Ksp

[gammmagirl]

Determine the mass in grams of lead (II) iodide that will dissolve in 500.0 mL of a solution containing 1.03 grams of lead (II) nitrate. Ksp of lead (II) iodide is 1.4 x 10-8.

1.03 g   x mol  =    .00622 mol
------      ------       -------------
0.5 L      331.2              L

PbI2 (s) ==> PbI2(aq) ==>Pb2+ +     2I-
                                      x +.00622     2x

                                  (.00622)(2x)^2=1.4 x 10^-8
                                     

x = 7.5 x 10^-4 M
7.5 x 10^-4 m/L x .5 L x 461.01 g/mole = .173 g

 

 

Edited March 30, 2018 by gammmagirl