A reaction mixture is formed by mixing 2.0 mL of 0.010 M Fe^3+ and 2.0 mL of 0.0
ID: 1016085 • Letter: A
Question
A reaction mixture is formed by mixing 2.0 mL of 0.010 M Fe^3+ and 2.0 mL of 0.010 M SCN^-, and the diluting the solution to a total volume or 25.0 mL. Calculate the initial concentrations of Fe^3+ and SCN^- under these conditions. Some experiments have indicated that this reaction has an equilibrium constant, K_f, of approximately 300. Use the value of K_f = 300 to calculate the equilibrium concentrations of all three species given the initial concentrations from problem 1. What would he the measured absorbance at 440 nm for this equilibrium mixture in a 1.0-cm cuvette? Use Beer's Law, with the extinction coefficient that is given in the Lab Discussion for FeSCN^2+ at 440 nm.)Explanation / Answer
1. V1S1=V2S2
i.e. 2 mL * 0.01 M = 2 * 10^-5 mol = 25.0 mL * S2 i.e. S2 = 8 * 10^-4 M
Hence, [Fe3+] initial = 8 * 10^-4 M
Likewise, [SCN-] initial = 8 * 10^-4 M
2.
Fe3+ + SCN - ---> Fe(SCN)2+
I 8 * 10^-4 8 * 10^-4 0
C -x -x +x
E (8 * 10^-4 - x) (8* 10^-4 - x) +x
Hence, Kf = [Fe(SCN)2+]eq/[Fe3+]eq * [SCN-]eq = 300
i.e. x/(8 * 10^-4 - x)^2 = 300
i.e. x = 300 (6.4 * 10^-7 -16 * 10^-4 x + x2) i.e. 300 x2 -1.48 x + 1.926 * 10^-4 = 0 Hence, x = 1.34 * 10^-4
So, [Fe(SCN)2+]eq = 1.34 * 10^-4 M
[Fe3+]eq = (8 * 10^-4 - 1.34* 10^-4) M = 6.66 * 10^-4 M
[SCN-]eq = 6.66 * 10^-4 M
c. Use Beer's law to calculate A(Absorbance) = € . c. l € = extinction coefficient at 440 nm = data not provided
c = [Fe(SCN)2+]eq = 1.34 * 10^-4 M, l = 1.0 cm
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