Posted by matthieu on Sunday, December 4, 2011 at 7:47pm.
You are working with (H^+)(OH^-) = Kw = 1E-14. This is the ion product of water which is a constant of 1E-14. So Pb(OH)2 will be the one that is pH depended BECAUSE that is the one with OH^- in it. As you change the pH, you are changing the H^+ (of course) so the OH must change, too. I think it is simpler to look at it as a Le Chatelier's Principle problem.
Pb(OH)2(s) ==> Pb^2+ + 2OH^-
Now, what happens as I increase H^+. That reacts with OH to form water and shifts the reaction to the right which has the effect of increasing the solubility of Pb(OH)2.
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