The following questions refer to the following system: 500.0 mL of 0.020 M Mn(NO
ID: 933582 • Letter: T
Question
The following questions refer to the following system: 500.0 mL of 0.020 M Mn(NO3)2 are mixed with 1.0 L of 1.0 M Na2C2O4. The oxalate ion, C2O4, acts as a ligand to form a complex ion with the Mn2+ ion with a coordination number of two.
Mn2+ + C2O42 The following questions refer to the following system: 500.0 mL of 0.020 M Mn(NO3)2 are mixed with 1.0 L of 1.0 M Na2C2O4. The oxalate ion, C2O4, acts as a ligand to form a complex ion with the Mn2+ ion with a coordination number of two. Reference: Ref 16-3 Find the equilibrium concentration of the Mn2+ ion.Explanation / Answer
The following questions refer to the following system: 500 mL of 0.02 M Mn(NO3)2 are mixed with 1.0 L of 1.0 M Na2C2O4. The oxalate ion, C2O4, acts as a ligand to form a complex ion with the Mn+2 ion with a coordination number of two.
Mn2+ + C2O42- goes to MnC2O4 K1 = 7.9 x 10^3
[Mn(C2O4)2]2- goes to MnC2O4 + C2O42- K2 = 1.26 x 10^-2
42. What is the equilibrium constant for the following formation:
Mn2+ + 2C2O42- goes to [Mn(C2O4)2]2-
(I know the answer is 6.3 x 10^5 , but why?)
43. Find the equilibrium concentration of the [Mn(C2O4)2]2- ion.
(I know the answer is 6.7 x 10^-3, but why?)
44. Find the equilibrium concentration of the Mn(C2O4) ion
(I know the answer is 1.3 x 10^-4, but why?)
ANSWER
42. When you add two equations, you multiply their respective K's to give you the overall K. Also, when you reverse an equation, K for the reverse equation is 1/K for the forward equation.
So, in order to combine these two equations to generate the one in the question, you have to reverse the second equation and then add it to the first one.
K(overall) = K1 X 1/K2 = 6.3 X 10^5
43. OK. When you combine the two solutions, your initial (prior to any reaction) concentrations of Mn2+ and C2O42- are:
[Mn2+] = 6.67X10^-3 M and [C2O42-] = 0.667 M. (I got these using M1V1=M2V2 using the initial concentrations and volumes and the final volume of the mixture.)
Since K for the reaction you are considering is a large number, you know that virtually all of the Mn2+ will be in the form of the complex ion. So, since the initial concentration of Mn2+ = 6.7 X 10^-3 M, that will be very very close to the final concentration of the complex ion.
(You also should see that the concentration of oxalate in the final solution will basically be unchanged, since there is so much more oxalate than there is Mn2+ in the initial solution.) 44. For this problem, start with the expression for the equilibrium constant K2. Since you know that the concentration of the complex ion is 6.7 X 10-3, and that the concentration of oxalate is basically, 0.67, just plug those into the expression for K2, and solve for the concentration of the MnC2O4. That will give you 1.3 X 10^-4.
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