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
I'm sorry. I have looked at this and just can't figure out where we are going. I can't decipher the data.
To calculate the [I-] in moles/Liter at equilibrium for each of the five solutions, we need to determine the concentration of I- based on the given experimental data. The concentration of I- can be obtained using the Beer-Lambert Law, which relates the absorbance of a solution to its concentration.
The Beer-Lambert Law states that the absorbance of a solution is directly proportional to the concentration of the absorbing species and the path length of the light through the solution. It can be expressed as:
A = εcl
where A is the absorbance, ε is the molar absorptivity (a constant that depends on the absorbing species and the wavelength of light used), c is the concentration of the absorbing species, and l is the path length of the light through the solution.
In this case, we have the absorbance values for each solution and we want to find the concentration of I-. The molar absorptivity (ε) is assumed to be constant.
Let's calculate the concentration of I- for each solution using the absorbance values and the Beer-Lambert Law:
Test tube no. 1:
A = 0.300
ε (constant) = ?
c (I-) = ?
Test tube no. 2:
A = 0.330
ε (constant) = ?
c (I-) = ?
Test tube no. 3:
A = 0.412
ε (constant) = ?
c (I-) = ?
Test tube no. 4:
A = 0.405
ε (constant) = ?
c (I-) = ?
Test tube no. 5:
A = 0.333
ε (constant) = ?
c (I-) = ?
To calculate the concentration of I-, we need to determine the value of ε, which can be obtained from the literature or experimental data. Once we have the value of ε, we can rearrange the Beer-Lambert Law equation to solve for c (I-):
c (I-) = A / (ε * l)
Substituting the given absorbance values and the path length of the light through the solution (l = 1 cm), we can calculate the concentration of I- for each solution.
Note: Once we have the [I-] for each solution, we can use them to calculate the [Pb2+] using the stoichiometry of the reaction and the formula of lead iodide (PbI2), as mentioned in the initial explanation. The [Pb2+] can then be used to calculate the solubility product (Ksp) for PbI2 using the expression Ksp = [Pb2+] * [I-]^2.