LAB: Photometric Determination of an Equilibrium Constant *Show All the work* Da
ID: 539058 • Letter: L
Question
LAB: Photometric Determination of an Equilibrium Constant
*Show All the work*
Data:
Solution
mL 0.200M Fe(NO3)3
mL 0.002M KSCN
mL 0.1M HNO3
Absorbance
A
5
1
14
0.36
B
5
2
13
0.79
C
5
3
12
1.14
D
5
4
11
1.68
E
5
5
10
2.27
Solution
mL 0.002M Fe(NO3)3
mL 0.002M KSCN
Absorbance
F
3
7
0.77
G
4
6
0.60
H
5
5
0.58
I
6
4
0.54
J
7
3
0.64
Questions:
For the calibration data include a table with solutions A-E, the volume of Fe3+, the volume of KSCN, total volume of Fe3+, KSCN and the absorbances values for each experimental run.
for the equilibrium data (solutions F-J) consider the chemical reaction: Fe3+(aq) + xSCN-(aq) <---> Fe(SCN)x ^(3-x)(aq)
Kc= [Fe(SCN)x ^(3-x)] / [Fe3+] [SCN-)]^x
where x is 1,2 or 3
in order to determine the value of x, complete the following table for each x value.
Soln
Initial Volume of Fe3+ (mL)
Initial volume of KSCN (mL)
Total volume (mL)
Initial [Fe3+] (M)
Initial [SCN-] (M)
abs
Fe(SCN)x^(3-x) (M)
Eq [Fe3+] (M)
Eq [SCN-] (M)
K
F
G
H
I
J
To determine the equilibrium concentrations of [Fe3+] and [SCN-] use the following two equations where x=1,2 or 3:
[Fe3+(aq)]eq =[Fe3+(aq)]Init - [Fe(SCN)x ^(3-x)]eq
and
[SCN-Fe3+]eq = x[SCN-]Init - [Fe(SCN)x ^(3-x)]eq
After determining the concentration at equilibrium, determine Kc, for each x value and reaction solution using:
Kc= [Fe(SCN)x ^(3-x)] / [Fe3+] [SCN-)]^x
Calculate for each x value the average K and standard deviation in K.
Solution
mL 0.200M Fe(NO3)3
mL 0.002M KSCN
mL 0.1M HNO3
Absorbance
A
5
1
14
0.36
B
5
2
13
0.79
C
5
3
12
1.14
D
5
4
11
1.68
E
5
5
10
2.27
Explanation / Answer
A) Calibration curve
[SCN-] is the limiting reagent
[FeSCN]2+ formed = [SCN-] present
Solution [FeSCN]2+ (M)
A 0.002 M x 1 ml/20 ml = 1 x 10^-4
B 0.002 M x 2 ml/20 ml = 2 x 10^-4
C 0.002 M x 3 ml/20 ml = 3 x 10^-4
D 0.002 M x 4 ml/20 ml = 4 x 10^-4
E 0.002 M x 5 ml/20 ml = 5 x 10^-4
Plot, [FeSCN]2+ on x-axis and absorbance on y-axis
slope = molar absorptivity = (1.14 - 0.79)/(3 x 10^-4 - 2 x 10^-4) = 4775 M-1.cm-1
B) Equilibrium constant K
Solution [FeSCN]2+eq (M) [Fe3+]initial (M) [Fe3+]eq (M)
F 0.77/4775 = 1.61 x 10^-4 0.002 M x 3 ml10 ml = 6 x 10^-4 6 x 10^-4 - 1.61 x 10^-4 = 4.39 x 10^-4
G 0.60/4775 = 1.25 x 10^-4 0.002 M x 4 ml10 ml = 8 x 10^-4 8 x 10^-4 - 1.61 x 10^-4 = 6.39 x 10^-4
H 0.58/4775 = 1.21 x 10^-4 0.002 M x 5 ml10 ml = 1 x 10^-3 1 x 10^-3 - 1.61 x 10^-4 = 8.39 x 10^-4
I 0.54/4775 = 1.13 x 10^-4 0.002 M x 6 ml10 ml = 1.2 x 10^-3 1.2 x 10^-3 - 1.61 x 10^-4= 8.47 x 10^-4
J 0.64/4775 = 1.34 x 10^-4 0.002 M x 7 ml10 ml = 1.4 x 10^-3 1.4 x 10^-3 - 1.61 x 10^-4= 1.24 x 10^-3
Solution [FeSCN]2+eq (M) [SCN-]initial (M) [SCN-]eq (M)
F 0.77/4775 = 1.61 x 10^-4 0.002 M x 3 ml10 ml = 1.4 x 10^-3 1.4 x 10^-3 - 1.61 x 10^-4 = 1.24 x 10^-3
G 0.60/4775 = 1.25 x 10^-4 0.002 M x 7 ml10 ml = 1.2 x 10^-3 1.2 x 10^-3 - 1.61 x 10^-4 = 8.47 x 10^-4
H 0.58/4775 = 1.21 x 10^-4 0.002 M x 6 ml10 ml = 1 x 10^-3 1 x 10^-3 - 1.61 x 10^-4 = 8.39 x 10^-4
I 0.54/4775 = 1.13 x 10^-4 0.002 M x 5 ml10 ml = 8 x 10^-4 8 x 10^-4 - 1.61 x 10^-4= 6.39 x 10^-4
J 0.64/4775 = 1.34 x 10^-4 0.002 M x 3 ml10 ml = 6 x 10^-4 6 x 10^-4 - 1.61 x 10^-4= 4.39 x 10^-4
Equilibrium constant
Solution F, K = (1.61 x 10^-4)/(4.39 x 10^-4)(1.24 x 10^-3) = 295.76
Similarly K value for other solutions can be calculated.
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