Table 2: Volumes for Unknown Equilibrium Solutions Solution Volume in mL of 0.00200 M Fe(NO3)3 Volume in mL of 0.00200 M NaSCN Volume in mL of 0.10 M HNO 3 E1 (blank) 5.00 0.00 5.00 E2 5.00 1.00 4.00 E3 5.00 2.00 3.00 E4 5.00 3.00 2.00 E5 5.00 4.00 1.00 E6 5.00 5.00 0.00 Table 1: Standard Solutions for FeSCN²+ Beer's Law Plot Solution [Fe3+] [SCN] S1 (blank) 4.0 x 10-2 M 0.0 M S2 4.0 x 10-2 M 4.0 x 10-5 M S3 4.0 x 10-2 M 8.0 x 10-5 M S4 4.0 x 10-2 M 12.0 x 10-5 M S5 4.0 x 10-2 M 16.0 x 10-5 M S6 4.0 x 10-2 M 20.0 x 10-5 M
Table 2: Volumes for Unknown Equilibrium Solutions Solution Volume in mL of 0.00200 M Fe(NO3)3 Volume in mL of 0.00200 M NaSCN Volume in mL of 0.10 M HNO 3 E1 (blank) 5.00 0.00 5.00 E2 5.00 1.00 4.00 E3 5.00 2.00 3.00 E4 5.00 3.00 2.00 E5 5.00 4.00 1.00 E6 5.00 5.00 0.00 Table 1: Standard Solutions for FeSCN²+ Beer's Law Plot Solution [Fe3+] [SCN] S1 (blank) 4.0 x 10-2 M 0.0 M S2 4.0 x 10-2 M 4.0 x 10-5 M S3 4.0 x 10-2 M 8.0 x 10-5 M S4 4.0 x 10-2 M 12.0 x 10-5 M S5 4.0 x 10-2 M 16.0 x 10-5 M S6 4.0 x 10-2 M 20.0 x 10-5 M
Chemistry for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 12.40PAE: Because carbonic acid undergoes a second ionization, the student in Exercise 12.39 is concerned that...
Related questions
Question
Solve for Unknown Concentrations using the Best Fit Line and Absorbances
y=mx+b
Example data (however use y= 5489.3x - 0.0821)
y = 4292.9 x - 0.021
Absorbance = 4292.9*concentration - 0.021
So, to find the unknown E concentrations for [FeSCN2+], would plug in the measured absorbance values and solve each for x. That gives the equilibrium concentrations for [FeSCN2+]
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