## Calculations of Solubility Product of Lead 2 Chloride

HELP! In this experiment I measured the initial mass of a piece of Al and then i put it into a beaker containing 100.0 mL of saturated PbCl2

The concentration I calculated for Pb is $$6.12 *10^-^5$$ mol/L
The concentration I calculated for Cl is $$1.224* 10^-^4$$ mol/L

Therefore my Ksp= [Pb] [Cl]^2
$$=[6.12*10^-^5] [1.224*10^-^8]^2$$
Ksp=9.17
This is what I calculated please verify

I need help answering this question if 100.0 ml of 0.02 mol/L lead 2 nitrate solution and 100.0 ml of 0.02 mol/L sodium chloride solution were combined would a precipitate of Lead 2 chloride be expected to form? I have to show all my work. But im not sure where to start.

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 G'day, Aisha. The first part of your post is a little confusing. How did you calculate [Pb2+] and [Cl-]? If these concentrations are indeed in a saturated solution then Ksp(PbCl2) = 9.16x10^(-13), just a couple of typos there. At 25 degrees C, the Ksp of PbCl2 is about 1.6x10^(-5), so perhaps the room was cold? If the second part is connected to the first part, then you are probably to find the ionic product (given by [Pb2+][Cl-]^2) and compare it to the Ksp value obtained earlier. Hint: for the same amount, doubling the volume halves the concentration. If IP > Ksp, a precipitate will form. ~ ~ ~ ~ ~ ~ I'm new here but I think this belongs in the Science Education forum.