4
$\begingroup$

Let's say I have molten potassium hydroxide and then I add some Calcium hydroxide. I imagine, since there are hydroxide groups floating around, it would be energetically neutral for the calcium ions to bind to a "new" hydroxide group, therefore they should be floating freely with the hydroxide groups, becoming solvated. I can't find any empirical data; is my reasoning correct and I can dissolve a hydroxide in any other molten hydroxide?

$\endgroup$
  • 1
    $\begingroup$ Well it would depend on case. In the one you mention, probably yes. $\endgroup$ – Mithoron Jul 31 at 18:11
  • $\begingroup$ The most robust way to answer this would probably be to find the corresponding phase equilibria diagram for $\ce{Ca(OH)2}$–$\ce{KOH}$ system. $\endgroup$ – andselisk Aug 2 at 3:54
  • $\begingroup$ @andselisk such a phase diagram may require deep digging to get to. Simply googling for a calcium hydroxide-potassium hydroxide phase diagram fails because other phase diagrams come up instead. A direct reference would be most welcome, thanks. $\endgroup$ – Oscar Lanzi Sep 1 at 23:02
1
$\begingroup$

Probably not so much calcium hydroxide, but alkali metal hydroxides mixing with each other in the liquid phase. The pairs $\ce{LiOH-NaOH}$ and $\ce{KOH-NaOH}$ are not only fully miscible, they have been studied as fuel cell electrolytes. The picture below, from the referenced article, shows results for the a range of electrolyte compositions that imply the extensive miscibility. The fuel cell itself is based on carbon oxidation to carbon dioxide.

Source

$\endgroup$
0
$\begingroup$

In order for the $\ce{Ca(OH)2}$ to dissolve, it should become solvated, which requires it to bind some of the OH- ions in molten $\ce{KOH},$ leaving some $\ce{K+}$ relatively bare. We already know that $\ce{Ca(OH)2}$ can be dehydrated easily, while $\ce{KOH}$ cannot (it boils at 1325 °C).

The heat of hydration of $\ce{K2O}$ is much greater than the heat of hydration of $\ce{CaO}.$ However, the heat of hydration of $\ce{Ca^2+}$ is greater than the heat of hydration of two $\ce{K+}.$ So, addition of a little water could solvate (hydrate) the calcium ion and make it more soluble.

A relatively easy way to test would be to take a concentrated solution of $\ce{KOH}$ (soluble ~50% in $\ce{H2O}$ @ 15 °C, 64% @ 100 °C) and add some $\ce{Ca(OH)2}.$ More $\ce{Ca(OH)2}$ is likely to dissolve than if no solvating moieties are present other than the $\ce{KOH}.$ Although $\ce{H2O}$ is not powerful enough to bring $\ce{Ca(OH)2}$ into solution, $\ce{OH-}$ ions might do the job — if the $\ce{K+}$ ions are not left bare. It is worth considering whether the water is the actual solvent, with $\ce{KOH}$ as a reactant

$$\ce{Ca(OH)2 + KOH -> Ca(OH)3- + K+},$$

or whether the $\ce{KOH}$ solution is simply a solvent.

$\endgroup$
  • $\begingroup$ Note that $“\ce{Ca^{++}}”$ is an obsolete notation and $“\ce{Ca^2+}”$ should be used instead. As you are already an experienced user, please use proper formatting. (re)Visit this page, this page and this one on how to format your posts better with MathJax and Markdown. Also, I'm not quite following the reasoning in this answer. Probably you can improve it if you add some references your assumptions are based upon. $\endgroup$ – andselisk Aug 2 at 3:52

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.