Ksp question - Nobel prize fir the person who solves it?

Question

Ksp for Pb(OH)2=8*10^-16. In presence of excess hudroxide, the formation of a soluble hydroxo complex of lead(II) may be significant. The equilibrium reaction is
Pb+2(aq) + 3OH- (aq) <=> Pb(OH)3 (superscript -2)

Consider a saturated solution of lead(II) hudroxide at pH 12.0. Calculate the molar concentration of free lead(II) ion, Pb+2

Answer 1

We can't solve it without the K for the complexation equilibrium.
By the way Pb(OH)3 should have charge -1. Either your charge is wrong or the stoichiometry of the reaction.
Please clarify these two points.
Just to give you a guideline

Your solution is saturated (but I assume there is no solid that can come to solution).
From the pH, you find [OH-] and then using the Ksp you determine the value of the Pb+2 "initial" concentration (let's call it C). Then

.. .. .. .. .. .. Pb(+2) + 3OH(-) <=> Pb(OH)3(-)
Initial .. .. .. .. C
React .. .. .. ..x
Produce .. .. .. .. . .. .. .. .. .. .. .. .. .. .. x
At equil. .. .. C-x.. .. .. 0.01 .. .. .. .. .. x

K=x/(((0.01)^-3)(C-x)) and solve for x
Then [Pb+2]=C-x
Note that in this case the final [Pb+2][OH-]^2< Ksp since some of the PB+2 will be quenched in the complex and thus the solution will not be saturated any more.

If you assume there is solid and you want to keep the solution saturated, then [Pb+2][OH-]^2 = Ksp and you don't need to bother with the other equilibrium ([OH-]=0.01 according to your formulation is the equilibrium concentration, so you can get from this equation the [Pb+2] regardless of the number of other equilibrium reactions the two ions are also involved).

Answer 2

I am not quite sure what you mean to say here. But perhaps in the equilibrium you meant to give the equation for Pb(OH)2. I am solving for that case.

Since for water, [H+][OH-]=10 ^ -14
and pH=12 is given (which means [H+]=10 ^ -12 ), we obtain
[OH-]=10 ^ -2

From the Ksp equation,
[Pb2+][OH-][OH-]=8 * 10 ^ -16

Use the value of [OH-] obtained above and solve for [Pb2+]

Notes:
1. If this was your homework, I am sad that you have asked someone else to do it.
2. If you 're an Indian MBBS aspirant, I can understand why you posted it here.