At room temperature, 80.0 ml of 0.125 M AgNO3(aq) and 20.0 ml of 0.500 M
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.
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.
(a) To calculate the initial concentrations of Ag+(aq) and Fe2+(aq), we can use the given volumes and molarities of the solutions.
For AgNO3(aq):
Volume = 80.0 mL = 0.0800 L
Molarity = 0.125 M
Initial moles of Ag+ = Molarity x Volume
= 0.125 M x 0.0800 L
= 0.010 mol
Initial concentration of Ag+(aq) = moles / volume
= 0.010 mol / 0.100 L (sum of both volumes)
= 0.100 M
For Fe(NO3)2(aq):
Volume = 20.0 mL = 0.0200 L
Molarity = 0.500 M
Initial moles of Fe2+ = Molarity x Volume
= 0.500 M x 0.0200 L
= 0.010 mol
Initial concentration of Fe2+(aq) = moles / volume
= 0.010 mol / 0.100 L (sum of both volumes)
= 0.100 M
Therefore, the initial concentrations of Ag+(aq) and Fe2+(aq) are both 0.100 M.
(b) To find the equilibrium constant Kc, we can use the concentrations at equilibrium. Given that the concentration of Fe3+ at equilibrium is 0.00505 M, we can assume that the concentrations of Ag+(aq) and Fe2+(aq) are both x (as they will not change upon reaction).
The balanced equation shows a 1:1 stoichiometric ratio between Ag+ and Fe2+ in the forward direction, so we can write the expression for Kc:
Kc = [Ag+][Fe3+]
--------------
[Fe2+]
Kc = (x)(0.00505 M)
----------------
(x)
Since the concentration of Fe3+ is given as 0.00505 M, we can substitute it into the equation:
Kc = 0.00505 M / x
(c) Diluting the resulting solution with water to 200.0 mL will not cause any chemical reaction to take place as there are no new reactants introduced. The diluted solution will simply have the same reaction components but at different concentrations.
To calculate the equilibrium concentrations of each reagent in this diluted solution, we need to consider the dilution factor. The final volume of the solution is 200.0 mL, and we have already determined the initial concentration of Ag+(aq) and Fe2+(aq) to be 0.100 M.
Using the dilution formula, we can calculate the new concentrations:
For Ag+(aq):
Initial concentration = 0.100 M
Initial volume = 100.0 mL
Final concentration = (Initial concentration x Initial volume) / Final volume
= (0.100 M x 100.0 mL) / 200.0 mL
= 0.0500 M
For Fe2+(aq):
Initial concentration = 0.100 M
Initial volume = 100.0 mL
Final concentration = (Initial concentration x Initial volume) / Final volume
= (0.100 M x 100.0 mL) / 200.0 mL
= 0.0500 M
Therefore, the equilibrium concentrations of Ag+(aq) and Fe2+(aq) in the diluted solution are both 0.0500 M.