gammagirl Posted March 30, 2018 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 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, 2018 by gammagirl correction Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted March 30, 2018 Share Posted March 30, 2018 Why is [Pb2+] defined as 0.00622 M at equilibrium? Are you familiar with how to use ICE tables? Link to comment Share on other sites More sharing options...
gammagirl Posted March 30, 2018 Author Share Posted March 30, 2018 yes should if be (x +.00622)(4x^2)? Idk how the solution to that math. Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted March 30, 2018 Share Posted March 30, 2018 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. Link to comment Share on other sites More sharing options...
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