# Solubility of solids in liquids

How does crystallisation happen and why?

I have been told that when a solid is dissolved in a liquid solvent, an equilibrium is set up between processes of dissolution and crystallisation. But if this is so then why does all solute dissolves (I am talking about an unsaturated solution), isn't it supposed to recrystallise simultaneously too? I am certainly missing a lot and most importantly what crystallisation is and how and why it happens?

I don't want to know anything about the separation technique using the principle of crystallisation. Just want to know the principle itself.

• I edited your question because it's a very good one, but I think your opening line blurred the emphasis. Feb 16, 2019 at 20:37
• @TryHard Thanks. It makes much more sense now. Feb 17, 2019 at 11:11

The principle of solubilization — in the ideal case — is closely related to that of melting. Think of solubilization as a two step process:

1. melt the solute;
2. mix the solute and the solvent.

Of course, step 1 involves raising the temperature to the melting point, or at least adding energy (in the form of heat) at the melting temperature in order to break attractive bonds between solute molecules. We will consider only the case where we are already at the melting point (an isothermal process), so you only need to add heat in order to turn the solid into liquid.

Step 2, on the other hand, is spontaneous if the solute and solvent mix ideally. You don't have to add heat (energy) for this step to proceed, assuming both substances are liquids and mix ideally.

Now what happens when you solubilize something is that you need to add less heat than you would have needed in step 1 to melt all of the solute, since some of the required (free) energy comes from the process of mixing.

By a similar line of argument, based on the properties of ideal solutions, you can relate (approximately) the solubility to the melting point and the heat of fusion as follows:

$$\log(x_2) = \frac{\Delta H^\circ_\mathrm{fus}}{R}\left(\frac{1}{T_\mathrm{fus}}-\frac{1}{T}\right)$$

Since $$\Delta H^\circ_\mathrm{fus}$$ is generally positive (heat is required to melt something) then increasing $$T$$ tends to increase the solubility of solids.

Crystallization is just the reverse of solubilization! And yes, there is always an equilibrium between soluble/insoluble solute, it's just that sometimes the soluble or insoluble part is negligible, depending on the values of $$\Delta H^\circ _\mathrm{fus}$$, $$T_\mathrm{fus}$$, and $$T$$.

I have been told that when a solid is dissolved in a liquid solvent, an equilibrium is set up between processes of dissolution and crystallisation.

Once a solid is mixed with a solvent, dissolution and crystallization reactions happen simultaneously. That part is correct. In some cases, an equilibrium is reached. However, in other cases the dissolution goes to completion, so this statement is a bit too general. It might be better to say "a competition between dissolution and crystallisation" ensues which sometimes leads to equilibrium.

At the beginning, when the concentration of solute is still low, the rate of crystallization is low. As more of the solid dissolves, the crystallization rate increases because the concentration of solute increases.

Under saturating conditions, at one point the rate of crystallization matches that of dissolution, and you arrive at an equilibrium. Under sub-saturating conditions (if the concentration does not reach the solubility limit), the crystallization rates never reaches that of the dissolution rate, and the amount of solid decreases until it is gone. The process never reaches equilibrium (it goes to completion, i.e. the reactant is depleted).

isn't it supposed to recrystallise simultaneously too?

Both in saturated solutions and in solutions that are not saturated, solute particles will collide (with the solid or with other solute particles). However, it is much easier for a solute particle to attach to a solid than for two solute particles to start a nucleus of crystallization. Thus, while solute particles do collide, they will not form a solid when concentrations are sub-saturated. In fact, you can make super-saturated solutions that lack solid because making crystallization nuclei requires higher solute concentration than adding solute particles to an existing solid.

Basically the process of recrystalisation is to dissolve all of your sample in a solvent which mainly dissolves the impurities and just fewer of your desired compound. If the sample is now dissolved in the refluxing solvent a clear solution results. If this solution is cooled down slowly, the formation of crystals starts and these crystals strongly prefer to form a crystal lattice (composed of one component only) in order to gain as much lattice enthalpy as possible. Therefore all impurities stay in solution and can be filtered off afterwards.

• You missed a several of points. (1) Recrystallization works best when there is a large difference between the solubility of the substance in hot and cold solvent. (2) The increase in purity depends on the substance, impurities and solvent. (3) Since the substance will be at least somewhat soluble in the cold solvent, some substance will remain in solution.
– MaxW
Feb 16, 2019 at 17:17