I've a solution 0.002M of Pb(NO3)2:
What is the minimum concentration of iodide ions (I-) necessary to observe the precipitation of PbI2?
What mass of KI should be added to observe the precipitation of PbI2 from the 25 mL volume lead nitrate solution?
................Pb(NO3)2 ==> Pb^2+ + 2NO3^-
I...............0.002..................0............0
C............-0.002.............0.002.......0.002*2
E.................0.................0.002.......0.004
..........PbI2 ==> Pb^2+ + 2I^-
Ksp = (Pb^2+)(I^-)^2
Look up Ksp and substitute.
You know (Pb^2+) from the Pb(NO3)2 equilibrium.
Solve for (I^-) in mols//L. Convert to mols/25 mL. Convert to grams KI/25 mL.
Post your work if you get stuck.
To determine the minimum concentration of iodide ions (I-) necessary to observe the precipitation of PbI2, we need to consider the solubility product constant (Ksp) of PbI2. In this case, PbI2 is the solid that will form when precipitation occurs, and its Ksp expression is as follows:
Ksp = [Pb2+][I-]2
Since we only have the concentration of Pb(NO3)2 and not the concentration of iodide ions (I-), we need to find a way to obtain the concentration of I-.
Assuming the dissociation of Pb(NO3)2 is complete, we can calculate the concentration of Pb2+ ions from the provided Pb(NO3)2 concentration. Pb(NO3)2 dissociates into Pb2+ and 2NO3- ions, so the concentration of Pb2+ will be twice the concentration of Pb(NO3)2:
[Pb2+] = 2 * 0.002M = 0.004M
Now we need to determine the minimum concentration of iodide ions (I-) needed to reach the Ksp value and initiate precipitation. By inspecting the Ksp expression, we can see that the concentration of iodide ions (I-) must be equal to the square root of the Ksp value divided by the concentration of Pb2+:
[I-] = sqrt(Ksp / [Pb2+])
To solve this, we need the value of Ksp for PbI2. The Ksp value for PbI2 is approximately 7.9 x 10^-9.
[I-] = sqrt(7.9 x 10^-9 / 0.004) = 4.99 x 10^-3 M
Therefore, the minimum concentration of iodide ions (I-) necessary to observe the precipitation of PbI2 is approximately 4.99 x 10^-3 M.
Now, let's calculate the mass of KI that should be added to observe the precipitation of PbI2 from a 25 mL volume of the lead nitrate solution.
First, we need to convert the volume of the solution from mL to liters:
Volume (L) = 25 mL / 1000 = 0.025 L
To calculate the number of moles of Pb(NO3)2 present in the solution, we multiply the concentration (in M) by the volume (in L):
Number of moles of Pb(NO3)2 = 0.002 M * 0.025 L = 5 x 10^-5 moles
Since Pb(NO3)2 dissociates into Pb2+ and 2NO3- ions, we have 5 x 10^-5 moles of Pb2+ ions.
Since the ratio of lead ions (Pb2+) to iodide ions (I-) in PbI2 is 1:2, we require twice the moles of iodide ions:
Number of moles of I- needed = 2 * 5 x 10^-5 moles = 1 x 10^-4 moles
Now, let's find the molar mass of KI, which is potassium iodide:
Molar mass of KI = molar mass of K + molar mass of I = 39.1 g/mol + 126.9 g/mol = 166 g/mol
Finally, we can calculate the mass of KI needed using the number of moles of I-:
Mass of KI = number of moles of I- * molar mass of KI = 1 x 10^-4 moles * 166 g/mol = 0.0166 g
Therefore, to observe the precipitation of PbI2 from the 25 mL volume of lead nitrate solution, you would need to add approximately 0.0166 grams of KI.