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4. Create Table 3 in your notebook. a.) Calculating the concentration of each diluted solution. • Calculate the initial concentration of Fe" and SCN, accounting for dilution using C₁V₁ C₂V₂, for each solution in Table 2. Show an example calculation for solution A for both the iron(III) and thiocyanate ions. Enter all calculated values into Table 3. b.) Creating the graph. ▪ Create a Beer's Law Plot (a.k.a. calibration curve) using the class average data from Table 1. Use Excel to create this graph. The concentration should be plotted on the x-axis and the absorbance on the y-axis. • Plot the best-fit line and report the equation of the line (y = mx + b) that relates conc. (x) to absorbance (y). This can be done by selecting Chart/Add Trendline from the menu. On the "Type" tab of the dialog box, select "linear". On the "options" tab, be sure to include the equation of the line and the R² value on the graph. Include all other items required for a good graph. c.) Determining "Final [FeSCN2] M" for Table 3 from the calibration curve Use the equation from your Excel graph to solve for concentration of FeSCN at equilibrium for each solution in Table 2 and enter the results in Table 3. (Remember that it is impossible for the conc. of FeSCN to be greater then the initial conc. of either reacting species.) Show a sample calculation for the [FeSCN2] from the equation of the best-fit line for Solution A.
Soln. mL of Fe (2.00×10 M) A B C D E Soln. A B C D 3.00 E 3.00 Note: All final volumes were 3.00 3.00 3.00 Table 2: Analysis of Solutions Used to Determine Ke Absorbance Values at 447 nm mL of SCN (4.00×10 M) 1.00 2.00 Initial [Fe] M (accounting for dilution) 3.00 4.00 5.00 25 0.025 0.017 0.037 930 0.400 0.051 0.031 0.053 012 0.084 0.050 0.078 0.110 0.065 0.106 1.180 0.915 0.142 0.080 0.129 0.137 ml made up with 0.500 M HNO3. 792 0.716 037 1.200 0.980 0.084 1.330 0.975 0.102 Table 3: Calculations for Determination of the Equilibrium Constant Final [FeSCN2] M (determined from calibration curve) Initial [SCN] M (accounting for dilution) 2.40x 10M 1.60 x10-1 M 2.40 x10-1M 3.20 x10-1 M 2.10x10-1M 4.80 x10-1 M 2.10 x10-1 M 6.40 x10-1 M 2.40 x10M 8.00 × 10-1 M Average Absorb. Kc (solve for Kc using an ICE table.)
Absorbance 0.3 0.25 0.2 0.15 0.1 0.05 0 0 Fe 3+ y = 3832.9x + 0.026 R2=0.9937 0.00001 0.00002 0.00003 0.00004 0.00005 0.00006 Concentration 0.00007