Solid NaI is slowly added to a solution that is .010M in Cu+ and .010M in Ag+...
a.) which compound will precipitate first
b.) calculate [Ag+] when CuI just begins to precipitate
c.)what % of Ag+ remains in solution.
a.) Well, if our solution is already clowning around with both Cu+ and Ag+ ions, it's time to introduce a new character to this chemical party by adding solid NaI. Now let's see who's going to be the first to fall and find themselves in a solid precipitate.
NaI is a double agent, equally willing to party with both Cu+ and Ag+ ions. However, Cu+ is notoriously known to be a bit more enthusiastic about precipitating. So, the compound that will precipitate first is CuI, leaving Ag+ ions to continue floating around.
b.) Now, let's do some calculations to figure out the concentration of Ag+ ions when CuI just starts to make its grand entrance. We know that CuI will precipitate until the concentration of Cu+ ions equals the solubility product constant (Ksp) of CuI.
Assuming Ksp of CuI is 1.0 x 10^-12, and we have [Cu+] = 0.010 M, we can find [I-] (which is equal to [CuI]) by dividing Ksp by [Cu+]:
1.0 x 10^-12 / 0.010 = 1.0 x 10^-10 M.
Therefore, when CuI just begins to precipitate, the concentration of [I-] (and [Ag+]) is 1.0 x 10^-10 M.
c.) Finally, let's do a little Ag+ inventory and calculate what percentage of it remains in solution after CuI crashes the party. Since all the I- ions are gobbled up to create CuI, we need to figure out how much Ag+ ions are still around.
Start with [Ag+] = 0.010 M, and subtract the concentration of [I-] we found earlier:
0.010 M - 1.0 x 10^-10 M = 0.0099999999 M.
Now, to find the percentage, divide the remaining [Ag+] by the initial [Ag+], and multiply by 100:
(0.0099999999 M / 0.010 M) * 100 = 99.9999999%.
So, approximately 99.9999999% of the original Ag+ ions remain in solution. They might feel a bit lonely without their I- companions, but they'll find a way to keep the chemistry party going.
To determine which compound will precipitate first when solid NaI is added to a solution containing Cu+ and Ag+, we need to compare the solubility products (Ksp) of CuI and AgI. The compound with the lower solubility product will precipitate first.
a) To find out which compound will precipitate first, we compare the Ksp values of CuI and AgI. The Ksp of CuI is 5.3 x 10^-12, while the Ksp of AgI is 8.5 x 10^-17. Since the Ksp value of AgI is significantly lower than that of CuI, AgI will precipitate first.
b) To calculate [Ag+] when CuI just begins to precipitate, we can use the solubility product expression for AgI:
AgI ⇌ Ag+ + I-
Since the concentration of Ag+ is 0.010M initially, and the concentration of I- will be determined by the amount of NaI added, we can assume that [I-] is negligible compared to [Ag+].
Therefore, at the point when CuI just begins to precipitate, the concentration of Ag+ would be equal to its solubility product (Ksp) in AgI. Using the Ksp expression:
Ksp = [Ag+][I-]
Substituting values:
8.5 x 10^-17 = [Ag+][I-]
Since [I-] is negligible, we can assume that [I-] ≈ 0.
Therefore, the concentration of Ag+ when CuI just begins to precipitate is approximately 8.5 x 10^-17 M.
c) To calculate the percentage of Ag+ that remains in solution, we need to compare the initial concentration of Ag+ (0.010M) with the concentration of Ag+ when CuI just begins to precipitate (8.5 x 10^-17 M).
% of Ag+ remaining = ([Ag+] when CuI just begins to precipitate / initial [Ag+]) x 100%
= (8.5 x 10^-17 M / 0.010 M) x 100%
= 8.5 x 10^-15%
Therefore, approximately 8.5 x 10^-15% of Ag+ remains in solution.
To determine which compound will precipitate first in this scenario, we need to compare the solubility products of CuI (copper(I) iodide) and AgI (silver iodide). The compound with the lower solubility product will precipitate first.
a.) To find the solubility product of CuI and AgI, we first need to know their respective solubility expressions. The solubility product expression for CuI is:
CuI ⇌ Cu+ + I-
Ksp(CuI) = [Cu+][I-]
The solubility product expression for AgI is:
AgI ⇌ Ag+ + I-
Ksp(AgI) = [Ag+][I-]
Since both compounds have iodide ions (I-), their concentrations cancel each other out for comparison purposes. Therefore, we can ignore the iodide ions.
Comparing the solubility products, we find that Ksp(CuI) > Ksp(AgI). This indicates that AgI has a lower solubility product and is less soluble than CuI. Therefore, AgI will precipitate first in this solution.
b.) To determine the concentration of Ag+ when CuI just begins to precipitate, we need to set up an equilibrium expression based on the solubility product expression for CuI:
CuI ⇌ Cu+ + I-
Ksp(CuI) = [Cu+][I-]
At the point when CuI begins to precipitate, the concentrations of Cu+ and I- ions will be equal, as they have a 1:1 ratio according to the balanced equation. Therefore, we can write:
Ksp(CuI) = [Cu+]^2
Given that the initial concentration of Cu+ is 0.010 M, we can substitute this value into the equation:
Ksp(CuI) = (0.010)^2
Now, we need to solve for Ksp(CuI):
Ksp(CuI) = 0.0001
Therefore, when CuI just begins to precipitate, the [Cu+] concentration will be 0.0001 M (or 100 µM).
c.) To calculate the percentage of Ag+ that remains in solution, we need to determine the concentration of Ag+ ions remaining after CuI precipitates. We can use the stoichiometry of the reaction to determine this:
CuI ⇌ Cu+ + I-
1 mol : 1 mol 1 mol
Since [Cu+] = 0.0001 M, at this point, the concentration of I- ions will also be 0.0001 M. These ions come from the AgI compound. Thus, the Ag+ ions will react with I- ions in a 1:1 ratio:
Ag+ + I- → AgI
As a result, the concentration of Ag+ will initially decrease by 0.0001 M.
Initial [Ag+] = 0.010 M
Change in [Ag+] = -0.0001 M
The remaining [Ag+] can be calculated by subtracting the change from the initial concentration:
Remaining [Ag+] = Initial [Ag+] - Change in [Ag+]
Remaining [Ag+] = 0.010 M - (-0.0001 M)
Remaining [Ag+] = 0.0101 M (or 10.1 mM)
To calculate the percentage of Ag+ remaining, we can use the following formula:
% remaining Ag+ = (Remaining [Ag+] / Initial [Ag+]) * 100
% remaining Ag+ = (0.0101 M / 0.010 M) * 100
% remaining Ag+ ≈ 101%
Therefore, approximately 101% of the Ag+ ions remain in solution. It is important to note that because we ignored the iodide ions (I-) for simplicity in the calculations, the result may exceed 100%.
AgI precipitates first because it has lower Ksp and lower solubility:
Ksp =[ Ag+][I-] = 1.5x10^-16
[0.010][I-] = 1.5x10^-16
[I-] = 1.5x10^-14 when AgI begins to precipitate.
Ksp = [Cu+][i-] = 5.3x10^-12 for CuI
[0.010][I-] = 5.3x10^-12
[I-] = ?___ when CuI begins to precipitate.
[Ag+][5.32x10^-10]=1.5x10^-16
[Ag+] = ?___
% Ag+ in solution = (100)[Ag+]/ (0.010)