[Lafleche]

Hi there, I'm very confused about one of the example we have to do in our assignment. We're ask to give the solubility of Ni(OH)2 in water, no problem there, and then in pH=7.5.

Now in High school, this would be very simple, find [OH-] and plug that in Kps=[Ni2+][OH-]^2.  Problem is, this gives you a solubility is 0.002M, which means the concentration of [OH-] is 0.004M which is >> then 3.16E-7M (from the pOH).  This indicates that the approximation can't be made because here we made the assumption that [OH-] from the solution pH=7.5 >> [OH-] from Ni(OH)2. 

Now the only thing I can think of is doing

Kps=[Ni2+](2[Ni2+]+3.16E-7)^2 but then we have a equation to the power 3 which is very hard to solve.  Any suggestions on how to make a different approximation that would be justified?

[AWK]

http://did.mat.uni-bayreuth.de/~wn/BLK/Poly/poly.html
put zero for quartic term

[Borek]

If pH is 7.5 there are some other substances in the solution - they can be everything (including a buffer). If there is a buffer present, there is no way to change the pH, so [OH^-] is constant. I would say there is not enough information to solve the problem without making some quite strong assumptions.

Difference between concentration of OH^- in pH 7.0 (pure water) and pH 7.5 is so small I don't see why the first question is "no problem" and the other is "problem". In both cases concentration of OH^- from Ni(OH)₂ is potentially changing by orders of magnitude.

[Lafleche]

@AWK I can't see the link?

@Borek, I agree, there's not enough info, I said buffer but I didn't mean too, all the question says is that it's a pH=7.5.  I believe that we can still just ignore the presence of OH like in water since it doesn't significantly change.  Even for water, because the constant for kps of Ni(OH)2 is so close to water we shouldn't ignore the dissolution of water, this is why its a multiple equilibrium, we can't assume the OH may be neglected.  I'll probably go see my prof or something for more explanations on it.