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Chemistry

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Question: When 1.97 grams of Pb(OH)2(s) is added to 1.00 liter of pure water at 25 C a saturated solution (some solid is present) is formed that has a pH of 8.97. Compute the Ksp value for Pb(OH)2 using the data.

How do i go about getting the Molarity from the grams given and pure water. i know that you can get the molarity for OH by miniusing it by 14.0 than taking the inverse log, but how do i get the Pb molarity.

  • Chemistry - ,

    Pb(OH)2 ==> Pb^+2 + 2OH^-
    Ksp = (Pb^+2)(OH^-)^2.
    pOH = -log(OH^-).
    You are given the pH, then 14-pH = pOH, and solve for (OH^-). Note carefully: Won't the (Pb^+2) be just 1/2 the (OH^-)?

  • Chemistry - ,

    i get that part, but how do i calculate with 1.97 grams of Pb(OH)2 (S), because i need the Pb molarity to calculate for Ksp and that value gives me Pb(OH)2 not just the Pb?

  • Chemistry - ,

    The 1.97 grams is extraneous information. You don't need it (as long as the problem says the solution is saturated AND there is a small amount of solid in the container). The (Pb^+2) IS 1/2 the (OH^-); i.e., 1/2*(OH^-)=(Pb^+2)
    So you know (OH^-) and you know (Pb^+2);
    (Pb^+2)(OH^-)^2 = Ksp and Voila! you're done. The problem you are having is that the 1.97 is there to let you know a saturated solution is formed and that is it's only function. It certainly doesn't have anything to do with the solubility BECAUSE it is a saturated solution. Anyway, by definition [Pb(OH)2] = 1.0000000 since solid is the normal state of Pb(OH)2.

  • Chemistry - ,

    So it would look like this than. the OH concentration is 9.334E-6 and the Pb^+2 is 9.334E-6/2 = 4.667E-6. So the Ksp for this problem is 4.356e-11

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