David L. answered • 03/06/14

Chemistry Tutor

If you set up the solubility equation for the lead(II) chloride, you'll have the aqueous Pb

^{2+}and Cl^{–}ions on the right side and the solid PbCl_{2}on the left.PbCl

_{2}(s) → Pb^{2+}(aq) + 2 Cl^{–}(aq)The ions will dissolve until their concentrations, when calculated in the K

_{sp}expression are equal to the K_{sp}value. You'll have to look up the K_{sp}value in your textbook or online, but the expression isKsp = [Pb

^{2+}][Cl^{–}]^{2}You are given [Pb

^{2+}], 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 K

_{sp}. If Q is smaller than Ksp, that means the ions aren't concentrated enough to equal K_{sp}and they will not precipitate. If Q is larger than K_{sp}, then there are too many ions in solution and the solid will precipitate out until Q = K_{sp}.I'll leave the details to you, but that's the basic setup. If you want more details, just ask.

Good luck!