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In class, I was taught that $\ce{NaOH}$ and $\ce{HCl}$ react completely (no equilibrium). I get the equation: $\ce{NaOH + HCl -> H2O + NaCl}$.

However, if I were to add a lot of $\ce{NaCl}$ to the solution, would some $\ce{NaOH}$ and $\ce{HCl}$ be formed according to Le Chatelier's principle?

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    $\begingroup$ RE: ... would some NaOH and HCl be formed according to le Chatelier's principle? // Not really. Your fundamental premise is wrong. NaOH, HCl, and NaCl are all essentially totally ionized in water. Each ion is surrounded by multiple layers of water molecules with water's dipole oriented towards the ion. $\endgroup$ – MaxW Mar 4 '18 at 22:59
  • $\begingroup$ MaxW is correct.However the above reaction is not an equilibrium reaction. $\endgroup$ – Chakravarthy Kalyan Mar 5 '18 at 8:16
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You yourself said that there is "no equilibrium", but then forgot that Le Chatelier's principle only applies to any system at equilibrium! Recall its definition:

Le Chatelier's principle: a principle stating that if a constraint (such as a change in pressure, temperature, or concentration of a reactant) is applied to a system in equilibrium, the equilibrium will shift so as to tend to counteract the effect of the constraint.

In your case, as also MaxW pointed out above, $\ce{NaOH}$, $\ce{HCl}$ and $\ce{NaCl}$ are all completely ionized in water (and their ions are hydrated/solvated by water molecules). There is no such thing as an "equilibrium" in this case, as the reaction, being an ionic reaction, completely proceeds in the forward direction.

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