Complete this prelab assignment (questions 1-3) on these sheets in the spaces provided. Combine this along with a scanned copy of your Lab Notebook and upload to CANVAS before the beginning of your Lab Section. *. In Part A of this experiment, you prepare 5 solutions of FeSCN2* [one of them is just a blank solution] according to the reaction below. Fe3+ (aq) + SCN- (aq) → FeSCN²+ (aq) We assume that the starting SCN determines the concentration of formed (because Fe3+ is in excess and SCN- is limiting). Calculate the concentration of FeSCN2* that forms for each of the solution (Beakers 1-4) and fill out the table below. Show your calculations beneath the table.
Complete this prelab assignment (questions 1-3) on these sheets in the spaces provided. Combine this along with a scanned copy of your Lab Notebook and upload to CANVAS before the beginning of your Lab Section. *. In Part A of this experiment, you prepare 5 solutions of FeSCN2* [one of them is just a blank solution] according to the reaction below. Fe3+ (aq) + SCN- (aq) → FeSCN²+ (aq) We assume that the starting SCN determines the concentration of formed (because Fe3+ is in excess and SCN- is limiting). Calculate the concentration of FeSCN2* that forms for each of the solution (Beakers 1-4) and fill out the table below. Show your calculations beneath the table.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
Related questions
Question
Correct numbers 1 and 3. Show all work/calculations:
![Complete this prelab assignment (questions 1-3) on these sheets in the spaces provided. Combine this
along with a scanned copy of your Lab Notebook and upload to CANVAS before the beginning of your
Lab Section.
*. In Part A of this experiment, you prepare 5 solutions of FESCN2* [one of them is just a blank
solution] according to the reaction below.
Fe3+ (aq) + SCN¯ (aq) → FeSCN²+ (aq)
We assume that the starting SCN determines the concentration of formed (because Fe3+ is in
excess and SCN" is limiting). Calculate the concentration of FESCN2* that forms for each of the
solution (Beakers 1- 4) and fill out the table below. Show your calculations beneath the table.
Beaker
0.200 M
0.0020 M
Concentration of
Number
Fe(NO3)3 (mL]
KSCN (mL]
H20 [mL]
FESCN2* (M)
1
5.0
5.0
40.0
0.0002
2
5.0
4.0
41.0
0.00016
5.0
3.0
42.0
0.00012
4
5.0
2.0
43.0
0.00008
5 (blank)
5.0
0.0
45.0
181
Show your work for question 1 here:
Total volume of all beakers = 50 mL
MSCN x VSCN
(FESCN2*] = (SCN'] =
values ok but,
you Should
NOT use
Total Volume
Beaker 1:
0.002 x 5
(FESCN2"] =
50
= 0.0002 M
%3D
Beaker 2:
0.002 x 4
(FESCN²*] =
50
= 0.00016 M
MiVi=M2V2
Beaker 3:
Show mol ratio
0.002 x 3
(FESCN2*] =
50
= 0.00012 M
Beaker 4:
0.002 x 4
(FESCN2"] =
50
= 0.00008 M
2. In Part A, you measure the absorbance of four FeSCN2* solutions and prepare a Beer's Law plot
(Absorbance vs. wavelength] similar to the one below.
Beer's Law Plot for FeSCN2+
0.6
y = 2417.5x + 0.0483
R? = 0.999
0.5
E
0.4
0.3
0.2
0.1
0.00005
0.0001
0.00015
0.0002
0.00025
Concentration (M)
You will use this plot as a standard curve in Part B to determine the concentration of an
unknown. If an unknown FeSCN2* solution gives an absorbance reading of 0.334, use the
equation of the best fit line shown above to determine its concentration. Show all work below.
Absorbance](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffbfefd57-6cd5-45bf-b890-c343c2fceb41%2Fe6a011ac-ec77-4c0f-99c2-94fb239bb83f%2F7rpjtv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Complete this prelab assignment (questions 1-3) on these sheets in the spaces provided. Combine this
along with a scanned copy of your Lab Notebook and upload to CANVAS before the beginning of your
Lab Section.
*. In Part A of this experiment, you prepare 5 solutions of FESCN2* [one of them is just a blank
solution] according to the reaction below.
Fe3+ (aq) + SCN¯ (aq) → FeSCN²+ (aq)
We assume that the starting SCN determines the concentration of formed (because Fe3+ is in
excess and SCN" is limiting). Calculate the concentration of FESCN2* that forms for each of the
solution (Beakers 1- 4) and fill out the table below. Show your calculations beneath the table.
Beaker
0.200 M
0.0020 M
Concentration of
Number
Fe(NO3)3 (mL]
KSCN (mL]
H20 [mL]
FESCN2* (M)
1
5.0
5.0
40.0
0.0002
2
5.0
4.0
41.0
0.00016
5.0
3.0
42.0
0.00012
4
5.0
2.0
43.0
0.00008
5 (blank)
5.0
0.0
45.0
181
Show your work for question 1 here:
Total volume of all beakers = 50 mL
MSCN x VSCN
(FESCN2*] = (SCN'] =
values ok but,
you Should
NOT use
Total Volume
Beaker 1:
0.002 x 5
(FESCN2"] =
50
= 0.0002 M
%3D
Beaker 2:
0.002 x 4
(FESCN²*] =
50
= 0.00016 M
MiVi=M2V2
Beaker 3:
Show mol ratio
0.002 x 3
(FESCN2*] =
50
= 0.00012 M
Beaker 4:
0.002 x 4
(FESCN2"] =
50
= 0.00008 M
2. In Part A, you measure the absorbance of four FeSCN2* solutions and prepare a Beer's Law plot
(Absorbance vs. wavelength] similar to the one below.
