Solubility of compounds in water

In summary, the strength of the bond between the hydrogen atom and the atom X in an acid determines whether it will dissociate completely or partially in water. Stronger acids have weaker bonds and are more likely to dissociate completely than weaker acids. This is due to the thermodynamic factors of bond energy and solvation energy. Additionally, entropy variations can also affect the dissociation of salts in solution.
  • #1
The_ArtofScience
83
0
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?
 
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  • #2
The_ArtofScience said:
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?
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
The dissociation of the [tex]COO-H[/tex] bond forms a fairly stable [tex]COO^{-}[/tex] anion, so it is acidic, however when compared to the stability of, say the conjugate base of [tex]HCl[/tex], the anion [tex]Cl^{-}[/tex] has a full noble gas configuration so it is much more stable and the [tex]H-Cl[/tex] is therefore much easier to dissociate.

Thermodynamically
[tex] HX \leftrightharpoons^{\deltaH} H^{+} + X^{-} [/tex]

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

[tex] H^{+} + H_{2}O \rightarrow^{\deltaH} H_{3}O^{+} [/tex]

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):

[tex]H_3O^+ + nH_2O \leftrightharpoons^{\deltaH} H_{3}O^{+}.nH_{2}O[/tex]

These combined effects, along with similar water based stabilization of the [tex]X^-[/tex] ion will both have negavie enthalpies. And in terms of thermodynamics the energy put into split the [tex]HX[/tex] 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:

[tex]X^- + H_3O^+ \leftrightharpoons HX + H_2O[/tex]

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 [tex]H^{+}[/tex] and [tex]Cl^-[/tex] 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 [tex]H_3O^+[/tex] rather than [tex]H^+[/tex] 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) [tex]H^+[/tex] notation.
 
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  • #4
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
I would explain that using:

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

(Van't Hoff Equation)
 
  • #6
AbedeuS said:
I would explain that using:

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

(Van't Hoff Equation)

Yes, but you should already know that

[tex]\Delta _r H^{\circ}}\ <\ 0[/tex]

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

[tex]\Delta _r S\ <\ 0[/tex]

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

[tex]\Delta _r G\ =\ 0[/tex]

we have

[tex]\Delta _r H\ =\ T\Delta _r S\ <\ 0.[/tex]
 
  • #7
Ah cool, didn't see it like that :D
 

Related to Solubility of compounds in water

1. What is solubility?

Solubility refers to the ability of a substance, known as the solute, to dissolve in a solvent, typically water. It is a measure of the maximum amount of solute that can dissolve in a given amount of solvent under specific conditions, such as temperature and pressure.

2. How is solubility determined?

Solubility can be determined experimentally by adding a known amount of solute to a solvent and observing whether it fully dissolves or if there is undissolved material remaining. The solubility can also be calculated using mathematical equations based on the properties of the solute and solvent.

3. What factors affect the solubility of compounds in water?

The solubility of compounds in water can be affected by several factors, including temperature, pressure, and the chemical properties of the solute and solvent. For example, increasing the temperature can often increase the solubility of solids in water, while increasing the pressure can increase the solubility of gases in water.

4. Why are some compounds more soluble in water than others?

The solubility of a compound in water depends on the nature of its chemical bonds and interactions with water molecules. Compounds with polar bonds or functional groups, such as -OH or -NH2, tend to be more soluble in water because they can form hydrogen bonds with water molecules. On the other hand, nonpolar compounds, such as oils, are not soluble in water because they do not have the ability to form hydrogen bonds with water.

5. How does solubility in water affect the properties of a compound?

The solubility of a compound in water can significantly impact its physical and chemical properties. For example, compounds that are highly soluble in water tend to have higher melting and boiling points compared to compounds that are less soluble in water. Additionally, solubility in water can influence the reactivity and bioavailability of a compound, making it an important factor to consider in various scientific fields, such as pharmacology and environmental science.

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