6
$\begingroup$

Theoretically, by dissolving a salt in water the melting point lowers, approximately 1.86 K kg/mol, making it more difficult to freeze water. However, the process of dissolution of certain salts is endothermic, lowing the temperature of the water. Is it possible that the water freezes due to that change of temperature although the melting point also lowers, or is it impossible?

For instance, consider the next case. In 1 kg of liquid water at 0 °C, we put a mol of KI. The melting point of water should be now about −1.86 °C. However, the enthalpy of dissolution of the KI is about 20 kJ/mol, meaning that the temperature of water should lower till −4.8 °C. So, theoretically, it should freeze, but does this happen in reality?

$\endgroup$
5

4 Answers 4

9
$\begingroup$

It is possible, in a significantly different way that you envision. Freezing point depression via the cryoscopic constant is an example of a colligative property, which holds only for relatively dilute solutions. Once the solutions get very concentrated, intermolecular interactions become more complicated and do not generalize, meaning different compounds will affect the solvent differently.

Therefore, with enough of the right solute, it is possible to actually increase the freezing point of water. Oscar Lanzi's example of clathrates is interesting, but if you want to use salts, then I present you tetrabutylammonium hydroxide. It is known to form stable hydrates with a defined composition and containing a large amount of water, such as $\ce{(C4H9)4N+OH^-.30 H2O}$. The 30-hydrate is a solid which melts at approximately 30 °C, containing 67.6% water by weight.

This specific compound, at this specific concentration, forms a particularly stable network of solvating water molecules. You can think of it as stabilizing the structure of ice, allowing it to occur above the normal melting point of pure water. Thus, in principle, if you add enough of the anhydrous salt to water, when you reach the right ratio, the solution will "freeze".

Many compounds actually display this behaviour of forming solid hydrates at certain defined compositions, but tetrabutylammonium hydroxide is unusual in that it forms solid hydrates with a very large amount of water.

$\endgroup$
2
$\begingroup$

Not a salt, but methane works by forming a different phase, https://en.wikipedia.org/wiki/Methane_clathrate. More than just a gee-whiz curiosity, methane clathrate was a big complicating factor in the Deepwater Horizon explosion in the Gulf of Mexico several years ago.

$\endgroup$
1
  • 2
    $\begingroup$ Although this is could be an interesting comment, I don't see how it answers the question. $\endgroup$
    – airhuff
    Commented Dec 9, 2017 at 21:17
-4
$\begingroup$

*My answer is theory based and contains mainly of text.

Your thought is correct, that salts do require heat to dissolve. This is observed when you add a large amount of table salt to your typical cup of water, the temperature drops (only slightly though).

The heat of the solution affects the solubility of the salt, so the cooler your water, the less salt you are able to dissolve it in.

And in the world, there is still no salt that can have the ridiculous high molar solubility to lower the solution to freezing temperatures.

$\endgroup$
1
  • $\begingroup$ I don see it at all. For example, in the case given (potassium iodide), the solubility of KI at 0ºC is 7mols/(kg of water). If we put one mol in a kg of water its temperature should be around -5ºC. I am not very sure, but considering that the solubility of KI at 20ºC is about 8mols/kg, chances are that a mol of KI can be dissolved in a kg of water at -5ºC $\endgroup$
    – maxbp
    Commented Dec 9, 2017 at 18:23
-5
$\begingroup$

I don't think it is possible. My reason being that the world has many oceans and seas even salt water lakes. All of which vary in different densities and different temperatures but some form ice packs, ice burgs, glaciers yet others don't freeze at all. Again it's down to the type of salt density & composite of that salt surely???

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.