Determination of the solubility product of PbI2. From the experimental data we obtain [I-] directly. To obtain Ksp for PbI2, we must calculate [Pb2+] in each equilibrium system. This is most easily done by constructing an equilibrium table. We first find the initial amount of I- and Pb2+ ion in each system from the way the mixture were made up. Knowing I- and the formula of lead iodide allows us to calculate [Pb2+], Ksp then follows directly.
PbI2 gives Pb2+(aq) + 2I- (aq)
Ksp = [Pb2+] [I-]^2
Data:
Test tube no. 1 2 3 4 5
mL 0.12 M Pb(NO3)2 5 5 5 5 saturated soln of PbI2
mL 0.03 M KI 2 3 4 5
mL 0.2 M KNO3 3 2 1 0
total volume in mL 10 10 10 10
absorbance of solution 0.300 0.330 0.412 0.405 0.333
[I-] in moles/Liter at equilibrium?
____
(Calculate for each of the five solutions)
Calculations
for each of the five solutions
initial no. of Pb2+?
___X 10^-5
initial no. of I-?
___X 10^-5
(no. moles I- at equilibrium)?
___X 10^-5
(no. moles I- precipitated)?
___X 10^-5
(no. moles Pb2+ precipitated)
___X 10^-5
(no. moles Pb2+
at equilibrium)
___X 10^-5
[Pb2+] at equilibrium)
____
Ksp PbI2
____
Calculations
For each of the five solutions:
1. Calculating initial numbers of Pb2+ and I- ions before equilibrium:
Solution 1:
Initial moles of Pb2+ = (5 mL)(0.12 M) = 0.6 mM = 6 x 10^-4
Initial moles of I- = (2 mL)(0.03 M) = 0.06 mM = 6 x 10^-5
Solution 2:
Initial moles of Pb2+ = (5 mL)(0.12 M) = 0.6 mM = 6 x 10^-4
Initial moles of I- = (3 mL)(0.03 M) = 0.09 mM = 9 x 10^-5
Solution 3:
Initial moles of Pb2+ = (5 mL)(0.12 M) = 0.6 mM = 6 x 10^-4
Initial moles of I- = (4 mL)(0.03 M) = 0.12 mM = 12 x 10^-5
Solution 4:
Initial moles of Pb2+ = (5 mL)(0.12 M) = 0.6 mM = 6 x 10^-4
Initial moles of I- = (5 mL)(0.03 M) = 0.15 mM = 15 x 10^-5
Solution 5:
Initial moles of Pb2+ and I- cannot be calculated directly because the mixture was made from a saturated solution of PbI2. However, the numbers can be calculated indirectly from the absorbance, after finding [I-] at equilibrium.
2. Finding [I-] at equilibrium can be done using the absorbance, which is a measure of the equilibrium concentration. To obtain [I-] at equilibrium, you need the relationship between absorbance and concentration. This is beyond the scope of the information provided, so we will assume you have those values or can obtain them experimentally.
3. Using absorbance values to obtain [I-] at equilibrium:
(Assuming you have the values using Beer-Lambert Law or standard curve)
4. Calculating the moles of I- precipitated and moles of Pb2+ precipitated:
For each solution, subtract the moles of I- at equilibrium from the initial moles of I-. Divide the moles of I- precipitated by 2 to obtain the moles of Pb2+ precipitated, as explained by the stoichiometry of the reaction.
5. Calculating the moles of Pb2+ at equilibrium:
For each solution, subtract the moles of Pb2+ precipitated from the initial moles of Pb2+. Divide this value by the total volume (10 mL) to obtain [Pb2+] at equilibrium.
6. Calculating Ksp:
Using the formula Ksp = [Pb2+][I-]^2, substitute the values of [Pb2+] and [I-] at equilibrium and calculate Ksp for each solution. Find an average Ksp value considering all the five solutions.
To calculate [I-] in moles/Liter at equilibrium for each of the five solutions, we need to use the given data and the stoichiometry of the reaction:
Test tube no. 1:
[I-] = 0.03 M / (10 mL + 5 mL) = 0.002 M
Test tube no. 2:
[I-] = 0.03 M / (10 mL + 5 mL) = 0.002 M
Test tube no. 3:
[I-] = 0.03 M / (10 mL + 5 mL) = 0.002 M
Test tube no. 4:
[I-] = 0.03 M / (10 mL + 5 mL) = 0.002 M
Test tube no. 5:
Since it is a saturated solution of PbI2, all the iodide ions in the KI solution react to form PbI2. Therefore, [I-] in this case is 0 M.
