Chemistry - Why does solubility of solids sometimes decrease with temperature increase?

Solution 1:

In short, for this group 2 hydroxide crystal, the dissolution and solvation process is an exothermic equilibrium. Increasing the temperature drives the reaction in the reverse, favouring crystallisation!

However, it would be nice to rationalise some of these terms, so I will lay out a very brief overview of general solubility factors so you can understand the process in a little more detail.

Solubility is a complex thermodynamic process that takes many factors into account. It can be understood as the process of dissolving a crystal in a solvent, and then forming a solvent cage around the solute (solvation). It is a function of three interactions at heart;

1. solute-solute
2. solute-solvent
3. solvent-solvent

Large solute-solute interactions hinder solubility and are associated with high lattice enthalpies (of crystalline solutes), these must be overcome in order to break up the solid and dissolve it. Solvent-solvent interactions must also be overcome in order to solvate the dissolved solute, therefore for a highly interacting solvent the solubility is reduced.

The formation of ordered solvent structures such as clathrin cages etc orders the solvent, and therefore reduces solvent entropy significantly. Therefore solvents with high degrees of freedom hinder solubility.

Solute entropy favours solvation, this reflects the higher solubility of flexible molecules, this is because in the solid they are all locked up, while in solution their rotational and vibrational modes are populated.

In addition to these if the solute has multiple ionic forms then the solubility will be a function of pH too, so must be held constant in practice. The surface profile of the solid too, as well as the extent of cavitation and porosity are also key factors (since they affect the number of solvent-solute interactions possible).

Importantly the type of solvent into which you are trying to dissolve your crystal in is also of fundamental importance! The rule "like dissolves like" is best understood by the (semi-empirical) general solubility equation, \begin{equation} \log S =\frac 12-0.05 (T_m-T)-\log P \end{equation}
A large partition coefficient (the ratio of how soluble the crystal is in organic solvent to aqueous solvent) reduces the solubility, (and mathematically since $\log x$ follows $x$ in limits).

To further complicate the situation there are several definitions of solubility that we can take. In practice we choose intrinsic aqueous solubility, mainly used for high throughput and speed in drug development.

Solubility is really an experimentalists tool for parametrising the extent of Gibbs energy for the dissolving-solvation process, which in turn is usually modelled by complex thermodynamic cycles via gas phase species or super cooled liquids. Modern research into solubility focuses on finding either molecular descriptors through random forest/QSPR or support vector models etc (which are all just number crunching techniques to find correlations between large sets of data) or through fundamental physics and chemistry of the process. \begin{equation} \log S=\frac{(-\Delta G_{sol}/RT)}{\ln 10} \end{equation}
So we understand solubility as an equilibrium constant for the Gibbs change of the solvation process (whatever particular cycle you use to get that data).

Now that we know something about solubility we can ask ourselves, under what conditions does increasing the temperature reduce solubility?

In practice the temperature can effect the entropic term, but also the enthalpies of the crystal lattices, the heats given out by the solute-solvent interactions etc. It is the balance of these process that determines the solubility of something.

In your case we are observing a calcium hydroxide crystal dissolve in aqueous solvent. As we increase the temperature the solubility decreases. There is an equilibrium between crystallisation and (dissolution + solvation). Increasing the temperature is shifting that equilibrium to favour the crystallisation rather than (as we crudely expect) in the forward. This is simply an enthalpic term, due to the solvation process being an exothermic process, increasing the temperature drives the equilibrium in the reverse by Le Chatlier's principle.

Hope that is a bit clearer, and that you now understand some of the ideas behind solubility. :)

Solution 2:

I would also like to add another answer which may give a better understanding of the application of Le Chateliers principle over here.
Dissolution of $\ce{Ca(OH)2}$ is an exothermic process ($\Delta_\mathrm{sol} H<0$) thus it can also be represented as:

$$\ce{Ca(OH)2(s) + H2O(l) <=> Ca^2+(aq) + 2OH- (aq) + \Delta}$$

At particular temperature a particular net amount of heat is evolved in the system during this process ,thus if the temperature was increased, the heat content of the system would increase and there would be stress on the right hand side of the equilibrium representation since heat is being evolved there,so by Le Chateliers principle the system would consume some of that heat by shifting the equilibrium to the left to reduce the stress,thereby producing more $\ce{Ca(OH)2}$ in solid state thus decreasing the solubility.

Conclusively

In general,if in a nearly saturated solution,the dissolution process is endothermic ($\Delta_\mathrm{sol} H>0$),the solubility should increase with rise in temperature and if it is exothermic ($\Delta_\mathrm{sol} H<0$) the solubility should decrease.These trends are also observed experimentally.