Test tube number 0.0020 M Fe(NO 3 ) 3 (mL) 0.0020 M SCN – (mL) H 2 O (mL) 1 3.00
ID: 1028755 • Letter: T
Question
Test tube
number
0.0020 M Fe(NO3)3
(mL)
0.0020 M SCN–
(mL)
H2O
(mL)
1
3.00
2.00
5.00
2
3.00
3.00
4.00
3
3.00
4.00
3.00
4
3.00
5.00
2.00
DATA TABLE
Parts I and II
Wavelength of Maximum Absorbance:
___________________
Concentration of [FeSCN]2+ after mixing
Absorbance
1
2.0*10^-4M
0.299
2
1.10*10^-4M
0.245
3
1.2*10^-4M
0.193
4
8.0*10^-5M
0.121
5
0.0M
0.026
Record the equation of the line for your calibration curve. (A = eC)
x= 1384x + .02194
Unknown Absorbance
0.043
Concentration Unknown
(show work in
Analysis Question #1)
3.1*10^-5
Part III
Test tube number
Absorbance
1
.038
2
.040
3
.044
4
.047
DATA ANALYSIS
1. (Part II) Show how you calculated the concentration of your unknown [SCN-] sample.
2. (Part III) Use the net absorbance values, along with the best fit line equation of the standard solutions in Part I to determine the [FeSCN2+] at equilibrium for each of the mixtures that you prepared in Part III. Complete the table below and give an example of your calculations.
.043/1384= 3.1*10^-5
Test tube number
1
2
3
4
[FeSCN2+]eq
3. (Part III) Complete the ICE charts for test tubes 1-4 in Part III. Once you have calculated all the equilibrium concentrations, calculate the Keq for each. Show an example calculation for Keq in part 5.
TEST TUBE 1
[Fe3+]
[SCN–]
D
[FeSCN]2+
Keq =
Initial
Change
Equilibrium
TEST TUBE 2
[Fe3+]
[SCN–]
D
[FeSCN]2+
Keq =
Initial
Change
Equilibrium
TEST TUBE 3
[Fe3+]
[SCN–]
D
[FeSCN]2+
Keq =
Initial
Change
Equilibrium
TEST TUBE 4
[Fe3+]
[SCN–]
D
[FeSCN]2+
Keq =
Initial
Change
Equilibrium
5. Calculate and show work for the average Keq.
Test tube
number
0.0020 M Fe(NO3)3
(mL)
0.0020 M SCN–
(mL)
H2O
(mL)
1
3.00
2.00
5.00
2
3.00
3.00
4.00
3
3.00
4.00
3.00
4
3.00
5.00
2.00
Explanation / Answer
Data analysis
1. Equilibrium [SCN-] = 0.043 x 8 x 10^-5/0.121 = 3 x 10^-5 M
2. Equilibrium FeSCN2+
Test Tube Equilibrium [FeSCN2+]
1 0.038/1384 = 2.74 x 10^-5 M
2 0.040/1384 = 2.89 x 10^-5 M
3 0.044/1384 = 3.18 x 10^-5 M
4 0.047/1384 = 3.40 x 10^-5 M
3. Keq, ICE chart
Test tube 1,
[Fe3+] + [SCN-] <====> [FeSCN2+]
I 6 x 10^-4 4 x 10^-4 -
C -2.74 x 10^-5 -2.74 x 10^-5 +2.74 x 10^-5
E 5.73 x 10^-4 3.73 x 10^-4 2.74 x 10^-5
Keq = (2.74 x 10^-5)/(5.73 x 10^-4 x 3.73 x 10^-4)
= 128.3
---
Test tube 2,
[Fe3+] + [SCN-] <====> [FeSCN2+]
I 6 x 10^-4 6 x 10^-4 -
C -2.89 x 10^-5 -2.89 x 10^-5 +2.89 x 10^-5
E 3.11 x 10^-4 3.11 x 10^-4 2.89 x 10^-5
Keq = (2.89 x 10^-5)/(3.11 x 10^-4 x 3.11 x 10^-4)
= 298.8
Similarly for test tube 3 and 4, calculations can be done.
[formulas,
final molarity = initial molarity x initial volume/total volume]
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