An aqueous solution of 0.25 M silver nitrate, AgNO3, and 0.25 M iron(II) nitrate, Fe(NO3)2, are allowed to come to equilibrium in the following chemical reaction: Ag+(aq) + Fe2+(aq) Fe3 + (aq) + Ag(s) If the equilibrium constant, Kc, for the reaction is 3.0 × 10-3, what is the equilibrium concentration of Fe3+?

Question

A. 1.2 × 10-2 M
B. 1.9 × 10-4 M
C. 7.5 × 10-4 M
D. 4.8 × 10-2 M

Answer 1

The problem isn't clear but I will assume that is 0.25M AgNO3 and 0.25M Fe(NO3)2 AFTER they are mixed; otherwise each will be diluted by the other.


......Ag+aq + Fe2+aq => Fe3+aq + Ag(s) I...0.25M...0.25M........0........0
C...-x........-x.........x........x
E..0.025-x...0.025-x.....x........x

Substitute the E line into Kc expression and solve for x. Note that Ag(s) is NOT a part of the Kc expression. Post your work if you get stuck.

Answer 2

I did exactly what you said to do and I got the answer wrong. I Got A but the answer was B. Please explain.

Answer 3

First, I made a typo and didn't see it (on both 0.25 numbers) and I'm glad you picked that up. Here is the problem.
You included (Ag)solid as part of the Kc expression which I warned you not to do.
That should be 3E-3 = (x)/(0.25-x)^2 and solve. You get a quadratic and the answer comes out x = 1.875E-4 M which rounds to 1.9E-4 M. Sorry about the typos but kudos to you for picking that up. Remember solids and pure liquids are not considered part of K expression (actually they can be but by definition they are 1.0).

Answer 4

You did it again Bob. WOW