Posted by Eliz on Monday, May 11, 2009 at 9:11pm.
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^-)?
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?
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.
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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