Given the reaction $\ce{NaCl~(s) <=> Na+~(aq) + Cl^{-}~(aq)}$. $\ce{NaCl}$ is added to water and equilibrium is reached.
Does the reaction shift to the left/right/not shift at all if more water is then added?
It seems to me that adding water would decrease the concentrations of $\ce{Na+}$ and $\ce{Cl-}$, and therefore the reaction would shift to the right. Is this correct?
The reaction $$\ce{NaCl~(s) <=>[\ce{H2O}] Na+~(aq) + Cl^{-}~(aq)}$$ is at equilibrium when approximately $359~\mathrm{g/L}$ sodium chloride are dissolved in water, i.e. the solution is saturated, or $c(\ce{Na+}) = c(\ce{Cl-}) = c_\mathrm{max}$.
If less $\ce{NaCl}$ is dissolved, it is only one phase, which is in equilibrium with itself and all solid salt is dissolved, therefore the equilibrium would be entirely on the right side. At that point adding water would not change the equilibrium at all, it would only dilute the whole system.
When the solution is saturated, there is still solid salt present, the equilibrium is not entirely on the right side. Adding water will disturb the equilibrium, i.e. $c_\mathrm{present}<c_\mathrm{max}$. Therefore more of the excess $\ce{NaCl}$ will dissolve, until the maximum concentration is reached again. If the solution is still saturated after that, nothing will have changed in your solution except for the volume.
If your solution is not saturated anymore, the equilibrium is entirely on the right side and the concentrations will decrease.
I think I may understand your question.
Adding water to $\ce{NaCl~(s)}$ causes it to ionize into $\ce{Na+}$ and $\ce{Cl-}$, now if you think about Le Chatelier's Principle, considering the equation: $\ce{NaCl~(s) + H2O -> Na+ + Cl-}$, the equilibrium will shift to the right, and the concentrations of $\ce{Na+}$ and $\ce{Cl-}$ will increase.
Now if you think about it realistically, since the salt is ionized with water, the more water you add, the more $\ce{Na+}$ and $\ce{Cl-}$ you will get, now if this was considered, it is essentially increasing the concentration of the ions.
However, usually when dealing with water, the system is considered to be in water. So you may see water on both sides of the reaction, in which case adding water would have no effect on equilibrium.
