Question

Table 1: Calibration Curve Data (4 pts. total) Tube Vol. of 0.200M Volume of Volume of Total Vol. Fe(NO3)3 0.00200M KSCN 0.500M HNO; Smi 10.5 mL 150 mL zob 20.5ml 19.98m2 J 2 5.00mL lll ML 14.05 mL 20.16 mil 5.00 ml 1.5 mL 13.5 mL 20 ml 4 5.00mL 2.0 mL 13.01 mL 20.01 mL 15 500mL I 2.50 ML 12.5 mL 20 ml Table 2: Calibration Curve Calculations and Percent Transmittance Recorded (1/2 pt. for each blank in first four columns, 1 pt. for each blank in final column) Tube Initial Molarity Initial Molarity Final Molarity Percent Transmittance Fe(NO3)3 KSCN Fe(SCN)2+ Observed Absorption Calculated .22 .001 M .ool M .ool M ₂001 M .001 M lxlos m 5.0 x10 SM 60.91. 2.22 x10 m 1, 19104 m 38.51. 3x10 m 1.5x10-4 m 40.37 4x100m 1.99x109 m 18.187. 5x10cm 2.5x16" M 19. 27. Table 3: Initial Data for Equilibrium Constant Determination (4 pts, total) Tube Vol. of 0.00200M Volume of Volume of 0.500M Total Vol. Fe(NO3)3 0.00200M KSCN HNO3 15.00 mL 11.03 mL I 4.94 AL 110.97 mL 125 .00 mL 2.02 ml 3.98 ML 11.00 mL 3 5.00 mL 3.05 ML 2.98 mL 11.00 ml 4 5.00 ML 3.93 ML 2.10 ML 11.03 mL 155.00 mL 4.97 ML 1.03 ML 110 mL Table 4: Calculations for Equilibrium Constant Determination and Transmittance Recorded: (1/2 pt. for each blank in first four columns, 2 pt. for each blank in final column) Initial Concentrations Equilibrium Results Tube Initial Molarity Initial Molarity Percent Transmittance Absorption *Equilibrium Fe(NO3)3 KSCN Observed Calculated Molarity Fe(SCN)2+ :31 ( lxlos m 2.06x10 m 77,1 % 2 1.x10' M 4.044106 m 49.2 % 3 1,105 M 6.1810 bm 45.01 [Lx 10 S M 7.86x10m 34.2% 11x10 S M 9.94x10m 19.1% 1.35 .47 .72 * Calculated from the equation of the best fit line. (20 total points for all ICE tables)

Need help finding equilibrium molarity

Answer 1

First of all, the calibration curve is not correct. Given the high excess of Fe(III) in this solutions, it can be considered that all of the SCN^- is converted to the product, so the concentration of the product is the same as the initial concentration of SCN^-.

The curve is (commas are decimal separators):

y = 182633x + 0,0221 R2 = 0,9971 Abs 0 0,000001 0,000002 0,000004 0,000005 0,000006 0,000003 C(M)

I've eliminated the middle point, which was clearly an outlier.

Now we can determine the concentration of FeSCN²⁺. For tube 1, given the absorbance (y in the calibration curve), you can determine the concentration (x in the calibration curve) as:

C = A – 0.0221 182633-1 0.11 -0.0221 - = 4.81010-7M

This concentration is the product of the 1:1 reaction of Fe(III) and SCN^-. Thus, the concentration of this species in equilibrium is the difference between their initial concentration and the concentration that reacted to yield the product (the "change" in the ICE table). In tube 1, the equilibrium concentration for Fe(III):

1x10^-5 - 4.81x10^-7 = 9.52x10^-6 M

For SCN^-: 2.06x10^-6 - 4.81x10^-7 = 1.58x10^-6 M

With these, we can calculate the equilibrium constant:

K = [FeSCN2+1 [Fe3+1SCN-1 4.81.210-7 9.52.c 10-6.1.58.c10-6

Repeating this process for all tubes, you get:

	Tube 1	Tube 2	Tube 3	Tube 4	Tube 5
Initial Fe(III)	1,00E-05	1,00E-05	1,00E-05	1,00E-05	1,00E-05
Initial SCN(-)	2,06E-06	4,04E-06	6,10E-06	7,86E-06	9,94E-06
A	0,11	0,31	0,35	0,47	0,72
FeSCN(II) eq	4,81E-07	1,58E-06	1,80E-06	2,45E-06	3,82E-06
Fe(III) eq	9,52E-06	8,42E-06	8,20E-06	7,55E-06	6,18E-06
SCN(-) eq	1,58E-06	2,46E-06	4,30E-06	5,41E-06	6,12E-06
K	32028	75961	50836	60089	101079