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+?
A. 1.2 × 10-2 M
B. 1.9 × 10-4 M
C. 7.5 × 10-4 M
D. 4.8 × 10-2 M
You did it again Bob. WOW
I did exactly what you said to do and I got the answer wrong. I Got A but the answer was B. Please explain.
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).
To solve this problem, we can use the equilibrium constant expression, which relates the concentrations of the reactants and products at equilibrium. The balanced chemical equation for the reaction is:
Ag+(aq) + Fe2+(aq) Fe3+(aq) + Ag(s)
The equilibrium constant expression, Kc, is given by:
Kc = [Fe3+][Ag(s)] / [Ag+][Fe2+]
We are given the value of Kc (3.0 × 10^−3) and the initial concentrations of Ag+ (0.25 M) and Fe2+ (0.25 M). Let's assume the equilibrium concentration of Fe3+ to be x M.
Substituting the given values into the equilibrium constant expression:
3.0 × 10^−3 = [x][1] / [0.25][0.25]
Simplifying the equation:
3.0 × 10^−3 = 4x
Dividing both sides by 4:
x = (3.0 × 10^−3) / 4
x = 7.5 × 10^−4 M
Therefore, the equilibrium concentration of Fe3+ is 7.5 × 10^−4 M.
The correct answer is option C: 7.5 × 10^−4 M.
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.