gammmagirl Posted March 30, 2018 Report Share Posted March 30, 2018 (edited) 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 LPbI2 (s) ==> PbI2(aq) ==>Pb2+ + 2I- x +.00622 2x (.00622)(2x)^2=1.4 x 10^-8 x = 7.5 x 10^-4 M7.5 x 10^-4 m/L x .5 L x 461.01 g/mole = .173 g Edited March 30, 2018 by gammmagirl Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.