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Author: Subject: Precipitation with Alcohol
hodges
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[*] posted on 21-4-2004 at 15:10
Precipitation with Alcohol


I keep seeing references to precipitating salts from water solution using alcohol. Why does this work? I understand that many inorganic salts are less soluable in alcohol than in water. But by adding alcohol to an already existing water solution, the total amount of liquid available to dissolve the salt is larger than it was before. Lets say 1 gram of a salt will dissolve in a quantity of water. Only 0.1 grams will dissolve in the same amount of alcohol. By adding the alcohol to the water, now 1.1 grams will dissolve, which is more, not less, than would originally dissolve in the water. Where is my logic wrong here?
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t_Pyro
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[*] posted on 21-4-2004 at 23:29


A compound dissolves by forming weak interactions or "bonds" with the solvent molecules. For ionic salts, the energy required to overcome the lattice energy of the solid is provided by the hydration energy- energy released when water molecules form dipole interactions with the ions. For covalent substances, solubility is mainly due to hydrogen bonding or weak van der waal forces.

In any case, if there is a chance that a stronger interaction can be formed, that course is taken. In the present case, for example, the dipole interaction between the ions in question and water molecules is weaker than the hydrogen bonds between alcohol and water molecules. Thus, if alcohol is added to the solution, the weaker dipole moments are broken, and the stronger hydrogen bonds form, thus precipitating the salt.

A similar precipitation occurs when a covalent substance dissolved in a mostly covalent solvent is added to water, and the solvent can form hydrogen bonds with water. Therefore, the covalent substance precipitates out, while the hydrogen bonds between the covalent solvent and water form. eg. ppt of iodine when a solution of iodine and acetone is added to water.
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[*] posted on 16-2-2010 at 23:29


" Lets say 1 gram of a salt will dissolve in a quantity of water. Only 0.1 grams will dissolve in the same amount of alcohol. By adding the alcohol to the water, now 1.1 grams will dissolve, which is more, not less, than would originally dissolve in the water. Where is my logic wrong here?"

Your logic is faulty. You should have noticed that NaCl is only slightly soluble in EtOH so by adding EtOH you lowered the solubility of NaCl in the water. That's a qualitative statement as opposed to an explanation. Basically you're increasing the entropy of the system and solution of NaCl is endothermic so if you heated up the solution the NaCl could remain dissolved depending on the amount of NaCl and the heat applied. Sonication is another way to add energy and might also get the salt back in solution. I haven't tried this.

All that said the above explanation of the hydrogen bonding between EtOH and water being stronger than the NaCl ionic interactions with water apply. I don't know why this is so but I haven't really thought about it. As an observation however, EtOH and water mix freely and the net volume of such mixture is lower than the sum of the two volumes. So the interaction between them seems quite strong.




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