# Why does the solubility of some salts decrease with temperature?

It is well known to us that the Solubility of solute in a solution increases with the increase in the temperature because, when the temperature increases the molecules of the solvent gain more kinetic energy. Thus the molecules move randomly and having greater distance from each other which is responsible for the large voids between them and it gives more space for the solute molecules between them to come in. However, there a few salts like cerium sulphate , lithium carbonate sodium carbonate monohydrate, etc. whose solubility decreases with the increase in temperature. How it is so ?

• May 27, 2016 at 18:33

A saturated solution is at equilibrium (rate of dissolution is equal to rate of crystallization) with some equilibrium constant $$K_1$$. If you change the temperature of the system at equilibrium, you will observe a different equilibrium constant $$K_2$$. Whether the equilibrium constant increase or decreases is described by the Van't Hoff equation: $$\ln \left(\frac{K_2}{K_1}\right) = \frac{-\Delta H}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$

So it depends on the sign of the enthalpy of reaction, in this case the enthalpy of dissolution.

It is well known to us that the Solubility of solute in a solution increases with the increase in the temperature because, when the temperature increases the molecules of the solvent gain more kinetic energy.

The statement about solubility is not always true, and the explanation leaves out something. A lot of things change when you increase the temperature. The solvent gains kinetic energy, the solute gains kinetic energy, and the solid gains kinetic energy. How this influences the solubility depends on the specific system, and is hard to predict.

However, there a few salts like cerium sulphate , lithium carbonate sodium carbonate monohydrate, etc. whose solubility decreases with the increase in temperature. How it is so ?

Now that we introduced the Van't Hoff equation, we know it must have to do with the enthalpy of dissolution. Usually, you expect the interactions in the solid that need to be broken to be stronger than the gains from solvating the solute. In these cases, it is apparently the opposite.

• Note that you can use "stretchy parentheses" with \left( and \right) pairs to enclose an entire fraction, e.g. \left(\frac{K_1}{K_2}\right) gives $$\left(\frac{K_1}{K_2}\right)$$ Also, you can combine different brackets (e.g. [ ) or ( }) and use a placeholder (invisible bracket) such as \left. to leave a bracket on one side, for example \left.\frac{K_1}{K_2}\right\}: $$\left.\frac{K_1}{K_2}\right\}$$ Jan 21, 2019 at 15:02
• @KarstenTheis enthalpy of dissolution for both, lithium carbonate and sodium hydroxide, is negative. But solubility of former decreases and that for later increases upon increasing of temperature. Sep 29, 2021 at 5:20
• @Apurvium Apparently, for a saturated solution of sodium hydroxide, dissolution is endothermic even though iti is exothermic at low concentrations (see Andrew's comment for Airhuff's answer). That would explain the distinct temperature dependency of solubilities of lithium carbonate vs sodium hydroxide. Sep 29, 2021 at 10:23
• Sep 29, 2021 at 10:25
• enthalpy of dissolution of infinitely dilute solution: diverdi.colostate.edu/C433/miscellanea/CRC%20reference%20data/… Sep 29, 2021 at 10:26

What you refer to, the decreased in a solid's solubility in a liquid with increased temperature, is frequently called retrograde or inverse solubility, and occurs when the dissolution of the solute is exothermic.

The explanation for this can be viewed as a manifestation of Le Chatelier's principle. Essentially, since evolved heat can be viewed as a product of an exothermic reaction, the addition of more heat (e.g. a higher temperature) is equivalent to adding a product to the product side of the chemical equation for dissolution, driving the equilibrium back toward the reactants, in this case toward the undissolved compound.

Take the exothermic dissolution of calcium sulfate in water for example:

$$\ce{CaSO4_{(s)} <--> Ca^{2+}_{(aq)} + SO4^{2-}_{(aq)} + heat}$$

In this case, the solubility decreases with increasing temperature because by increasing the temperature you are adding heat to the product side. Inversely, if you pull heat from this system, e.g. cool it, you drive the equilibrium toward the products side and solubility is increased.

• I've heard this argument many times before, but I don't know if it's very good. For example, sodium hydroxide dissolves very exothermically in water, but nevertheless its solubility increases very steeply with temperature (from 418 g/L at 0 °C to 3370 g/L at 100 °C according to Wikipedia). Rather than enthalpy of dissolution, in most cases it is presumably the entropy of dissolution that dictates whether a compound has increased or decreased solubility with increasing temperature. Feb 18, 2017 at 5:11
• @NicolauSakerNeto, I certainly didn't expect the Le Chatelier's principle argument to go without opposition ;) And it's not a perfect model, as your exception points out. But I have a problem with the entropy argument also. You would intuitively think that NaOH to Na+ and OH- would have a smaller entropy change than $\ce{CaSO4}$ to $\ce{Ca^{2+}}$ and $\ce{SO4^{2-}}$, based on the lower degree of symmetry of d $\ce{SO4^{2-}}$ as compared to OH-. So shouldn't $\ce{CaSO4}$ solubility increase with temperature if entropy change controls the temperature dependence of solubility? Feb 18, 2017 at 5:44
• The solvation of species with high charge density forces water molecules in their vicinity to adopt a more strict subset of conformations, which is entropically less favourable. Both the anion and cation in $\ce{CaSO4}$ are divalent, and thus may fall in this category. Meanwhile, $\ce{NaOH}$ contains only monovalent species and thus restricts the conformation of solvating water molecules less (though I am sure that the hydroxide ion in water is special due its Grotthuss-type mobility, which likely results in a particularly diffuse solvation shell and even less restricted conformations). Feb 18, 2017 at 7:57
• More generally, the solubility is governed by Gibbs free energy, which has both enthalpic and entropic components. In the example in this answer, the enthalpic component is dominant. The entropy continues to dominate in the NaOH counter-example though. Mar 6, 2018 at 17:42
• @NicolauSakerNeto I realize this is an old post, but I thought it useful to point out that NaOH dissolution is only exothermic at very low concentrations. As soon as you add enough that the pH goes up, the exothermic reaction $\ce{HO- + H+ -> H2O}$ essentially stops happening, and the dissolution switches to endothermic. Try making a saturated solution of NaOH in water. It will get quite cold. May 8, 2019 at 20:21

As others noted, the effect of temperature is governed completely by $$\Delta H$$ of dissolution. The reason that dissolution can have either positive or negative $$\Delta H$$ is because the favorability is determined by the Gibbs free energy $$\Delta G = \Delta H - T \Delta S$$. If the $$\Delta S$$ term is negative (before multiplication by -T) and $$\Delta H$$ is also negative, dissolution is favorable only if $$|\Delta H|>|T\Delta S|$$.

It is important to note also that the magnitude of $$\Delta G^\circ$$ does NOT correlate with the solubility, since the change in concentration per change in $$\Delta G^\circ$$ is not constant. That is why $$\Delta S^\circ$$ does not determine the effect of temperature.

This can be seen by noting that the important value is $$K$$. Since $$\Delta G^\circ=-RT\ln K=\Delta H^\circ - T\Delta S^\circ$$, we can rearrange and find that $$\ln K = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T}\right)+\frac{\Delta S^\circ}{R}$$, so a plot of $$\ln K$$ vs $$\frac{1}{T}$$ (which is the temperature dependence of solubility) has a slope whose sign is determined only by $$\Delta H^\circ$$.

• Can we say that, when we dissolve something, $ΔG_{sys}<0$ till saturation point and then $ΔG_{sys}>0$? Oct 3, 2021 at 10:45