# 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

____