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This question already has an answer here:

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?

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marked as duplicate by Klaus-Dieter Warzecha, Todd Minehardt, Jon Custer, airhuff, M.A.R. ಠ_ಠ Feb 22 '17 at 19:31

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    $\begingroup$ Also, there are quite a few different substances out there. Some solubilities increase with temperature, some decrease, some do not change at all. $\endgroup$ – Ivan Neretin Feb 22 '17 at 6:44
  • $\begingroup$ What textbook are you using? $\endgroup$ – MaxW Feb 22 '17 at 6:48
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    $\begingroup$ It is true. This sort of fact can be looked up. At 0 C 35.65g NaCl/100 grams H2O, at 100 C 38.99 grams NaCl/100 grams H2O. en.wikipedia.org/wiki/Solubility_table#S $\endgroup$ – MaxW Feb 22 '17 at 6:56
  • $\begingroup$ @MaxW i didn't understand what you are saying,it will be good if you answer it in some detail. $\endgroup$ – Fawad Feb 22 '17 at 6:57
  • $\begingroup$ Cadmium selenate is a salt for which the solubility decreases with increasing temperature. See Wikipedia table noted above. $\endgroup$ – MaxW Feb 22 '17 at 7:01
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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

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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.

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  • $\begingroup$ I take that NaCl dissociates because it breaks into Na+ & Cl ions which go to the OH- & H+ ions of water. Is that right ? Plus how does sugar dissociate in water. I was particularly weak in all this & particularly organic even an year back when we were giving jee $\endgroup$ – Shashaank Apr 1 '17 at 11:48
  • $\begingroup$ Yes, in case of NaCl, Na attracts the O of H20 and Cl attracts the two H of H20. Sucrose is also a slightly polar molecular. Check this out : middleschoolchemistry.com/lessonplans/chapter5/lesson4 $\endgroup$ – Mitchell Apr 1 '17 at 11:57
  • $\begingroup$ Generally the water molecules come toward the NaCl crystal and break it apart. $\endgroup$ – Mitchell Apr 1 '17 at 11:57
  • $\begingroup$ Sucrose is less polar than salt that's why it's tough to prepare a sugar solution. Right ? Also the the solubility has to increase if I hear the solution ? $\endgroup$ – Shashaank Apr 1 '17 at 12:00
  • $\begingroup$ Yes you are correct. Usually when you heat the solution the solublility increases because the dissolution process in endothermic. According to Le chatelier's principle, if you increase the temperature reaction proceeds towards the endothermic side in order to prevent the temperature change to take place. Fun fact : Although the dissolution process is endothermic, for NaCl the lattice enthalpy and the hydration enthalpy are almost the same in magnitude, so we don't observe a significant change in the surrounding temperature. $\endgroup$ – Mitchell Apr 1 '17 at 14:41
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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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