2.00x10^-1 M Fe(NO3)3 4.00x10^-4 M NaSCN 0.10 M HNO3 Total mL A1 Table V Total V
ID: 1018284 • Letter: 2
Question
2.00x10^-1 M Fe(NO3)3
4.00x10^-4 M NaSCN
0.10 M HNO3
Total mL
A1 Table
V Total
Vi /Vf / Vtotal
Vi / Vf /Vtotal
S1
2.00
0/ 0 /0
0.10 / 8.19 / 8.09
10.09
S2
2.00
1.06 / 2.00 / 0.94
0.21 / 7.10 / 6.89
9.83
S3
2.00
0.10 / 2.20 / 2.10
0.40 / 6.39 / 5.99
10.09
S4
2.00
0 / 3.01 / 3.01
0 / 4.99 / 4.99
10.00
S5
2.00
0.20 / 4.00 / 3.80
0.01 / 4.30 /4.29
10.09
S6
2.00
0.01 / 5.00 / 4.99
4.30 / 7.60 / 3.30
10.29
A2 Table
Blank
S2
S3
S4
S5
S6
ABS
100
78.9
61.5
50.3
38.7
33.4
1. Prepare a Beer's Law plot an include the equation and best fit line. Force the y-intercept through zero.
2. Set up 3 ICE boxes (E1, E3, and E5) that contain initial concentrations of all species and equilibrium equations of FeSCN^2+ based on the slpe of the line from the Beer's Law plot. Set up species. Using the calculated equilibrium concentrations, find the equilibrium constant for each solution. Calculate the average equilibrium constant for the solutions.
2.00x10^-1 M Fe(NO3)3
4.00x10^-4 M NaSCN
0.10 M HNO3
Total mL
A1 Table
V Total
Vi /Vf / Vtotal
Vi / Vf /Vtotal
S1
2.00
0/ 0 /0
0.10 / 8.19 / 8.09
10.09
S2
2.00
1.06 / 2.00 / 0.94
0.21 / 7.10 / 6.89
9.83
S3
2.00
0.10 / 2.20 / 2.10
0.40 / 6.39 / 5.99
10.09
S4
2.00
0 / 3.01 / 3.01
0 / 4.99 / 4.99
10.00
S5
2.00
0.20 / 4.00 / 3.80
0.01 / 4.30 /4.29
10.09
S6
2.00
0.01 / 5.00 / 4.99
4.30 / 7.60 / 3.30
10.29
Explanation / Answer
(a) We need to work with a number of tables here. First, we calculate the initial concentrations of Fe(NO3)3 and NaSCN in the solution.
The initial concentrations are found out using the following equations:
[Fe3+]0 = (volume of Fe3+ stock added)*(concentration of Fe3+ stock)/(total volume of solution)
[SCN]0 = (volume of SCN- stock added)*(concentration of SCN- stock)/(total volume of solution)
Note that we take only the volume of SCN- added (we do not include the initial and final volumes; we only take the difference).
A1
Vol. of 2.00*10-1 M Fe3+ (mL)
Vol. of 4.00*10-4 M SCN- (mL)
Vol. of 0.10 M HNO3 (mL)
Total volume (mL)
[Fe3+]0 (M)
[SCN-]0 (M)
S1
2.00
0.00
8.09
10.09
0.0396
0.000
S2
2.00
0.94
6.89
9.83
0.0407
4.329*10-5
S3
2.00
2.10
5.99
10.09
0.0396
8.325*10-5
S4
2.00
3.01
4.99
10.00
0.0400
1.204*10-4
S5
2.00
3.80
4.29
10.09
0.0396
1.506*10-4
S6
2.00
4.99
3.30
10.29
0.0389
1.940*10-4
Next, we need to find out the concentrations of [Fe(SCN)2+] by noting the reaction equation as
Fe3+ (aq) + SCN- (aq) --------> [Fe(SCN)2+] (aq)
We note that moles of Fe3+ consumed = moles of [Fe(SCN)2+] produced.
Again, moles of SCN- consumed = moles of [Fe(SCN)2+] produced.
Looking at the above table, we see that Fe3+ is present in large excess in comparison to SCN- Thus, the concentration of [Fe(SCN)2+] will be governed by the moles of SCN- consumed. Thus, we construct the table for Beer’s law as
A2 table
[Fe(SCN)2+]0 (*105) (M)
Absorbance
Blank
0
-
S2
4.329
33.4
S3
8.325
38.7
S4
12.040
50.3
S5
15.060
61.5
S6
19.400
78.9
Absorbance increases with increase in concentration; hence, I have written the values that way. Also note that (0,0) will be a point on the graph (since the absorbance is zero when the concentration is zero). Next, we plot the graph in excel.
We have drawn the best fit line possible. The slope of the plot is 4.2207 (ans)
A1
Vol. of 2.00*10-1 M Fe3+ (mL)
Vol. of 4.00*10-4 M SCN- (mL)
Vol. of 0.10 M HNO3 (mL)
Total volume (mL)
[Fe3+]0 (M)
[SCN-]0 (M)
S1
2.00
0.00
8.09
10.09
0.0396
0.000
S2
2.00
0.94
6.89
9.83
0.0407
4.329*10-5
S3
2.00
2.10
5.99
10.09
0.0396
8.325*10-5
S4
2.00
3.01
4.99
10.00
0.0400
1.204*10-4
S5
2.00
3.80
4.29
10.09
0.0396
1.506*10-4
S6
2.00
4.99
3.30
10.29
0.0389
1.940*10-4
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