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-
                                      .00622     2x^2
                                  (.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, 20187 yr by gammagirl
correction

Why is [Pb²⁺] defined as 0.00622 M at equilibrium? Are you familiar with how to use ICE tables?

yes

should if be (x +.00622)(4x^2)? Idk how the solution to that math.

22 minutes ago, gammagirl said:

yes

should if be (x +.00622)(4x^2)? Idk how the solution to that math.

 

But it isn’t that. How would it make sense to have a higher [Pb²⁺] than the initial concentration of PbI₂? Go through the ICE table again.