What are the concentrations of Cu^2+, NH3, and Cu(NO3)4^2+ at equilibrium when 18.8 g of Cu(NO3)2 is added to 1.0 L of a .400 M solution of aqueous ammonia? Assume that the reaction goes to completion and forms Cu(NH3)4^2+.
I think you made a typo with that Cu(NO3)4^2. I assume you meant [Cu(NH3)4]^2. 18.8 g Cu(NO3)2 = ?M. That's mols = grams/molar mass. Let's call that about 0.1 but you should go through and confirm that.
........Cu^2+ + 4NH3 ==> [Cu(NH3)4]^2
I....... 0.1....0.400.......0
C.......-0.1...-0.400......0.1
E.........0.......0........0.1
We make the ICE table above. You will need to look up Kf for [Cu(NH3)4]^2+ and you will find it is a huge number which means the reaction will go essentially to completion. Now you turn the reaction around and work the problem as IF it were a weak acid (it isn't of course) problem like this.
.........[Cu(NH3)4]^2 ==> Cu^2+ + 4NH3
I.........0.1..............0.......0
C..........-x..............x.......x
E.........0.1-x............x.......x
Now substitute the E line into Kf expression for the complex and solve for x, then evaluate 0.1-x.
To determine the concentrations of Cu^2+, NH3, and Cu(NO3)4^2+ at equilibrium, we need to consider the reaction that takes place between Cu(NO3)2 and NH3 to form Cu(NH3)4^2+.
The balanced chemical equation for the reaction is:
Cu(NO3)2 + 4 NH3 -> Cu(NH3)4^2+ + 2 NO3^-
Based on the equation, we can see that 1 mole of Cu(NO3)2 reacts with 4 moles of NH3 to produce 1 mole of Cu(NH3)4^2+.
To determine the moles of Cu(NO3)2, we need to convert the given mass (18.8 g) to moles. The molar mass of Cu(NO3)2 can be calculated as follows:
Molar mass of Cu = 63.55 g/mol
Molar mass of N = 14.01 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of Cu(NO3)2 = (63.55 g/mol) + 2 * [(14.01 g/mol) + (3 * 16.00 g/mol)]
= 63.55 g/mol + 2 * (14.01 g/mol + 48.00 g/mol)
= 63.55 g/mol + 2 * 62.01 g/mol
= 187.57 g/mol
Using the molar mass of Cu(NO3)2, we can calculate the number of moles:
Moles of Cu(NO3)2 = mass / molar mass
= 18.8 g / 187.57 g/mol
= 0.100 mol
Since the reaction goes to completion and 1 mole of Cu(NO3)2 reacts with 4 moles of NH3, we can conclude that the number of moles of Cu(NH3)4^2+ formed is also 0.100 mol.
The volume of the solution is given as 1.0 L. The concentration of Cu(NH3)4^2+ can be calculated by dividing the number of moles by the volume:
Concentration of Cu(NH3)4^2+ = moles / volume
= 0.100 mol / 1.0 L
= 0.100 M
Since 1 mole of Cu(NO3)2 produces 1 mole of Cu(NH3)4^2+, the concentration of Cu^2+ is also 0.100 M.
However, the concentration of NH3 is not given directly. We need to determine it based on the amount of NH3 consumed in the reaction. Since 1 mole of Cu(NO3)2 reacts with 4 moles of NH3, the moles of NH3 consumed will be:
Moles of NH3 consumed = 4 * moles of Cu(NO3)2
= 4 * 0.100 mol
= 0.400 mol
To find the concentration of NH3, we divide the moles consumed by the volume of the solution:
Concentration of NH3 = moles consumed / volume
= 0.400 mol / 1.0 L
= 0.400 M
Therefore, at equilibrium, the concentrations are as follows:
Concentration of Cu^2+ = 0.100 M
Concentration of NH3 = 0.400 M
Concentration of Cu(NO3)4^2+ = 0.100 M
To determine the concentrations of Cu^2+, NH3, and Cu(NO3)4^2+ at equilibrium, we need to use the stoichiometry of the reaction and the initial conditions provided.
The balanced equation for the reaction is:
Cu(NO3)2 + 4NH3 -> Cu(NH3)4^2+ + 2NO3^-
First, let's calculate the number of moles of Cu(NO3)2:
Molar mass of Cu(NO3)2 = atomic mass of Cu + (mass of N03 * 2)
Molar mass of Cu(NO3)2 = (63.55 g/mol) + ((14.01 g/mol + 16.00 g/mol * 3) * 2) = 187.55 g/mol
Number of moles of Cu(NO3)2 = mass / molar mass
Number of moles of Cu(NO3)2 = 18.8 g / 187.55 g/mol = 0.1 mol
Since the reaction goes to completion, all the Cu(NO3)2 will react to form Cu(NH3)4^2+. Therefore, the concentration of Cu(NH3)4^2+ at equilibrium will be equal to the total volume of the solution, which is 1.0 L.
Now, let's calculate the concentration of Cu^2+ at equilibrium. From the balanced equation, we know that 1 mole of Cu(NO3)2 forms 1 mole of Cu^2+ and 4 moles of NH3. Therefore, the concentration of Cu^2+ at equilibrium will be equal to the initial concentration of Cu(NO3)2, which is 0.400 M.
Finally, to calculate the concentration of NH3 at equilibrium, we need to consider that 1 mole of Cu(NO3)2 reacts with 4 moles of NH3. The initial concentration of NH3 is 0.400 M, and since the reaction goes to completion, the concentration of NH3 at equilibrium will be:
NH3 concentration = initial NH3 concentration - (4 * Cu(NO3)2 concentration)
NH3 concentration = 0.400 M - (4 * 0.400 M) = -0.600 M.
Note that the negative concentration of NH3 indicates that it has been consumed completely in the reaction and there is none left at equilibrium.
In summary, the concentrations at equilibrium are:
Cu^2+ = 0.400 M
NH3 = 0 M (completely consumed)
Cu(NH3)4^2+ = 1.0 L