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# Determination of an Equilibrium Constant

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Introduction

Determination of an Equilibrium Constant & Analysis of an Iron Tablet Khan Salinder Results and Analysis Part One: Prepare and Test Standard Solutions Table One Beaker Number 0.0020 M SCN- (L) 0.200 M Fe(NO3)3 (L) [FeSCN2+] in Equilibrium Absorbance 1 0.000 0.005 0.00M 0.000 2 0.002 0.005 8.0 x 10-5 M 0.273 3 0.003 0.005 1.2 x 10-4 M 0.396 4 0.004 0.005 1.6 x 10-4 M 0.527 5 0.005 0.005 2.0 x 10-4 M 0.652 Sample Calculation (Beaker 2): Calculating [FeSCN2+] when Reaction "Complete" Fe3+ (aq) ...read more.

Middle

Calculation (Tube 2): Calculating Kc Value Calculation of [ FeSCN2+] In Equilibrium y=3256.1x + 0.0049 0.111= 3256.1x + 0.0049 0.1061=3256.1x x = 3.26 x 10-5 [FeSCN2+] = 3.26 x 10-5 Calculation of [Fe3+] & [SCN-] In Equilibrium Fe3+ + SCN-?FeSCN2+ Kc = [FeSCN2+] [Fe3+] [SCN-] [ ] Fe3+ + SCN- --> FeSCN2+ Calculation of Kc Value Kc = [FeSCN2+] = 3.26 x 10-5 . [Fe3+] [SCN-] [1.97 x 10-3][1.97 x 10-3 ] Kc= 8.400 Average Kc Value = 15.99 Part Three: Prepare and Analyze Stock Fe Solution from Iron Tablets Table Three Test Tube Number Fe3+ (L) ...read more.

Conclusion

[Fe3+]= 9.47 x 10-3 [ ] Fe3+ + SCN- --> FeSCN2+ I x 2.0 x 10-3 0 C - 3.03 x 10-4 - 3.03 x 10-4 + 3.03 x 10-4 E =9.47 x 10-3 = 1.70 x 10-3 = 3.03 x 10-3 x= 9.77 x 10-3 M= Concentration of Fe3+ in stock solution n= MV 2.93 x 10-5 mol Fe3+ x 55.7g = 1.63 x 10-3 g Fe3+ = (9.77 x 10-3) (0.003) 1 mol Fe3+ n=2.93 x 10-5 mol Fe3+ Average Mass of Fe3+ = 2.32 x 10-3 g Fe3+ in Iron Tablets ...read more.

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