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Homework Help: Calculations of Solubility Product of Lead 2 Chloride

  1. Dec 18, 2005 #1
    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 [tex] 6.12 *10^-^5 [/tex] mol/L
    The concentration I calculated for Cl is [tex] 1.224* 10^-^4 [/tex] mol/L

    Therefore my Ksp= [Pb] [Cl]^2
    [tex] =[6.12*10^-^5] [1.224*10^-^8]^2 [/tex]
    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.
    Last edited: Dec 18, 2005
  2. jcsd
  3. Dec 18, 2005 #2
    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.
    Last edited: Dec 18, 2005
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