smaerd - 16-10-2010 at 21:19
Okay so say I place KCl and Na(NO3) into a flask with pure distilled water.
The obvious reaction equation is: KCl + Na(NO3) -> K(NO3) + NaCl
After doing the ionic equation, each ion is considered a spectator ion. So there is no net-ionic equation. My lab book says that this implies that no
reaction has taken place.
This is kind of confusing for me. Do these ion's exchange to form the products and keep changing within themselves to also include the reactants? Kind
of like a chemical equilibrium equation(which we haven't covered in class yet)? Or do they just remain as they were but as ions in solution?
For my experimental data we placed 8.738grams of NaNO3 and 7.664grams of KCl into 25mLs of water. We were then asked to find the molarity of the
compounds(NaNO3, KCl, NaCl, and KNO3) in 25mL of water. My approach was to use the above equation(KCl + Na(NO3) -> K(NO3) + NaCl).
I found the reagent in excess(being KCl). Which would imply that the molarity for NaNO3 is 0Mols/Liter of solution because it would have been used
entirely in the reaction, where "no reaction has taken place".
I did the rest of the stoichiometry and math behind it all(and the rest of the lab) but I'm worried my approach is incorrect. If no reaction has taken
place wouldn't that imply there are no products present in the solution? Or is "No reaction" terminology for none of the products exist outside of
being ions(they don't form a gas, or a precipitate, etc)?
How should I go about looking at this? I am not asking for answers, please don't give me them, I want to learn. I am only asking how
I should approach this. Keep in mind that this is material my professor never covered and is not included in the book nor the lab-book.
Thank you in advance!
DDTea - 16-10-2010 at 22:01
You're correct: no reaction has happened except dissolution of the compounds and solvation of the ions. So the two relevant reactions here are
really:
KCl<sub>(s)</sub> --> K<sup>+</sup><sub>(aq)</sub> + Cl<sup>-</sup><sub>(aq)</sub>
NaNO<sub>3(s)</sub> --> Na<sup>+</sup><sub>(aq)</sub> +
NO<sub>3</sub><sup>-</sup><sub>(aq)</sub>
Both of these salts dissolve pretty much completely in water due to their extremely high solubility. I'm not sure how anal your assignment is and if
they'll really care about the miniscule differences in solubility of these ionic compounds. If they do, then you'll want to consider to what extent
all of these potential compounds will fall out of solution, thus affecting the concentration of the ions remaining in aqueous solution.
If the assignment is NOT looking for that level of detail, though, then to speak of the molarity of KNO<sub>3</sub>,
NaNO<sub>3</sub>, KCl, and NaCl is pretty nonsensical. It would make more sense to describe the molarity of the corresponding cations and
anions individually.
smaerd - 16-10-2010 at 22:33
That's what I was thinking too. If the compounds do not exist in solution, only their ions do, how can you calculate their molarity? It doesn't make
much sense.
Well the first part of the lab was just doing some basic ionic reactions and figuring out their equations, and taking notes about the reactions, etc.
Then the second part was calculating the molarity of these compounds in the solution. The third part is about crystallization, lowering the temp, what
crashed out, raising the temp, what crashed out, lowering it again, what crashed out, etc. So it's not really looking for any intense level of detail
concerning solubility, it's just Chem 1 
.
Thank you for taking your time and replying to this thread!
[Edited on 17-10-2010 by smaerd]
psychokinetic - 17-10-2010 at 00:01
I'm glad you seem to have figured this out. Perhaps mention to the question writer the confusability in the question?
I get it a LOT in biochemistry.