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I'm having a difficulty understanding the following quote from Wikipedia - Dissociation:
Acetic acid is extremely soluble in water, but most of the compound dissolves into molecules, rendering it a weak electrolyte.
To what (non-ionic) molecules can acetic acid be dissolved? If the answer is $\ce{CH3COOH}$ then in what way is it extremely soluble, if it dissolved to itself?
There are two questions being asked here:
Why is aqueous acetic acid a weak electrolyte?
When acetic acid is dissolved in water there is an equilibrium reaction: $$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$
Since acetic acid is a weak acid, the equilibrium position lies well to the left, with only a small fraction of the acetic acid molecules reacting to form ethanoate and hydronium ions. The presence of this small amount of ions results in aqueous acetic acid being a weak electrolyte.
Why is acetic acid highly soluble in water?
Since the vast majority of acetic acid molecules do not dissociate when a sample is dissolved in water, the solubility has to do with the interactions between acetic acid molecules and water molecules. Water has a network of hydrogen bonds between molecules in its liquid phase and so when a substance dissolves in water this bonding is disrupted. However, acetic acid is able to form many new hydrogen bonds to water molecules and so this results in a highly favourable interaction, leading to the high solubility of acetic acid in water.
I think the Wikipedia page you quote is poorly worded. What on earth does it mean to ‘dissociate into molecules’? Nothing, in my opinion. A better wording is discussed below.
But first, let’s discuss what actually happens when acetic acid is dissolved in water. As you may know, glacial acetic acid consists mainly of $\ce{H3CCOOH}$ molecules that associate to form hydrogen bonding networks. There is practically no ionisation in glacial acetic acid, i.e. the autoprotonation equilibrium $(1)$ is leaning very strongly to the reactants’ side.
$$\ce{2 H3CCOOH <<=> H3CCOOH2+ + H3CCOO-}\tag{1}$$
So before dissolution, we are dealing with molecules of acetic acid. If we add these into water, most of them just stay being molecules; only a small percentage ionises in water according to reaction $(2)$. The acidity constant shown in the equation is a measure of how many molecules are dissociated; it depends on the concentration.
$$\begin{gathered}\ce{H3CCOOH <<=> H3CCOO- + H3O+}\\ K_\mathrm{a} = 10^{-4.76} = 1.74 \times 10^{-5}\end{gathered}\tag{2}$$
So upon dissolution we have:
Only the latter are charged compounds and thus only they contribute to the solution’s conductivity. Since they are few in number, conductivity is low.
Getting back to the original quote. It seems to imply that dissolving acetic acid in water ‘turns it’ into molecules; which is wrong: the acetic acid molecules remain the same all through. A better wording would emphasise that they do not change. For example:
Acetic acid is extremely soluble in water, but most of the dissolved compound remains as molecules, rendering it a weak electrolyte.
Acetic acid is extremely soluble in water, but only a small fraction is dissociated into ions, rendering it a weak electrolyte.
