[hopeful]

I am just learning of Le Chatelier's principle and was wondering if this reation that I thought would work.

NaCl(s)-->Na⁺(aq)+Cl^-(aq)

(Note: enough NaCl would be dissolved for the water to be completely saturated)

Now if I was to add HCl (which dissociates to H⁺Cl^-) to the right side would solid NaCl form in the bottom of the container (due to le chatelier's), or what would exactly happen seeing as once the HCl is added you now have a stress of Cl^- in the system?

Thank you

[Donaldson Tan]

NaCl(s)-->Na⁺(aq)+Cl^-(aq)

Your chemical equation isn't entirely valid.

In a saturated solution,
NaCl(s) <-> Na⁺(aq)+Cl^-(aq)

Now if I was to add HCl (which dissociates to H⁺Cl^-) to the right side would solid NaCl form in the bottom of the container (due to le chatelier's)

Yes, you are right.