A piece of Cu wire is placed into 1.00L of a saturated solution of silver acetate, AgCH3COO. When all the Ag+ has reacted, 2.00g of Cu has been used.
A) Write the net ionic equation for this reaction between Cu and Ag+.
B) Calculate the Ksp of AgCH3COO.
Thank you!
1. Cu + 2Ag^+ ==> 2Ag + Cu^2+
2. mols Cu = grams/molar mass = approx 0.03 BUT that's an estimate and you should do it more accurately.
mols Ag = twice mols Cu = approx 0.06
........CH3COOAg ==> CH3COO^- + Ag^+
I.........solid.......0..........0
C.........solid.......x..........x
E.........solic.......x..........x
From above (Ag^+) = 0.06 and that's x. Substitute into the Ksp expression and solve for Ksp.
Ksp = (Ag^+)(CH3COO^-)
Ksp = ?
A) To write the net ionic equation for this reaction, we first need to identify the reactants and products involved. In this case, the reactants are Cu (copper) and Ag+ (silver ion), and the product formed is Ag (silver).
The balanced chemical equation for the reaction between Cu and Ag+ can be expressed as follows:
Cu + 2AgCH3COO -> Cu(CH3COO)2 + 2Ag
The net ionic equation removes the spectator ions that do not participate in the reaction. In this case, the spectator ions are CH3COO- (acetate) and Cu2+ ions. Therefore, the net ionic equation is:
Cu + 2Ag+ -> Cu2+ + 2Ag
B) To calculate the solubility product constant (Ksp) of AgCH3COO, we need to know the concentrations of the reactant ions and the product ions at equilibrium.
From the given information, we know that 2.00g of Cu has reacted. To convert this mass to moles, we use the molar mass of Cu. The molar mass of Cu is approximately 63.55 g/mol.
Number of moles of Cu = mass / molar mass
Number of moles of Cu = 2.00g / 63.55 g/mol ≈ 0.0315 mol
Since Cu has a 1:1 stoichiometric ratio with Ag+ in the reaction, the number of moles of Ag+ ions consumed is also 0.0315 mol.
Now, we need to determine the initial concentration of Ag+ ions in the saturated solution. Since the solution is saturated, it means that it is in equilibrium with solid AgCH3COO. Therefore, the concentration of Ag+ ions is equal to the solubility of AgCH3COO.
Let's assume the solubility of AgCH3COO is x mol/L. Therefore, the concentrations of Ag+ ions and CH3COO- ions in the saturated solution are both x mol/L.
Using the stoichiometry of the balanced equation, we can say that 2x mol/L of solid AgCH3COO are formed for every 1 mole of Cu consumed.
Now, we can construct the solubility equilibrium expression for AgCH3COO:
Ksp = [Ag+]^2[CH3COO-]^2
Since the concentrations of Ag+ and CH3COO- are 2x and x mol/L respectively:
Ksp = (2x)^2(x)^2 = 4x^4
Finally, substituting the value of Ag+ concentration (2x) with the known value of consumed moles of Cu (0.0315 mol):
Ksp = 4(0.0315)^4 ≈ 1.29 x 10^-7
Therefore, the solubility product constant (Ksp) of AgCH3COO is approximately 1.29 x 10^-7.