A beaker contain 300.ml of a 0,20 M Pb(NO3) solution. If 200 ml of 0,20M solution og MgCl2 is added to the beaker, What will be the concentration of Pb2+ ions in the resulting solution

Answer 1

I'm sure you meant Pb(NO3)2
Pb(NO3)2 + MgCl2 ==> PbCl2 + Mg(NO3)2

The key to this problem is knowing that PbCl2 is a ppt. First you must determine which is the limiting reagent.

mols Pb(NO3)2 = M x L = approx 0.06
mols MgCl2 = M x L = approx 0.04

If we used 0.06 mols Pb(NO3)2 and all of the MgCl2 we needed, now much PbCl2 would ppt? That's 0.06 x (1 mol PbCl2/1 mol Pb(NO3)2) = 0.06 x 1/1 = 0.06 mols PbCl2.
How much PbCl2 would ppt if we used 0.04 mols MgCl2 and all of the Pb(NO3)2 we needed. That's 0.04 x (1 mol PbCl2/1 mol MgCl2) = 0.04 x 1/1 = 0.04
So MgCl2 is the limiting reagent and that makes it an easier problem.
Thus 0.04 mols PbCl2 will be formed (the problem doesn't ask for that), you will have no MgCl2 left over, but you will have 0.06-0.04 - 0.02 mols Pb(NO3)2 left over. What will the concn be? That's M Pb^2+ = mols Pb^2+/L solution which should be 0.02 mols/0.5L = ?