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Homework Help: HELP solubility product determine solubility of lead 2 chloride mol/L and g/L

  1. Dec 19, 2005 #1
    use you accepted value for the solubility product of lead 2 chloride to determine the solubility of lead 2 chloride in mol/L and g/L

    the accepted value is ksp=1.2*10^-5

    Ksp=[Pb] [Cl]^2

    1.2*10^-5 = (x) (2x)^2

    x=0.014 mol/L

    0.014 mol/L * 276.1 g/mol
    =3.98 g/L

    I dont know if I did this right can someone please tell me ASAP
  2. jcsd
  3. Dec 20, 2005 #2
    What did you represent with x? [Pb2+], right?

    The solubility of PbCl2 is the concentration of Pb2+ at equilibrium (Cl- has a higher concentration); the molar mass of Pb2+ is 207g/mol.
    Last edited: Dec 20, 2005
  4. Dec 20, 2005 #3
    The result in the initial post seems accurate to me. The only problem I see is that solubility is traditionally -x (not just x), because it represents the amount of PbCl_2 *dissolved* per liter. Your instructor may or may not care about this technicality. Also, your value used for the molar mass was 276.2. Using a periodic table I found online I got a value of 278.14, and so I would have reported the answer as:

    - .014 mol/L * 278.14 g/mol = - 3.9 g/L

    I think your answer for the bottom problem had a typo in it either way because when you multiply .014 * 276.1 you get 3.86 and not 3.98 like you are reporting.

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