# What an aqueous solution really contains?

I want to know what an aqueous solution is. For example, let's consider a $1~\mathrm{M}$ aqueous solution of $\ce{MgCl2}$. Does it also contain $\ce{MgOH2}$? Though it is a simple question, I feel it is important to understand something as vital to chemistry as this.

• It's alright, everyone has trouble with this at first. :) Aqueous means that the specific substance has been dissolved in $\ce{H2O}$. – Asker123 Apr 28 '15 at 0:44
• Then aqueous solution of MgCl2 means MgCl2 is dissolved in water – On the way to success Apr 28 '15 at 0:46
• Not really a simple question at all, but a topic that can cause difficulty. Good question! – user15489 Apr 28 '15 at 2:40
• water as solvent --> aqueous solution – Freddy Apr 28 '15 at 16:15

Uh... I think the insight you need is about chemical equilibrium.

An aqueous solution is exactly the mixture formed when something dissolved in water. This is easy to understand, but here is a more difficult question: what's the component of this mixure?

To answer this is not easy.

In a glass of pure water, there are still a little $$\ce {H+}$$ (to say presicely, $$\ce {H3O+}$$) and $$\ce {OH-}$$. This is because sometimes proton ($$\ce {H+}$$) will "jump" from one $$\ce {H2O}$$ molecular to another one.

$$\ce {2H2O <=> H3O+ + OH-}$$.

(Since we usually do not use the expression $$\ce {H3O+}$$, this process can be rewrited as $$\ce {H2O <=> H+ + OH-}$$ to avoid disruption.)

This phenomenon (called the autoionization of the water) is very weak. The concentrations of $$\ce{H+}$$ and $$\ce{OH-}$$ in "pure water" is $$\ce{[H+]=[OH^{-}]=}1.0 \times 10^{-7} M$$ when the temperature is 25°C (298K).

Then you put something in this glass of water. As other people mentioned, most of covalent compounds, such as glucose, will not break apart in water since the intermolecular force (covalent bond) do not break. Water moleculars may surround those covalent compounds like a cage to make them dissolved. So there won't be other things.

But sometimes it breaks: when acid compounds are dissolved into water, like $$\ce {HCl}$$ and $$\ce {H2SO4}$$, they will give their hydrogen atoms as proton to water:

$$\ce {HCl + H2O -> H3O+ + Cl-}$$

$$\ce {H2SO4 + H2O -> H3O+ + HSO4-}$$

This is because they are more acidic than $$\ce {H3O+}$$, here "acid" means "who gives proton(s)", while "base" means "who accepts proton(s)" (This kind of definion is called as Brønsted-Lowry theory, and as you see, in autoionization of water, one of the $$\ce {H2O}$$ is acid and another is base).

Noticed that acid like $$\ce {HCl}$$ and $$\ce {H2SO4}$$ will nearly definitely give their $$\ce {H+}$$ to water compounds (they are too “willing” to do it), we call those acid as strong acid. But some acid are not: protons sometimes jump back.

For example, $$\ce{HSO4-}$$ will meet an equilibrium:

$$\ce{HSO4- + H2O <=> H3O+ + SO4^2-}$$

This kind of acid will be called as weak acid. In an aqueous solution of a weak acid, most of the weak acid compounds keep unchanged.

Yet we haven't mentioned what you concerned: how about aqueous solution of ionic compounds? They usually break their intermolecular forces (ionic bond) via hydration, i.e. combination with water compound. Water compound have some charges on them, so they will attract ions and break them apart. Especially, we will see bases we are familiar with -- those who will ionized as cations and $$\ce {OH-}$$ -- first.

In fact, the structure of hydrated cation will be complex ("$$\ce {H3O+}$$" is also not the actual situation), so we usually just write them as unhydrated form, such as $$\ce {Fe^2+}$$ or $$\ce {Na+}$$. As you know, there are weak bases and strong bases.

So bases like $$\ce {NaOH}$$ are quite "willing" to accept protons:

$$\ce {NaOH -> Na+ + OH-}$$

But $$\ce {Fe(OH)3}$$ may not, and it also meet an equilibrium:

$$\ce {Fe(OH)3 <=> Fe^3+ + 3OH-}$$

Again, in an aqueous solution of a weak acid, most of the weak base sturcture keep unchanged. And for this reason, weak bases are usually insoluable in water, since water "have no good idea" to deal with the whole relatively large ion sturcture. But in fact, the word "insoluable" means that there is another equilibrium like this:

$$\ce {Fe(OH)3_{(aq)} <=> Fe(OH)3_{(s)}}$$

and the solid form dominates.

So here we meet the hardest part: what about adding some salts into water, like $$\ce {MgCl2}$$ you mentioned?

So far you have learned about ionization of the salt, and you know most of salt will completely ionized. But how about those ions?

A balance like $$\ce {Mg(OH)2 <=> Mg^2+ + 2OH-}$$ shows that $$\ce {Mg(OH)2}$$ is a weak base. But if we see it as $$\ce {Mg^2+ + 2OH- <=> Mg(OH)2}$$ it means $$\ce {Mg^2+}$$ is a weak acid. And if you remember I have said in autoionization of water, the water compound can be both base and acid (we say this kind of substances are amphoteric). Then we can know that there will be a new equilibrium like this:

$$\ce {Mg^2+ + 2H2O <=> Mg(OH)2 + 2H+}$$

(We usually call this as hydrolyzation of salt.)

And because $$\ce {Mg^2+}$$ is a weak base, you may expect that there will be not a lot $$\ce {Mg(OH)2}$$ and it's still soluable. Also there will be not a lot $$\ce {H+}$$ so the solution is just a bit more acidic then water.