Now let's move on to the calculations for each solution:
Test tube no. 1:
Initial no. of Pb2+ = (0.12 M) × (5 mL / 10 mL) = 0.06 M
Test tube no. 2:
Initial no. of Pb2+ = (0.12 M) × (5 mL / 10 mL) = 0.06 M
Test tube no. 3:
Initial no. of Pb2+ = (0.12 M) × (5 mL / 10 mL) = 0.06 M
Test tube no. 4:
Initial no. of Pb2+ = (0.12 M) × (5 mL / 10 mL) = 0.06 M
Test tube no. 5:
Since it is a saturated solution of PbI2, the initial no. of Pb2+ is negligible (assumed to be 0 M).
Now, let's calculate the moles of I- at equilibrium:
(no. moles I- at equilibrium) = [I-] × (10 mL + 5 mL) × 0.001 = [I-] × 0.015
(no. moles I- precipitated) = 0.002 M × (5 mL / 10 mL) × 0.001 = 0.001 M × 0.001 = 1 × 10^-6 M
(no. moles Pb2+ precipitated) = 1 × 10^-6 M / 2 = 0.5 × 10^-6 M
(no. moles Pb2+ at equilibrium) = initial no. of Pb2+ - moles Pb2+ precipitated
= (0.06 M - 0.5 × 10^-6 M)
[Pb2+] at equilibrium = (no. moles Pb2+ at equilibrium) / (total volume in Liters) = (0.06 M - 0.5 × 10^-6 M) / 0.015 = ___
Ksp PbI2 = [Pb2+] × ([I-])^2 = ___ × (0.002 M)^2 = ___
To determine the solubility product (Ksp) of PbI2, we need to calculate the concentration of Pb2+ ([Pb2+]) in each equilibrium system given the experimental data.
First, let's calculate the initial amount of I- and Pb2+ ions in each system. The initial amount of I- can be calculated by multiplying the volume of 0.03 M KI solution (in mL) by its concentration (in moles/Liter) and dividing by 1000 to convert mL to Liters.
Initial amount of I- = (Volume of 0.03 M KI) * (0.03 moles/Liter) / 1000
For each system:
- Test tube no. 1: Initial amount of I- = (2 mL) * (0.03 moles/Liter) / 1000
- Test tube no. 2: Initial amount of I- = (3 mL) * (0.03 moles/Liter) / 1000
- Test tube no. 3: Initial amount of I- = (4 mL) * (0.03 moles/Liter) / 1000
- Test tube no. 4: Initial amount of I- = (5 mL) * (0.03 moles/Liter) / 1000
- Test tube no. 5: Initial amount of I- = 0 (as it is the saturated solution of PbI2)
Next, we need to calculate the moles of I- at equilibrium. This can be done by multiplying the initial amount of I- by the absorbance of the solution and dividing by the absorbance of the saturated solution. The absorbance of the saturated solution can be obtained by measuring it experimentally.
Moles of I- at equilibrium = (Initial amount of I-) * (Absorbance of the solution) / (Absorbance of the saturated solution)
For each system, calculate the moles of I- at equilibrium using the corresponding initial amount and absorbance value.
Next, we can calculate the moles of I- precipitated by subtracting the moles of I- at equilibrium from the initial amount of I-.
Moles of I- precipitated = Initial amount of I- - Moles of I- at equilibrium
Similarly, we can calculate the moles of Pb2+ precipitated by multiplying the moles of I- precipitated by the stoichiometric ratio of Pb2+ to I- in the balanced equation.
Moles of Pb2+ precipitated = 2 * (Moles of I- precipitated)
Now, we can calculate the moles of Pb2+ at equilibrium by subtracting the moles of Pb2+ precipitated from the initial amount of Pb2+.
Moles of Pb2+ at equilibrium = Initial amount of Pb2+ - Moles of Pb2+ precipitated
Finally, to calculate the concentration of Pb2+ at equilibrium ([Pb2+]), divide the moles of Pb2+ at equilibrium by the total volume of the solution in Liters.
[Pb2+] at equilibrium = (Moles of Pb2+ at equilibrium) / (Total volume in mL / 1000)
Repeat these calculations for each system and record the values of [Pb2+] at equilibrium.
Once you have the [Pb2+] at equilibrium for each system, you can directly calculate the solubility product (Ksp) of PbI2 by multiplying the [Pb2+] by the square of [I-] (which is the initial amount of I- in moles/Liter).
Ksp = [Pb2+] * [I-]^2
Therefore, to answer the specific question, you need to perform these calculations for each of the five solutions using the given data and formulas provided in the question.