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Am fairly new to chemistry and apologize if the question and the description are too vague.If you think I lack basic background knowledge to deal with something like this, please point me to a resource, a book or anything that can help me in this regard. So in case this question is off-topic, please suggest where I should begin- at least starting point for my research will be helpful. However, in that case, if possible, please be specific, cause its too much of an effort to read entire chemistry textbooks to find out the answer to this one question.

Would like to know how to seperate NaI dissolved in water. Please suggest a way to do so without heating and evaporation. Like, what compound should I add to the solution so that the NaI can react with that resulting in solid products, maybe in the form of lumps/signicantly sized suspended particles that can be easily filtered out.

Additionally I would like to know how to find out, given a solvent (or any liquid compound), whether NaI dissolves in it or not. For eg: say that I have methanol, will NaI dissolve in it? For methanol i know the answer and its a yes. But how do i test solubility for any other compound. Of course chemists should not have to carry out practical experiments to find out this(it must probably be part of a known set of facts(about a particular compound) to chemists or they will be able to deduce it from some other known facts. How will a chemist answer this question?)

And finally, once I have any solution with NaI dissolved, how to separate it without evaporation(but by some chemical reaction). For eg: how to separate NaI from methanol? how to figure out the reactants I'll need in case the solvent is known?

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    $\begingroup$ Why are you so set against evaporation? It's by far the easiest method. $\endgroup$ – bon Sep 9 '15 at 13:21
  • $\begingroup$ evaporation can be risky if the solvent is flammable or toxic. hence the issue $\endgroup$ – chemdoc Sep 9 '15 at 13:27
  • $\begingroup$ alright, if someone can describe that evaporation is safe, then am open to that idea. $\endgroup$ – chemdoc Sep 9 '15 at 13:39
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    $\begingroup$ You were talking of water solution. Water is neither flammable nor toxic. Will you take my word for it, or should I find the proper reference? $\endgroup$ – Ivan Neretin Sep 9 '15 at 13:45
  • $\begingroup$ in the last part of the description, i have pondered upon what's to be done in case the solvent is methanol for instance. is it safe enough to carry out in a lab, if its methanol? $\endgroup$ – chemdoc Sep 9 '15 at 13:51
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I summarize some answers given in comments.

Evaporation is really one of the standard methods in any combination of solid (cristalline, amorph, polar, unpolar) and solvent (polar, unpolar, flammable...). But you must not think of it as the pot heated on a flame. In modern synthesis you often already work with a nitrogen or noble gas atmosphere in a closed apparature and use varying pressure for controlled destillation. So it is no problem to just reduce the pressure and evaporate your (organic) solvent. In your special case you even can think of a pot heated on a flame, there is no risk and disadvantage. It is simply an effective method.

Do you know the concept of polarity? A good rule of thumb is that polar substances mix with polar ones but not with unpolar ones. So water dissolves salt but not paraffin wax. Benzene on the other side dissolves parrafin wax but no salt and water and benzene do not mix with themselves.

=================Edit (Addition of Information, because of question in comment)

So let's see how to regain the solvent. First of all there is not only one boiling temperature. The boiling temperature is always a function of pressure. Approximately it can be described by the Clausius-Clapeyron-Equation. $$ T_2(p_2)=(\ln(\frac{p_1}{p_2})k+\frac{1}{T_1})^{-1} $$ Here $k$ is some constant depending on the specific solvent and $T_1$ and $p_1$ is a corresponding pair of known boiling temperature and pressure. For example $100^\circ$C and 1bar for water. In general the boiling temperature goes down when pressure goes down. (Water boils faster in the Himalaya)

So if you have a typical organic solvent like aceton with a boiling temperature of $56^\circ$C under 1bar it evaporates already at $20^\circ$C when applying a pressure of e.g. 200mbar. (it is not the exact number) But it also condenses again when cooling down. Using this you can make it liquid again and regain it. Compare this image of a typical distill. http://www.ssc.education.ed.ac.uk/BSL/pictures/distill.jpg

Summarizing: you use a pump to get into your desired range of boiling temperature and cool the gas in the condenser again to a temperature below the boiling temperature. Depending on your system you have boiling temperatures below $0^\circ$C and cool with liquid nitrogen in the condenser or normal temperature range and cool with water or...

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  • $\begingroup$ a little bit of an advanced answer for me at the moment. will need to learn more. one very important thing i forgot to mention is that its the solvent that am intending to recover and keep. not the NaI $\endgroup$ – chemdoc Sep 9 '15 at 16:41
  • $\begingroup$ May I ask if you are in school and what class. I can write more, but I do not want to be boring explaining things you already know ;) $\endgroup$ – mcocdawc Sep 9 '15 at 16:54
  • $\begingroup$ nope. am a working professional . software engineer. Wish i had paid more attention to my chem lectures when i was in school :-) $\endgroup$ – chemdoc Sep 9 '15 at 17:05
  • $\begingroup$ And I wish I will be interested into other subjects this way, when starting to work. ;) Unfortanetly I only know good introductory textbooks in German. But you could look for something like chemistry for biology/medical... students. I think that you will learn polarity and other basic concepts more efficiently with such a textbook instead of forums. You will get a overall picture to ask the right questions. $\endgroup$ – mcocdawc Sep 9 '15 at 19:56
  • $\begingroup$ So I will assume school knowledge in chemistry and advanced knowledge in maths for editing the answer. $\endgroup$ – mcocdawc Sep 9 '15 at 20:08

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