Molar Solubility given Ksp

Question

What is the molar solubility (in mol/L) of Pb(OH)₂ (Ksp = 2.54⋅10^−16) in a solution buffered at 11.59. 

I am getting confused with setting this question up, I think. Keep getting it marked incorrect.

Answer 1

Hi Rebecca A.,

You set up like any Ksp problem, with the coefficients for ions transformed into powers of the molarity. So:

Ksp = 2.54*10^-16 = [Pb2+]^1[OH-]^2 = [Pb(2+)]^1*[10^(14-11.59)]^2

Quick reasoning: p[OH-] = 14-11.59 because as pH->14, then p(OH)->0 ([OH] = 1 M) -- right?

So where did you go off track -- failing to get the exponents, or forgetting that buffering means that you should just USE that value for the [OH-], it's not generated by dissociation of Pb(OH)₂ ?

So what do I mean, "not generated by dissociation of the lead hydroxide"? Well, supposing IF you had just thrown lead hydroxide into water, then both the [Pb2+] and the [OH-] values would involve expressions of the same variable, which we usually just call "x". In fact, if [Pb2+] = x , then [OH-] would have to be 2x, since those were the stoichiometric relationship of ion numbers in the lead hydroxide formula. But, in this problem, 2x for the [OH-] has been swept into the given buffered [OH-] level, so it doesn't appear in the Ksp expression. If, on the other hand, you had just STARTED with an aqueous, unbuffered medium of pH = 11.59, then you WOULD have to add in the 2x to the [OH-] expression, at least until you eyeballed the expression and found that you could disregard it (pKsp/3 ~ 5.4 but p(OH-) ~ 2.4 so the hydroxide already present is much in excess of what the lead hydroxide would contribute!).

-- Cheers, --Mr. d.