7
$\begingroup$

The Wikipedia article about aqueous solutions claims that:

An aqueous solution is a solution in which the solvent is water

So basically, is writing $\ce{HCl(aq)}$ equal to: $\ce{HCl(l) + H2O(l)}$?

$\endgroup$
  • 3
    $\begingroup$ By definition An aqueous solution is a solution in which the solvent is water. So the Wikipedia article isn't "making a claim," rather it is a truth by definition. // Conversely an non-aqueous solution is any solvent but water. $\endgroup$ – MaxW Feb 26 '17 at 17:48
9
$\begingroup$

$\ce{HCl{(l)}}$ is not the correct notation in this case. Arrhenius acids and bases (chemicals that interchange $\ce{H^+}$ with water to form $\ce{H_3O^+}$ or $\ce{OH^-}$) break apart in water, forming ions. In your case, there is an equilibrium $\ce{HCl{(aq)} + H_2O{(l)}<=> H_3O^+{(aq)} + Cl^-{(aq)}}$ which very strongly lies to the right; virtually all of the $\ce{HCl{(aq)}}$ is broken apart in this way (which is why $\ce{HCl}$ is known as a strong acid).

Now, not everything dissolved in water has to dissociate into ions. For example, glucose $\ce{(C_6H_12O_6{(s)})}$ dissolves in water to form $\ce{C_6H_12O_6{(aq)}}$. Weak acids are acids that do not entirely dissociate into ions (for example, $\ce{CH_3COOH{(aq)} + H_2O{(l)} <=> CH_3COO^-{(aq)} + H_3O^+{(aq)}}$ still has some undissociated $\ce{CH_3COOH{(aq)}}$ in solution.)

A final point is that some dissolved species can, in fact, be solvents on their own. In this case, the definition of whether the species is aqueous or liquid is not well-defined and usually depends on the context. For example, a 1:4 mixture of $\ce{MeOH}$ and $\ce{H_2O}$ is typically written as $\ce{MeOH{(l)} + {H_2O{(l)}}}$, but a solution where a small amount of $\ce{MeOH}$ (maybe used as a reactant) would be denoted $\ce{MeOH{(aq)}}$. This has to do with the definition of the thermodynamic activities of the species and if you have questions about this, there's probably someone else on here who is a better expert than I to explain.

| improve this answer | |
$\endgroup$
5
$\begingroup$

The example you give is not quite right. $\ce{HCl}$ is a gas at standard temperature and pressure. It is also a strong acid in water, so a more correct way to express the dissolution of $\ce{HCl}$ would be:

$$\ce{HCl{(g)} + H2O{(l)} <=> H3O+{(aq)} + Cl^-{(aq)}}$$

A simpler example would be glucose, which readily dissolves in water without dissociating into different species:

$$\ce{C6H12O6{(s)} + H2O{(l)} <=> C6H12O6{(aq)} + H2O{(l)}}$$

When a compound dissociates like a salt for example, we write it like this:

$$\ce{NaCl{(s)} + H2O{(l)} <=> Na+{(aq)} + Cl^-{(aq)} + H2O{(l)}}$$

Because water is simply acting as the solvent, you could leave it out of the last two examples, and the fact that water is the solvent would be designated by the $\ce{(aq)}$ designation of the dissolved species.

| improve this answer | |
$\endgroup$
2
$\begingroup$

No HCl would dissolve in the water and for an aqueous solution to exist the solvent must be water. HCl(aq) + H2O(l) is correct. Never a liquid with a liquid without it dissociation of the HCl.

| improve this answer | |
$\endgroup$
2
$\begingroup$

No. $\ce{HCl}$ is a gas at STP (standard pressure and atmosphere). As such, you can store it in pressurized cylinders, but commonly it is stored as a concentrated aqueous solution $\ce{HCl(aq)}$. In the gas phase, $\ce{HCl}$ is composed of covalently bonded molecules, H bonded to Cl.

In water, the compound ionizes virtually completely to $\ce{Cl-}$ and $\ce{H+}$ (with the understanding that the latter is shorthand for the more complicated reality - the proton is essentially never found alone (except in outer space, or at extreme temperatures (eg. inside the Sun)).

$\ce{HCl(aq)}$ is equivalent to $\ce{Cl- + H+ + H2O}$ , where it is understood that both ions are hydrated (surrounded by oriented water molecules).

| improve this answer | |
$\endgroup$
  • $\begingroup$ I'd quibble that you should make the point that HCl(l) is the liquefied HCl gas. It has a boiling point of −85.05 °C at atmospheric pressure. $\endgroup$ – MaxW Feb 25 '17 at 23:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.