Solubility of NaCl in water [duplicate]








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[Why does solubility of solids sometimes decrease with temperature increase?](/questions/31020/why-does-solubility-of-solids-sometimes-decrease-with-temperature-increase)

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I learned that solubility of substance increases with increase of temperature. But I came to know solubility of NaCl doesn't increase when temperature is increased . So what is the reason for this phenomena?


[![](https://i.stack.imgur.com/xRdnJ.jpg)](https://i.stack.imgur.com/xRdnJ.jpg)




Solubility of solid in a liquid depends on the nature of the solid.


It depends on whether the dissolution process is exothermic or endothermic. Using Le Chatelier's principle, if the temperature is increased for an exothermic process then dissolution decreases and if the temperature is increased for an endothermic process then dissolution increases.


I am going to stick to the case of $NaCl$, if you want to know more about Le Chatelier's principle, click on the link below and look for the 2nd answer <https://physics.stackexchange.com/questions/311409/why-do-most-gases-dissolve-easier-at-colder-temperatures/311429#311429>


[![enter image description here](https://i.stack.imgur.com/yinHn.gif)](https://i.stack.imgur.com/yinHn.gif)


There are some solids for whom change in temperature doesn't affect the degree of dissolution.


There are some substance which on heating, change on a particular temperature from one polymorphic form to another (like $\alpha$ to $\beta$ as in the case of ammonium nitrate) or from one hydrated form to another ($CaCl\_2.6H\_2 O -> CaCl\_2.4H\_2 O$) or from hydrated to anhydrous form ($Na\_2SO\_4.10H\_2O -> Na\_2SO\_4$).


Generally, dissolution process in a solvent is a disintegration process. This process needs energy. In such cases, energy is absorbed. But in some cases, besides the process of breaking or ionization, there is hydrate formation. During hydration heat is evolved. The net result is that heat is either evolved or absorbed. 


There are also cases in which heat of separation of ions in just equal to the heat of hydration and there is very little heat effect as in the case of $NaCl$. The heat of solution of $NaCl$ is very small as the heat of ionization is nearly equal to the heat of hydration. Therefore, temperature doesn't affect the dissolution of $NaCl$ as the heat of hydration almost equalizes heat of dissolution.


 5

One way to understand solubility is to start with the Clausius -Clapyron equation (CC), which despite approximations involved is a good description the vapour pressure over various solids and liquids. For the ideal solution


$$ \ln\left (\frac{p\_2}{p\_1}\right) = \frac{-\Delta H\_{vap}}{R}\left( \frac{1}{T\_2}-\frac{1}{T\_1} \right )$$
where *p* is the vapour pressure, *T* the temperature and $\Delta H$ the enthalpy change between states $1$ and $2$. 


The heat of sublimation is $\Delta H\_{sub}=\Delta H\_{vap}+\Delta H\_{fus}$
and calculating the pressure over the pure solid form of the solvent gives 
$$ \ln\left (\frac{p\_2}{p^\*\_1}\right) = \frac{-\Delta H\_{sub}}{R}\left( \frac{1}{T\_2}-\frac{1}{T\_1} \right )$$


and so subtracting these two equations gives


$$ \ln\left (\frac{p^\*\_1}{p\_1}\right) = \frac{\Delta H\_{fus}}{R}\left( \frac{1}{T\_2}-\frac{1}{T\_1} \right )$$


If it is assumed that Raoult’s Law applies then $ p\_1=xp^\*\_1$ where *x* is the mole fraction of the solute. Substituting into the last equation gives
$$ \ln(x)= -\frac{\Delta H\_{fus}}{R}\left( \frac{1}{T\_2}-\frac{1}{T\_M} \right )$$
where now $T\_M$ is the melting temperature of the pure solute. As $\Delta H/T\_M$ is a constant then the mole fraction and hence solubility in solution is  

$$ \ln(x) \propto \frac{-\Delta H\_{fus}}{R}\left( \frac{1}{T\_2} \right )$$


which shows that the mole fraction at temperature *T* varies as
$$x\_T \propto \exp \left (- \frac{\Delta H\_{fus}}{RT}\right )$$


and so the solubility rises with increase in temperature. Different species will rise more or less slowly depending on their heat of fusion as $\Delta H\_{fus}/R$. 


Using data for NaCl shows that the mole fraction hardly varies between $200$ to $400$ K, whereas there is a huge increase for $\ce{NaNO3}$ under the same conditions.


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