Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: PistolSlap on April 20, 2012, 01:21:44 AM
-
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. :)
-
But I have a problem that the Qsp formula uses M not mols
You should calculate concentrations of Fe2+ and OH- first.
-
But I have a problem that the Qsp formula uses M not mols
You should calculate concentrations of Fe2+ and OH- first.
Can you give me a hand with this?
-
How you calculate a molar concentration after dilution?
-
How you calculate a molar concentration after dilution?
Do you need to start your own thread?
-
AWK tries to help you. To solve the question you need to calculate concentrations of ions after the solutions were mixed, that's equivalent to calculating concentrations after dilution.
So, do you know how to calculate concentration of ions after the dilution?