same spectrophotometer was not used for all solutions. Another error could have occurred from the cuvette was not the same for all solutions and the cuvette not being clean. A more major error is that for the concentration vs. Absorbance Graph the point (.105, .00005) is not correct. The point should be (.105,.00003) this would of changed the equation of the trend line which in result would change all the concentrations, resulting in changing all the K values.