Chemistry
posted by Laura
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
____
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