If 75.0mL of a 0.0010M aqueous solution of Fe(NO3)2 was added to 100.0mL of a 0.00050M aqueous solution of NaOH would a precipitate form?
How would I do this?
I thought I would first draw out the molecular and net ionic equations:
Fe(NO3)2 + 2NaOH --> Fe(OH)2 + 2NaNO3
Fe2+ (aq) + 2OH- (aq) --> Fe(OH)2 (s)
Then I would use the stoiciometry to determine the moles of ions present --
0.00050M NaOH x 2OH-/1NaOH = 0.001mol OH-
and do the same for Fe2+, then plug them into my Qsp = [Fe2+][OH-]2 and see how it compares to the Ksp Fe(OH)2 = 4.87×10-17.
But I have a problem that the Qsp formula uses M not mols, and I also need to account for the various volumes of solution, which I haven't done, so I'm obviously really messing up somewhere.
Can someone please explain this in great detail to me what I do and why? I've got a final in 2 days and need to know this. Thanks.