1
$\begingroup$

Does "easily hydrated" imply highly soluble in water too?

Small ions like $\ce{Li^+}$ and $\ce{Be^{2+}}$ are said to be easily hydrated but does it also imply they are highly soluble?

I looked up some books and they say that compounds of Lithium like $\ce{Li_2O}$ and $\ce{LiF}$ are sparingly soluble.Why?

$\endgroup$
  • 1
    $\begingroup$ This may be a pathological example, but superabsorbent polymers hydrate very easily, swelling to many times their original volume, yet they can remain insoluble in water. $\endgroup$ – Nicolau Saker Neto Jul 10 '16 at 8:03
  • 3
    $\begingroup$ Many ceramics hydrate well, but are not soluble. The difference is the ceramic is happy to add OH groups in various places, but that does not imply that it wants to break up into constituent molecules surrounded by water. It is not a transitive property. $\endgroup$ – Jon Custer Jul 10 '16 at 11:30
3
$\begingroup$

Note that you are considering two different things: easily hydrated cations and soluble compounds. Solubility is complicated, and compounds containing the same cations may well differ in solubility-- lithium oxide reacts violently with water to form lithium hydroxide, for example, whereas lithium fluoride is only sparingly soluble.

What causes the difference between the two compounds? For one, the $\text{O}^{2-}$ anion is an extremely strong base, and readily reacts with water to form $2\,\text{OH}^-$; likely the high enthalpy of reaction provides a thermodynamic driving force for the dissolution of lithium oxide. If we then consider lithium fluoride, no large driving force exists to favor dissolution. While $\text{F}^-$ may react with water to form $\text{HF}$ and $\text{OH}^-$, $\text{F}^-$ is only weakly basic and such a reaction would not be as exothermic as that with lithium oxide.

We must then consider smaller driving forces that were drowned out when considering lithium oxide. Even though $\text{Li}^+$ and $\text{F}^-$ are both small, highly charged species with large charge densities that will interact favorably (with regards to enthalpy) with the polar water molecules in solution, their charge densities will also cause water molecules to cluster around the ions, forming hydration shells and hence decreasing entropy; this is thermodynamically unfavorable. Furthermore, the electropositivity of $\text{Li}^+$ and the electronegativity of $\text{F}^-$, combined with their small ionic radii, lead to a large lattice energy for $\text{LiF}$, further disfavoring dissociation. These effects all contribute to the sparing solubility of $\text{LiF}$.

Related question.

| improve this answer | |
$\endgroup$
0
$\begingroup$

Hydration is a characteristics of an ion. Solubility is dictated by a set of anions and cathions. Consider Na$^+$ in Na2CO3 (soluble) and NaHCO3 (poorly soluble) and Ca$^+$ in CaCO3 (insoluble) Ca(HCO3)2 (soluble). Anions alone or cations alone are not enough to predict solubility.

Also note the case of a proton exchange membrane. It has moieties that are hydrated well, but it is not soluble at all.

| improve this answer | |
$\endgroup$

Your Answer

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