The ksp of PbI2 can be calculated from the data below (similar to Tube #2): 5.00

Question

The ksp of PbI2 can be calculated from the data below (similar to Tube #2): 5.00 mL of 0.0120 M Pb(NO2)2 are mixed with 3.00 mL 0.0300 M KL And diluted to a final volume of 10.00 mL with KNO3 solvent. A yellow precipitate of PbI2 forms. The precipitate is separated from the supernatant solution which is analyzed for[T]. The concentration of T in the test solution is calculated from the graph to be 1.82Times10-3 M. Calculate the following: Initial [Pb+2] (after mixing but before precipitating) Initial [T] (after mixing but before precipitating) Equilibrium [T] = (graph [T] Times 10.0mL/3.00 mL) [T] reacted to form precipitate [pb+2] reacted to form precipitate Equilibrium [Pb+2] Ksp of PbI2

Explanation / Answer

a) Total volume after mixing = 5.00 + 3.00 = 8.00 mL

Therefore initial [Pb2+] = (5.00 x 0.120)/ 8.00 = 0.0075 M = 7.5 x 10-3 M

b) Initial [I-] = (3 x 0.0300)/ 8.00 = 0.01125 M = 11.25 x 10-3 M

c) Equilibrium [I-] = 1.82 x 10-3 x(10/3) = 6.0667 x 10-3 M

d) [I-] reacted to form precipitate = 0.0300 M - 6.0667 x 10-3 M = 0.0239 M

e) Number of mole of I- reacted = 0.0239 M x (3 x 10-3 L) = 0.0717 x 10-3 mol

therefore number of moles of Pb2+ reacted to form PbI2 = 0.0717 x 10-3 / 2 =0.03585 x 10-3 mol

therefore [Pb2+] reacted = 0.03585 x 10-3 mol / 5.00 x 10-3 L = 0.00717 M

f) Equilibrium [Pb2+] = 0.0120 M - 0.00717 M = 0.00483 M

g) [Pb2+] in supernatant solution = Equilibrium [Pb2+] x 5 / 10 = 0.002415 M

therefor Ksp of PbI2 = {[Pb2+] in supernatant solution}x {[I-] in supernatant solution}2

= 0.002415 x (1.82 x 10-3)2 = 7.999 x 10-9

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