I Data Report Sheet: Spectrophotemetric Determination of an Equilibrium Constant, Part A: Determination of a Calibration Carve: Std. 2 Stnd. 3 Stnd. 5 Stnd. 6 (H) 0.000127 |Fe(SCN)"] Column A 90 0 00060000 0 0 0.432 0.480 0.633 0.831 0.887 Absorbance Column E y%3D 4339.8x + 0.0258 Equaion of Calibration Line: Correlation Coefficient: 4339,8 Slope (br): Part B: Determination of K Expt. 1 Expt. 2 Expt. 3 Expt. 4 Expt. 5 0.00165 0.00165 0.00165 0.00165 0.00165 (Table I) 0.00033 66000 0 0.00132 0.00165 990000 (Table 1) (Fe(SCN)"leqsil. 0.290 0.524 0.803 1.024 Column F |Fe"leqail (Fe LFe(SČNY L ISCNJequi ((SCN FCS L 148.6799 135.1678 142.2659 141.8466 134.2843 Average K:140 448 6. Save Share II

Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter16: Solubility And Precipitation Equilibria
Section: Chapter Questions
Problem 37P
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I
Data Report Sheet: Spectrophotemetric Determination of an Equilibrium
Constant,
Part A: Determination of a Calibration Carve:
Std. 2
Stnd. 3
Stnd. 5
Stnd. 6
(H)
0.000127
|Fe(SCN)"]
Column A
90 0 00060000 0 0
0.432
0.480
0.633
0.831
0.887
Absorbance
Column E
y%3D 4339.8x + 0.0258
Equaion of Calibration Line:
Correlation Coefficient:
4339,8
Slope (br):
Part B: Determination of K
Expt. 1
Expt. 2
Expt. 3
Expt. 4
Expt. 5
0.00165
0.00165
0.00165
0.00165
0.00165
(Table I)
0.00033
66000 0
0.00132
0.00165
990000
(Table 1)
(Fe(SCN)"leqsil.
0.290
0.524
0.803
1.024
Column F
|Fe"leqail
(Fe LFe(SČNY L
ISCNJequi
((SCN FCS L
148.6799
135.1678
142.2659 141.8466
134.2843
Average K:140 448
6.
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Share
II
Transcribed Image Text:I Data Report Sheet: Spectrophotemetric Determination of an Equilibrium Constant, Part A: Determination of a Calibration Carve: Std. 2 Stnd. 3 Stnd. 5 Stnd. 6 (H) 0.000127 |Fe(SCN)"] Column A 90 0 00060000 0 0 0.432 0.480 0.633 0.831 0.887 Absorbance Column E y%3D 4339.8x + 0.0258 Equaion of Calibration Line: Correlation Coefficient: 4339,8 Slope (br): Part B: Determination of K Expt. 1 Expt. 2 Expt. 3 Expt. 4 Expt. 5 0.00165 0.00165 0.00165 0.00165 0.00165 (Table I) 0.00033 66000 0 0.00132 0.00165 990000 (Table 1) (Fe(SCN)"leqsil. 0.290 0.524 0.803 1.024 Column F |Fe"leqail (Fe LFe(SČNY L ISCNJequi ((SCN FCS L 148.6799 135.1678 142.2659 141.8466 134.2843 Average K:140 448 6. Save Share II
Expert Solution
Step 1

Equation of the calibration curve is y = 4339.8x + 0.0258 where

y = Absorbance

x = [FeSCN2+]

 

Experiment 1

y = 0.290

So,

0.290 = 4339.8x + 0.0258

x = 0.000061 M

Now,

Fe3+ + SCN- FeSCN2+

  Fe3+ SCN- FeSCN2+
Initial 0.00165 0.00033 0
Change -x -x x
Equilibrium 0.00165-x 0.00033-x x

x = 0.000061 M

 

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