Beer's Law Plot for FeSCN2+
0.6
y = 2417.5x + 0.0483
R? = 0.999
0.5
E
0.4
0.3
0.2
0.1
0.00005
0.0001
0.00015
0.0002
0.00025
Concentration (M)
You will use this plot as a standard curve in Part B to determine the concentration of an
unknown. If an unknown FeSCN2* solution gives an absorbance reading of 0.334, use the
equation of the best fit line shown above to determine its concentration. Show all work below.
Absorbance
![You will use this plot as a standard curve in Part B to determine the concentration of an
unknown. If an unknown FeSCN2* solution gives an absorbance reading of 0.334, use the
equation of the best fit line shown above to determine its concentration. Show all work below.
y = absorbance = 0.334
y = 2417.5x + 0.0483
y - 0.0483
X =
0.334 - 0.0483
2417.5
2417.5
x = 1.182 x 10-4 M
Unknown [FESCN²*] = 1.182 x 10-4 M
%3D
182
3. In Part C, you prepare four solutions (starting concentrations are given blow) in test tubes and
you will measure the Absorbance after the solutions have reached equilibrium. Shown below is
an example of just ONE solution, with absorbance data shown as well. Using this data and the
standard curve from question 2 on the previous page, calculate Kf.
SHOW ALL WORK BELOW!
Starting Concentrations
Test Tube
0.00200 M Fe(NO3)3 [mL] 0.0020 M KSCN [mL]
H20 [mL]
1
3.0
2.0
5.0
Absorbance Data
(FESCN²*] at
Test Tube
Absorbance at Amax
equilibrium
Kf
1
0.118
2.88 x 10-4 M
121
Given absorbance: y = 0.118
y – 0.0483
X =
0.118 – 0.0483
2417.5
2417.5
x = 2.883 x 10'5 M
(FESCN2*] = 2.88 x 10$ M
(Fe**) =;
MFex VFe
0.002 x 3
= 0.006 M
Total Volume
MSCN X V SCN
10
0.002 x 2
[SCN'] =
= 0.0004 M
%3D
Total Volume
10
ICE Table:
Fe3+
SCN-
FESCN2+
0.006
0.0004
C
-X
-X
+X
E
0.006 – x
0.004 - x
[FESCN2*]
Kf =
[Fe3+][SCN¯]
0.118 = 2417.5+ 0.0483
2417.5x = 0.334 – 0.0483 = 0.0697
(0.006 – x)(0.004 – x)
0.0697
X =
2417.5
= 2.88 x 10-5 M
At equilibrium,
(FESCN2*] = 2.88 x 105 M
(Fe3*] = 0.0060 – x = 0.0060 – 2.88 x 105 = 0.0059712
(SCN'] = 0.0040 – x = 0.0040 – 2.88 x 10$ = 0.0039712 not Sure what
went wrong.-.
2.88 x 10-5
Kf =
(0.0059712 x 0.0039712)
Kf = 121](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffbfefd57-6cd5-45bf-b890-c343c2fceb41%2Fe6a011ac-ec77-4c0f-99c2-94fb239bb83f%2Fo60bf2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:You will use this plot as a standard curve in Part B to determine the concentration of an
unknown. If an unknown FeSCN2* solution gives an absorbance reading of 0.334, use the
equation of the best fit line shown above to determine its concentration. Show all work below.
y = absorbance = 0.334
y = 2417.5x + 0.0483
y - 0.0483
X =
0.334 - 0.0483
2417.5
2417.5
x = 1.182 x 10-4 M
Unknown [FESCN²*] = 1.182 x 10-4 M
%3D
182
3. In Part C, you prepare four solutions (starting concentrations are given blow) in test tubes and
you will measure the Absorbance after the solutions have reached equilibrium. Shown below is
an example of just ONE solution, with absorbance data shown as well. Using this data and the
standard curve from question 2 on the previous page, calculate Kf.
SHOW ALL WORK BELOW!
Starting Concentrations
Test Tube
0.00200 M Fe(NO3)3 [mL] 0.0020 M KSCN [mL]
H20 [mL]
1
3.0
2.0
5.0
Absorbance Data
(FESCN²*] at
Test Tube
Absorbance at Amax
equilibrium
Kf
1
0.118
2.88 x 10-4 M
121
Given absorbance: y = 0.118
y – 0.0483
X =
0.118 – 0.0483
2417.5
2417.5
x = 2.883 x 10'5 M
(FESCN2*] = 2.88 x 10$ M
(Fe**) =;
MFex VFe
0.002 x 3
= 0.006 M
Total Volume
MSCN X V SCN
10
0.002 x 2
[SCN'] =
= 0.0004 M
%3D
Total Volume
10
ICE Table:
Fe3+
SCN-
FESCN2+
0.006
0.0004
C
-X
-X
+X
E
0.006 – x
0.004 - x
[FESCN2*]
Kf =
[Fe3+][SCN¯]
0.118 = 2417.5+ 0.0483
2417.5x = 0.334 – 0.0483 = 0.0697
(0.006 – x)(0.004 – x)
0.0697
X =
2417.5
= 2.88 x 10-5 M
At equilibrium,
(FESCN2*] = 2.88 x 105 M
(Fe3*] = 0.0060 – x = 0.0060 – 2.88 x 105 = 0.0059712
(SCN'] = 0.0040 – x = 0.0040 – 2.88 x 10$ = 0.0039712 not Sure what
went wrong.-.
2.88 x 10-5
Kf =
(0.0059712 x 0.0039712)
Kf = 121
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