12
$\begingroup$

Water molecules and various salt molecules are very different. However, it seems very difficult to separate the two. Once a salt is dissolved in water, an energy or chemical intensive method (like boiling) is required to separate the salt back out again. Why is this?

$\endgroup$
  • 4
    $\begingroup$ This question appears to be off-topic because it is about chemistry rather than physics. Would chemistry.SE be a better place for this question? $\endgroup$ – Danu Jul 3 '14 at 13:26
  • 7
    $\begingroup$ I don't think this is off-topic here. It might get more answers or exposure at Chemistry, but I think it's perfectly on-topic here. My suggestion would be to wait a bit (a day or two) and if there's no love here, ask to have it migrated. $\endgroup$ – tpg2114 Jul 3 '14 at 14:01
  • 3
    $\begingroup$ @Danu: Whilst it is certainly on-topic for the chemistry SE, I believe it is also on-topic for the physics SE. In addition, in many aspects, physics subsumes chemistry. $\endgroup$ – JamalS Jul 3 '14 at 14:10
  • 2
    $\begingroup$ This question describes a physical (rather than chemical) reaction as being of interest, seems on topic to me. $\endgroup$ – Kyle Oman Jul 3 '14 at 14:32
  • 2
    $\begingroup$ In fact, it may even be more physics than chemistry in the sense that the reason for salt and water to be hard to separate is partly the same as why it is hard to separate different gases in a mixture. $\endgroup$ – Ruslan Jul 3 '14 at 16:56
7
$\begingroup$

From The Feynman Lectures on Physics, Vol. I [1]:

If we put a crystal of salt in the water, what will happen? Salt is a solid, a crystal, an organized arrangement of “salt atoms.” [...] Strictly speaking, the crystal is not made of atoms, but of what we call ions. An ion is an atom which either has a few extra electrons or has lost a few electrons. In a salt crystal we find chlorine ions (chlorine atoms with an extra electron) and sodium ions (sodium atoms with one electron missing). The ions all stick together by electrical attraction in the solid salt, but when we put them in the water we find, because of the attractions of the negative oxygen and positive hydrogen for the ions, that some of the ions jiggle loose.

enter image description here
Figure 1-6

In Fig. 1–6 we see a chlorine ion getting loose, and other atoms floating in the water in the form of ions. This picture was made with some care. Notice, for example, that the hydrogen ends of the water molecules are more likely to be near the chlorine ion, while near the sodium ion we are more likely to find the oxygen end, because the sodium is positive and the oxygen end of the water is negative, and they attract electrically.

Feynman has done well in explaining you the process in atomic point of view. Now comes your complexity of separating salt from water in a salt solution. During the process of boiling, the intermolecular forces will be broken between water molecules, and also between ions and water molecules.

Water molecules ($\mathrm{H_2O}$) being less massive ($18.01528(33)$) than the other two individual ions ($\mathrm{Na}$, $22.98976928(2)$ – $\mathrm{Cl}$, $35.45(1)$), flies off easily leaving sodium and chlorine ions. These ions once again attract each other to form crystals. In other words, energy is required to break the intermolecular forces and release ions from prison to join their partners.

Reference

  1. Feynman Lectures on Physics. Vol. 1, pp. 1–6 (numbers may vary depending on edition).
$\endgroup$
  • $\begingroup$ If the energy required to break the ionic bonds is so high, how come we don't see that heat produced when the bonds are being formed, aka the salt being dissolved? $\endgroup$ – woojoo666 Jul 15 '14 at 17:00
  • $\begingroup$ @woojoo666, who says you don't see heat produced? Have you ever tried dissolving salt in water in a calorimeter? I haven't done it with salt, but I did it in a high-school science class with sodium hydroxide. We dissolved anhydrous NaOH at room temperature in distilled water at room temperature in a styrofoam cup, and the temperature rise was substantial. I assume that the same thing (but probably less dramatic) would have happened if we'd used NaCl instead. $\endgroup$ – Solomon Slow Feb 3 '16 at 17:54
  • $\begingroup$ @james large wouldnt you expect the opposite? $\endgroup$ – lalala Feb 8 '18 at 21:03
  • $\begingroup$ @lalala, um, no. It takes energy to separate salt from water. I expect the reverse process (dissolving salt in water) to release the same amount of energy. $\endgroup$ – Solomon Slow Feb 8 '18 at 21:19
  • $\begingroup$ @jameslarge well, my intuition was different, so I opened up a new question on this. You might want to share your knowledge on this point. physics.stackexchange.com/questions/385341/… $\endgroup$ – lalala Feb 9 '18 at 15:49
4
$\begingroup$

In short, they are hard to separate, because even though the molecules are very different, they have properties that attract them to each other.

Water is a polar molecule. The oxygen molecule oxidizes the two hydrogen molecules, creating a positive charge on the hydrogen side, and a negative charge on the oxygen side.

Meanwhile, salt is composed of sodium, a positive ion, and chlorine, a negative ion. The charges on the water molecule attract the oppositely charged ions and pull them off of the salt crystal, effectively breaking the salt apart at the molecular level.

Then energy that it takes to separate the particles back again is essentially what is needed to counteract these attractive forces.

$\endgroup$
0
$\begingroup$

You can see that the enthalpy of hydration is a two step proccess of solvation and reverse crystallization. The $\Delta H_\mathrm{hydr}$ is actually positive, so you have to give energy just to dissolve the $\mathrm{NaCl}$ in water. In order to separate the water from the $\mathrm{NaCl}$, you need to account for the enthalpy of evaporation of water and the enthalpy of reverse hydration:

Enthalpy of evaporation of water

The temperature is in Kelvins, so going at around the $300\,\rm K$ mark you can see the $\Delta H_\mathrm{evap}$ is $44\,\rm KJ/mol$. Substracting the $\Delta H_\mathrm{hydr}$, you get a lower $\Delta H$, meaning that the salt actually lowered the energy required to boil off the water.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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