Question

• Determining k in Beer's Law 2. Determining the equilibrium constant Test Tube No. Absorbance Test Tube No. Absorbance 0,060 이79 0.253 0319 0.400 0.120 0.268 0.361 0.461 0.695 Post Laboratory Questions 1. Complete the following table and calculate k-value for the Beer's Law. All concentrations should be in M. [Fe(SCN)2) Absorbance CA Test Tube No. Diluted Fe(NO3)2 (mL) 0 . 2 3 4 s Reference 1.0 2.0 3.0 40 5.0 1.0x105 20x10-5 3.0x105 4,0x105 Soxo5 0.080 0.179 0255 0,319 0.400 8.0x07 3.82X606 7,59x106 1,276x105 2,0x105

please correct me If I did anything wrong. im very confused on this.

Answer 1

Stock solution of Fe(NO3)3 has concentratiom of 0.0025 M and its 4 mL is diluted to 100 mL of water. Thus diluted Solution has conc =

M1V1 =M2V2

0.0025 * 4 = M2 * 100

M2 = 0.0001 M

This concentraion is used to make the solution of test tube 1-5.

Test Tube #	Vol of Fe(NO3)3 (mL)	No of Moles of Fe3+	Vol of 1 M KSCN (mL)	No of Mole of SCN-	Vol of HNO3 (mL)	Total Vol	[FeSCN2+] Formed, M
1	1	0.0000001	5	0.005	4	10	0.00001
2	2	0.0000002	5	0.005	3	10	0.00002
3	3	0.0000003	5	0.005	2	10	0.00003
4	4	0.0000004	5	0.005	1	10	0.00004
5	5	0.0000005	5	0.005	0	10	0.00005

Calculations for above table -

mole = Molarity * Vol (L)

Mole in TT#1 = 0.0001 * 0.001 = 0.0000001 moles

simiiarly for all other test tubes also.

For KSCN moles also similarly as above.

NOw , as the rxn is 1:1 equivalent hence the no of moles of Fe3+ = o of moles of FeSCN2+

In this way, [FeSCN2+] = 0.0000001 / Total volume (L)

[FeSCN2+] = 0.0000001 / (10/1000)

[FeSCN2+] = 0.0000001 / 0.01 = 0.00001 M in TT#1

Similarly for all other test tubes also.

On plotting Calibration curve we got -

B (y-intercept) = 0.0122 +/- 0.0104670912865036

A (slope) = 7,800 +/- 315.59467676119

& Graphi - o x Calibration Curve 0.451 0.35 Abs 0.05 1e-05 2e-05 3e-05 Conc (M) 4-05 5e-05 6e-05

Hence K = 7,800 (Slope of plot) = ε

Determination of Keq -

Test Tube #	Vol of 0.0025 M Fe(NO3)3 (mL)	No of Moles of Fe3+	Vol of 0.0025 M KSCN (mL)	No of Mole of SCN-	Vol of HNO3 (mL)	Total Vol (mL)	initial [Fe3+] (M)	initial [SCN-] (M)	[FeSCN2+] Formed, M	Final [Fe3+] (M)	Final [SCN-] (M)
6	1	0.0000025	1	0.0000025	5	7	0.0003571	0.0003571	0.00001538	0.0003417	0.00034172
7	1	0.0000025	2	0.0000050	4	7	0.0003571	0.0007143	0.00003435	0.00032275	0.00067995
8	1	0.0000025	3	0.0000075	3	7	0.0003571	0.0010714	0.00004628	0.00031082	0.00102512
9	2	0.0000050	2	0.0000050	3	7	0.0007143	0.0007143	0.00005910	0.0006552	0.000655
10	2	0.0000050	3	0.0000075	2	7	0.0007143	0.0010714	0.00008910	0.0006252	0.0009823

From Beers-Lambert Law-

A = ε C l

For TT#6-

0.120 = 7,800 * C * 1

C = 0.00001538 M

Similiarly for TT#7-10 also.

NOw,

$Keq = \frac{[FeSCN^{2+}]}{final[Fe^{3+}]Final[SCN^-]}$

Using above formula -

Keq of TT#6	131.709
Keq of TT#7	156.5249
Keq of TT#8	145.2478
Keq of TT#9	137.6701
Keq of TT#10	145.0824

Average Keq =

143.2469

Cheers, I solved everything for you.