All you have to do is find the concentration of OH- and solve for pH (pOH= -log [OH-] nad then 14-pOH=ph
[OH-} = Ksp/[M+}
Suppose that two hydroxides, MOH and M'(OH)2, both have Ksp = 1.0 10-12 and that initially both cations are present in a solution at concentrations of 0.0036 mol · L-1. Which hydroxide precipitates first?
I know MOH ppt's first...but am not sure how to approach this second part of the problem.
At what pH does the first hydroxide precipitate, when solid NaOH is added?
2 answers
I agree with Trixie. I think the pH is about 4 (note that the ppt is forming WHEN the pH is ACID).