At room temperature, 80.0 ml of 0.125 M AgNO3(aq) and 20.0 ml of 0.500 M

Question

Fe(NO3)2(aq) are mixed together, generating the following equilibrium system
Ag+
(aq) + Fe2+(aq) Ag(s) + Fe3+(aq).
At equilibrium, the concentration of Fe3+ is 0.00505 M.
(a) (4 pts) Calculate the initial concentrations of Ag+
(aq) and Fe2+(aq) immediately upon mixture of
the two solutions (you may disregard any changes in volume due to mixing).
(b) (4 pts) What is the value of the equilibrium constant Kc at room temperature for this system?
(c) (6 pts) Dilute the resulting solution with water to 200.0 ml. What chemical reaction, if any, will
take place? Calculate the equilibrium concentrations of each reagent in this diluted solution.

Answer 1

A. (Ag^+) = 0.125 x (80/100) = 0.100
(Fe^2+) = 0.500 x (20/100) = 0.100
B.
.......Ag^+ + Fe^2+ ==> Ag(s) + Fe^3+
I...0.100....0.100.......0.......0
C.....-x......-x.......solid.....x
E..0.100-x..0.100-x....solid.....x
The problem tells you x is 0.00505; therefore, at equilibrium
(Ag^+) = 0.1-0.00505
(Fe^2+) = 0.1-0.00505
(Fe^3+) = 0.00505

Kc = (Fe^3+)/(Ag^+)(Fe^2+)
Substitute the numbers above and solve for Kc.

C.
If the solution is diluted to 200 mL, then the initial concn of Ag^+ = 0.1/2 = 0.05
Initial (Fe^2+) = (0.1/2) = 0.05

So now the equilibrium is
........Ag^+ + Fe^2+ ==> Ag(s) + Fe^3+
I.....0.05.....0.05.......0.......0
C......-x......-x........solid....x
E.....0.05-x...0.05-x...solid......x

Substitute into Kc exprssion along with Kc from part B and solve for x and 0.05-x. Remember Kc doesn't change with dilution.