Question

LAB: Photometric Determination of an Equilibrium Constant

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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.

Answer 1

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.