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^-)?
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.
4 answers
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 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
3 answers