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You have 95 mL of a solution that has a lead(ll) concentration of 0.0012 M. Will PbCl2 precipitate when 1.20 grams of solid NaCl is added?

This is a solubility problem and I am a little confused on how to set it up.
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1 Answer

If you set up the solubility equation for the lead(II) chloride, you'll have the aqueous Pb2+ and Cl ions on the right side and the solid PbCl2 on the left.  
PbCl2 (s) → Pb2+ (aq) + 2 Cl (aq)
The ions will dissolve until their concentrations, when calculated in the Ksp expression are equal to the Ksp value.  You'll have to look up the Ksp value in your textbook or online, but the expression is
Ksp = [Pb2+][Cl]2
You are given [Pb2+], and you can figure out the molarity of NaCl (aq) from the mass and the volume.  (Usually it's okay to assume that the solid NaCl doesn't add anything to the 95 mL.)  Because NaCl is a strong electrolyte, it will dissolve completely and [Cl] will be equal to your calculated molarity of NaCl.
You can figure out the reaction quotient, Q, which has the same mathematical expression as the Ksp.  If Q is smaller than Ksp, that means the ions aren't concentrated enough to equal Ksp and they will not precipitate.  If Q is larger than Ksp, then there are too many ions in solution and the solid will precipitate out until Q = Ksp.
I'll leave the details to you, but that's the basic setup.  If you want more details, just ask.  
Good luck!