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.
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
____
1 answer