A reaction mixture is formed by mixing 2.0 mL of 0.010 MFe^3+ and 2.0 mL of 0.01
ID: 1016092 • Letter: A
Question
A reaction mixture is formed by mixing 2.0 mL of 0.010 MFe^3+ and 2.0 mL of 0.010 MSCN^-, and the diluting the solution to a total volume of 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 front problem 1. What would be 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 nmExplanation / Answer
Solution :-
Q1) Calculating the concentrations of the diluted solutions
Formula
M1V1=M2V2
Calculating the concentration of the Fe^3+
M2 = M1V1/V2
M2 = 0.010 M * 2.0 ml / 25 ml
M2 =0.0008 M
Calculating the concentration of the SCN^-
M2 = M1V1/V2
M2 = 0.010 M * 2.0 ml / 25 ml
M2 =0.0008 M
Q2 Kf= 300
Calculating the concentration of the species at the equilibrium
Intially all the ions combine to form FeSCN^2+ because the Kf is very large
At equilibrium some of the FeSCN^2+ dissociates
FeSCN^2+ ------ > Fe^3+ + SCN^-
0.0008 M 0 0
-x +x +x
0.0008-x x x
Since the equation is dissociation equation therefore Kf is inversed
(1/Kf) = [Fe^3+][SCN^-]/[FeSCN^2+]
(1/300) = [x][x]/[0.0008-x]
0.00333 = x^2 / 0.0008-x
Solving for the x we get
X= 0.000666 M
Using the value of x we can find the equilibrium concentrations
[Fe^3+] = x = 0.000666 M
[SCN^-] = x= 0.000666 M
[FeSCN^2+] = 0.0008 M –x
= 0.0008 - 0.000666 M
= 0.000134 M
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