ud Calculationsa For each test tube solution enter the initial concentration of
ID: 526364 • Letter: U
Question
ud Calculationsa For each test tube solution enter the initial concentration of Fe and scN! the equilibrium concentration of FesCN 2, into the ICE table given. Complete entries for the rest of the table and calculate the K value for each of the tables. The values of K should be confined to a narrow range to reflect constancy, Comment on the quality of your work in this regard and calculate the average K. Standard Solutions 5.0 mL of 0.200M Fe(NO) 0 mL of 0.002M KSCN; and orange co 4.0 mL of M HNOs. Total volume of the mixture is 10 mL concentration of Fe(Non, is 0.001 M (Use Mavi Mrva Concentration of KSCN is 0.0002 M Concentration of IFesCN2 is (Use limiting reactant theory and equation) e (aq) (SCN) (and) co (F (aq) Since (Fe")> (SCN i) in the solution, essentially all (SCN has been oonverted to (FeSCN according to the Le Chatelier's Principle. Thus the equilibrium concentration of (FesCN is equal to the initial (SCNH) concentration which is calculated as: 0.002M x 1.0 mL/100 ml -2.0 x 10 M standard Sample: Absorbance of standard solution FesCN2 is Plot a linear graph of concentration of [FescN x-axis) vs. Absorbance (y-axis) with line passing through ongin Equation of Best Fit y -mx, where y is absorbance and x is concentration Equation is Slope of the equation (m) isExplanation / Answer
From the data
[SCN-] = [FeSCN2+]eq from standard solution = 0.0002 M
absorbance of standard = 0.74
Calculation of [FeSCN2+]eq in test tubes 1-3
Test tube 1,
[FeSCN2+]eq = 0.66 x 0.0002/0.74 = 0.00018 M
Test tube 2,
[FeSCN2+]eq = 1.44 x 0.0002/0.74 = 0.00039 M
Test tube 3,
[FeSCN2+]eq = 2.42 x 0.0002/0.74 = 0.00065 M
ICE table,
Test tube 1
Fe3+ + SCN- <=========> FeSCN2+
I 0.001 0.0002 -
C -0.00018 -0.00018 +0.00018
E (0.001-0.00018) (0.0002-0.00018) 0.00018
So,
Keq = 0.00018/(0.001-0.00018)(0.0002-0.00018) = 10975.61
Similarly Keq for other test tubes using ICE chart can be done.
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