hen excess Ag metal is added to 0.246 L of 1.290 M Fe3+(aq) at 298K, the following equilibrium is established:
Ag+(aq) + Fe2+(aq) Ag(s) + Fe3+(aq)
(1) If the equilibrium concentration of Ag+ is 0.504 M, what is the value of K?
K =
(2) Using the value of K from part (1), calculate the new equilibrium concentrations of the three species in solution, if the volume of the solution is increased to 0.563 L by the addition of pure water?
[Ag+] =
[Fe2+] =
[Fe3+] =
Simple
To solve these equilibrium problems, we need to use the concept of equilibrium constants. In this case, we are given an equilibrium reaction and the initial concentrations of some species.
(1) To find the value of K, we can use the equilibrium concentrations of Ag+. We can set up an expression for K using the equilibrium concentrations of Ag+, Fe2+, and Fe3+:
K = [Ag+][Fe2+]/[Ag][Fe3+]
Given that [Ag+] = 0.504 M, we need to find the equilibrium concentrations of [Ag], [Fe2+], and [Fe3+].
To find the equilibrium concentrations, we need to consider the stoichiometry of the reaction. From the balanced equation:
Ag+(aq) + Fe2+(aq) → Ag(s) + Fe3+(aq)
We can see that for every 1 Ag+ ion, we need 1 Ag solid to be in equilibrium. So, the equilibrium concentration of Ag is equal to the concentration of Ag+ (0.504 M). Thus, [Ag] = 0.504 M.
To find the equilibrium concentrations of [Fe2+] and [Fe3+], we need to use the initial concentration of Fe3+ and the stoichiometry of the reaction. Since excess Ag is added, we can assume that the concentration of Fe3+ remains unchanged. So, [Fe3+] = 1.290 M.
Now, we can substitute these values into the expression for K:
K = [Ag+][Fe2+]/[Ag][Fe3+]
= (0.504 M)[Fe2+]/(0.504 M)(1.290 M)
= [Fe2+]/(1.290 M)
So, the value of K for this reaction is [Fe2+]/1.290 M.
(2) Given that the volume of the solution is increased to 0.563 L by adding pure water, the total volume of the solution is now 0.563 L + 0.246 L = 0.809 L.
To find the new equilibrium concentrations, we can use the concept of dilution. The number of moles of the species remains the same before and after dilution. Therefore, the concentration can be determined using the equation:
moles = concentration × volume
First, let's find the new concentration of Ag+:
moles of Ag+ before dilution = [Ag+] × volume
= 0.504 M × 0.246 L
moles of Ag+ after dilution = moles of Ag+ before dilution
= moles of Ag+ before dilution/total volume
= (0.504 M × 0.246 L)/0.809 L
Now, let's find the new concentrations of [Fe2+] and [Fe3+] using the equation:
moles of Fe2+ before dilution = [Fe2+] × volume
= [Fe2+] × 0.246 L
moles of Fe3+ before dilution = [Fe3+] × volume
= 1.290 M × 0.246 L
moles of Fe2+ after dilution = moles of Fe2+ before dilution/total volume
= (moles of Fe2+ before dilution)/(0.809 L)
moles of Fe3+ after dilution = moles of Fe3+ before dilution/total volume
= (moles of Fe3+ before dilution)/(0.809 L)
Finally, the new equilibrium concentrations are:
[Ag+] = moles of Ag+ after dilution/total volume
[Fe2+] = moles of Fe2+ after dilution/total volume
[Fe3+] = moles of Fe3+ after dilution/total volume
Calculate these values using the given information and you will obtain the new equilibrium concentrations for [Ag+], [Fe2+], and [Fe3+].