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
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

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  1. I'm sorry. I have looked at this and just can't figure out where we are going. I can't decipher the data.

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