# Is NaCl in water considered “solution” despite of NaCl's having some precipitation?

Is $\ce{NaCl}$ in water aqueous? For a certain amount of $\ce{NaCl}$, it can dissolve in water but if you add more $\ce{NaCl}$ it can't dissolve any more.

So, if in a certain reaction $\ce{NaCl}$ is formed with water:

$$\ce{HCl(aq) + NaOH(aq) -> H2O(l) + NaCl(aq)}$$

My question is does $\ce{NaCl}$ in the previous reaction totally dissolve in the water formed? Or in other words is this equation correct?

$$\ce{HCl(aq) + NaOH(aq) -> H2O(l) + NaCl(aq) + NaCl(s)}$$

Another question is why do we write (l) beside water rather than (aq) since it's part of the aqueous solution?

• I have improved the formatting of your post using $\LaTeX$. For more information on how to do this yourself please see here and here. – bon Jun 28 '15 at 16:09

When we are considering $\ce{HCl(aq)}$ and $\ce{NaOH(aq)}$. the system contain sufficient amount amount of solvent water molecules in situ already present and this is more than sufficient to hydrate the produced $\ce{NaCl}$ in the reaction. Hence the first equation
$$\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}$$
• I have improved the formatting of your post using $\LaTeX$. For more information on how to do this yourself please see here and here. – bon Jun 28 '15 at 16:11