In: Chemistry
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 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.