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.
For which salt in each of the following groups will the solubility depend on pH?
Pb(OH)2
PbCl2
could you explain too?
thanks!
1 answer