Addition: You can also think this comes from these two balances:

1. $$\ce{H2O <=> H+ + OH-}$$ (autoionization).
2. $$\ce{Mg(OH)2 <=> Mg^2+ + 2OH-}$$. in further calculation you may meet in future, this kind of idea can make you avoid some mistakes.

So you may expect in a $$1M$$ aqueous solution of $$\ce {MgCl2}$$, there will be a lot of $$\ce {Cl-}$$ ($$2M$$) and $$\ce {Mg^2+}$$ (a bit less than $$2M$$), a little $$\ce {H+}$$ and a little $$\ce {Mg(OH)2}$$, and there will be $$\ce {OH-}$$ which is negligible in the most case. (Of course there is very much water!)

While you can see in the case of $$\ce {NaCl}$$ there won't be $$\ce {NaOH}$$ and $$\ce {HCl}$$ compounds in the solution: because $$\ce {NaOH}$$ and $$\ce {HCl}$$ are "strong", there won't be a hydrolyzation equilibrium. But yes, you can say there are $$\ce {H+}$$ and $$\ce {OH-}$$, as you have known, there are $$1.0 \times 10^{-7} M$$ of them. However, we usually not concern about this. And as you see, all those $$\ce {H+}$$ and $$\ce {OH-}$$ in an aqueous solution, in effect, come from water compounds. (This is why there will be "strong" acid and base: because they are all giving/receiving protons from water, even if they actually have different ability, compared to water, they are behaving undistinguishably. This is called as leveling effect of solvent.)

There are a lot of things I haven't mentioned here, because I think you are not able to calculate something about equilibrium, and maybe you are not familiar with thermodynamic idea yet. But here I just hope what I said is interesting enough for you.

Well, a simple conclusion:

To find out what will be there in an aqueous solution, you may consider:

1. If the solute ionize: Is it a covalent compound? Is it acid?
2. If yes, whether the ions are weak acid or base (in Brønsted-Lowry theory)? Remember, a cation which can be formed from a weak base will be weak acid; while an anion which can be formed from a weak acid will be weak base. (Or you may use an awful word: they are conjugate acid base pair!)

I think these two step are enough for you for the time being.

• Ahaha! I think you really like explaining things! Keep up the good work! (But including a conclusion is very recommended with long answers) – M.A.R. Apr 28 '15 at 16:10
• Good job on taking the time to properly format your answer with LaTeX. Just a tip: you can write equilibrium arrows using \ce{<=>} instead of \ce{<->}, so you get the correct $\ce{<=>}$ instead of $\ce{<->}$. – Nicolau Saker Neto Apr 28 '15 at 20:09
• @Asydot>" ionic compounds? They usually break their** intermolecular** forces" or Intramolecular? – Adnan AL-Amleh Jan 26 '19 at 0:10

Aqueous means that the specific compounds has dissolved in $\ce{H2O}$ or in other words, it dissociated into it's ions. When it is dissolved:

1. Ionic Compounds: $\ce{NaCl}$ for example dissociates in water to $\ce{Na+}$ and $\ce{Cl-}$ ions. Intramolecular breaking.

2. Covalent Compounds: Covalent compounds do not break apart but they do break their intermolecular force.

And one more thing, SOLIDS, LIQUIDS, GASSES aren't usually aqueous. I'm saying usually because there is always some exception to this and someone will point that out. Hopefully this helps, comment for more information.

• Then an aqueous solution of solution of NaCl is a mixture of Na+, cl-,OH-and H+. – On the way to success Apr 28 '15 at 3:10
• Then do Na+, cl-,OH-and H+ mixed and form NaOH and H2O? – On the way to success Apr 28 '15 at 3:11
• Compare energy released in dissociation of NaCl and energy required to form NaOH/HCl . You will have your answer. – Gowtham Apr 28 '15 at 5:29
• Sorry for this, but I don't understand the solid, liquid, gas thing ... If you are talking about the notations s, l, g and aq, then there shouldn't be exception. We use s, l, g for roughly description of the state of substances, but aq is accurate; it is, I think, refer to a phase. If you see aq and l in the same chamical equation, then they must be in different phases. – Asydot Apr 28 '15 at 19:42
• Exactly, that's what I meant. – Asker123 Apr 28 '15 at 19:45

Aqueous means your solvent for the solution is water. It's a term to differentiate from other solution like alcoholic or etheric, etc. On the sol chem part of your question, a molecule can take many different form in solution and their amount can usually be calculated with a constant and equilibrium equation, but for $\ce{MgCl2}$, it is considered that most of the salt that dissolve turn into $\ce{Mg^{2+}}$ and $\ce{2 Cl-}$.

An aqueous solution is a solution in which the solvent is water. It is usually shown in chemical equations by appending (aq) to the relevant chemical formula. For example, a solution of table salt, or sodium chloride (NaCl), in water would be represented as NaCl(aq).

Source: wikipedia

• So does it means NaOH is reacted with water and formed NaOH? – On the way to success Apr 28 '15 at 1:38
• Not on this case, because $\ce{NaOH}$ is a very strong base and won't be formed in solution. But if you dissolve, for example, $\ce{NH3}$ in water (so, an aqueous solution of ammonia), you'll have the equilibrium $\ce{NH3 + H2O <=> NH4OH}$. – Molx Apr 28 '15 at 3:15
• You can't directly copy paste from the wikipedia. If you do then you need to mention it as a source and add directly pasted stuff in blockquote. Please consider the editing. – Freddy Apr 28 '15 at 16:21