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 equilibrium Test Tube Absorbance at Amax Kf 1 0.118 2.88 x 10-4 M
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 equilibrium Test Tube Absorbance at Amax Kf 1 0.118 2.88 x 10-4 M
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
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![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
equilibrium
Test Tube
Absorbance at Amax
Kf
1
0.118
2.88 x 10-4 M
Given absorbance: y = 0.118
y - 0.0483
X =:
2417.5
0.118 – 0.0483
2417.5
x = 2.883 x 10$ M
(FESCN2*] = 2.88 x 10$ M
MFex VFe
0.002 x 3
(Fe**) =
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
-X
-X
+x
0.006 -x
0.004 - x
[FESCN2+]
Kf =
[Fe3+][SCN-]
(0.006 – x)(0.004 - x)
0.118 = 2417.5 + 0.0483
2417.5x = 0.334 - 0.0483 = 0.0697
0.0697
X =
= 2.88 x 10-5 M
2417.5
At equilibrium,
(FESCN2*] = 2.88 x 10S M
(Fe3*] = 0.0060 - x = 0.0060 –- 2.88 x 105 = 0.0059712
[SCN'] = 0.0040 – x = 0.0040 - 2.88 x 10S = 0.0039712
%3D
not sure what
wernt 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%2Fd170f95e-0b96-4e65-8efa-ee670c09edb6%2F7sf1vwr_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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
equilibrium
Test Tube
Absorbance at Amax
Kf
1
0.118
2.88 x 10-4 M
Given absorbance: y = 0.118
y - 0.0483
X =:
2417.5
0.118 – 0.0483
2417.5
x = 2.883 x 10$ M
(FESCN2*] = 2.88 x 10$ M
MFex VFe
0.002 x 3
(Fe**) =
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
-X
-X
+x
0.006 -x
0.004 - x
[FESCN2+]
Kf =
[Fe3+][SCN-]
(0.006 – x)(0.004 - x)
0.118 = 2417.5 + 0.0483
2417.5x = 0.334 - 0.0483 = 0.0697
0.0697
X =
= 2.88 x 10-5 M
2417.5
At equilibrium,
(FESCN2*] = 2.88 x 10S M
(Fe3*] = 0.0060 - x = 0.0060 –- 2.88 x 105 = 0.0059712
[SCN'] = 0.0040 – x = 0.0040 - 2.88 x 10S = 0.0039712
%3D
not sure what
wernt wrong -.
2.88 x 10-5
Kf =
(0.0059712 x 0.0039712)
Kf = 121
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