Solubility of compounds in water

1. Feb 24, 2008

The_ArtofScience

Hi

I'm not too familiar with why certain compounds tend to disassociate completely/partially in the water solvent. In one experiment from lab I remembered that HCL dissolved completely whereas CH3COOH (acetic acid) only partially dissolved. Weak acids generally don't hydrolyze completely while stronger acids do- my question here is why is that true? Does this in anyway have to do with electronegativity differences?

2. Feb 25, 2008

lightarrow

If the acid is HX, it has to do with the strenght of the bond between hydrogen atom and the atom X to which is bond, compared to the strenght of the bond between H+ and O in H3O+ and X- with H2O.
So, the bond energy between H and Cl in HCl is much lower than the bond of H+ and O in H3O+ plus the bond energy between X- and H2O, for example (in this case what counts more is the first, that is H+ with H2O bond energy). For other kinds of acids, analogue considerations.

3. Feb 25, 2008

AbedeuS

The dissociation of the $$COO-H$$ bond forms a fairly stable $$COO^{-}$$ anion, so it is acidic, however when compared to the stability of, say the conjugate base of $$HCl$$, the anion $$Cl^{-}$$ has a full noble gas configuration so it is much more stable and the $$H-Cl$$ is therefore much easier to dissociate.

Thermodynamically
$$HX \leftrightharpoons^{\deltaH} H^{+} + X^{-}$$

The enthalpy required to break this bond should be positive, I.E. energy must be supplied to break it.

$$H^{+} + H_{2}O \rightarrow^{\deltaH} H_{3}O^{+}$$

The formation of the hydronium ion in solution will have a negative enthalpy change, it will release energy.

A further interaction is also noted, the hydronium (solvated proton) ion will not exist on its own in solution and will be surrounded by water in order to stabilize the positive charge (make less positive):

$$H_3O^+ + nH_2O \leftrightharpoons^{\deltaH} H_{3}O^{+}.nH_{2}O$$

These combined effects, along with similar water based stabilization of the $$X^-$$ ion will both have negavie enthalpies. And in terms of thermodynamics the energy put in to split the $$HX$$ should be outweighed by the energy gained from solvation and hydronium ion production. So in theory it would seem ok to suggest that all acids would just want to go the path of thermodynamics and become completely dissociated. However the reverse reaction has an effect.

The anion in solution, if not completely stable will readily protinate to reform the acid:

$$X^- + H_3O^+ \leftrightharpoons HX + H_2O$$

Although in the instantaneous addition of the acid to water (i.e. at t=0) the reaction for dissociation of the acid will be completely forward, as soon as the concentration of conjugate base starts to increase, the reverse reaction will start to take place. And so we reach an equilibrium, where the reverse reaction happens at the same rate as the forward reaction and we, the observers, see no change in pH. The reverse reaction for a strong acid is so thermodynamically weak that it barely takes place, weras for an extremely weak acid, such as ethanoic acid, it has quite noticable stopping power on the forward reaction.

As a side note, trying to get an "acidic" gas like HCl to dissociate into a $$H^{+}$$ and $$Cl^-$$ in space, with no water molecules around to bind, is thermodynamically unfavourable due to the production of a "Naked" proton rather then a "Solvated" hydronium ion. This is a quite common textbook argument for why chemists tend to write $$H_3O^+$$ rather than $$H^+$$ although it taken me about 2 weeks of organic synthesis lectures to get bored of the convention and stick to the easier (yet "Technically" innacurate) $$H^+$$ notation.

Last edited: Feb 25, 2008
4. Feb 26, 2008

lightarrow

Good post.
However, in general, I wouldn't omit to mention entropy variations, at least for ion solvation, it's important to explain some "anomalous" behavior of solutions, like the fact some salts dissolve better at lower temperatures.

5. Feb 26, 2008

AbedeuS

I would explain that using:

$$ln(\frac{K_2}{K_1}) = -\frac{\Delta _r H^{\circ}}{R}(\frac{1}{T_2}-\frac{1}{T_1})$$

(Van't Hoff Equation)

6. Feb 27, 2008

lightarrow

Yes, but you should already know that

$$\Delta _r H^{\circ}}\ <\ 0$$

and that's not so easy to understand, a priori, since you should compute the difference between solvation and lattice "dissolution" enthalpies; instead it's easier to understand that

$$\Delta _r S\ <\ 0$$

(at least, it's easier to me) then, from the fact that at equilibrium

$$\Delta _r G\ =\ 0$$

we have

$$\Delta _r H\ =\ T\Delta _r S\ <\ 0.$$

7. Feb 29, 2008

AbedeuS

Ah cool, didn't see it like that :D