# Why is water evaporated from the ocean not salty?

I am wondering about this. When salty water in the ocean evaporates we are getting the clean distilled water. Why is that? I was trying to think on this and maybe the comparative size/mass of water molecules to the size of different salts molecules plays a role here, but it is not likely. The size/mass of e.g. alcohol molecules is much higher but they still evaporate quite well. Could you explain intuitively?

• I think that it's better to think in terms of energy than in terms of size/mass in this case. Salt dissolves into water as charged Na+ and Cl- ions. When water molecules evaporate it's difficult for individual Na+ or Cl- ions to latch onto them for a ride up into the atmosphere because that would greatly increase the electrostatic energy associated with the water molecule by a lot. So the water evaporates while the Na+ and Cl- ions tend to be left behind. – Samuel Weir Oct 15 '16 at 18:36
• Samuel, could you elaborate? This is a very interesting aspect of evaporation. I mean that I do understand that the more water is evaporated the more ions of Na and Cl are binding together again and form crystals. But I still do not see how is it difficult for individual Na or Cl ions to latch onto them for a ride... Probably this is because I do not really understand how the clean water evaporates. Is it evaporating as a whole molecules or it also breaks into some ions/atoms? – Alexander Arendar Oct 15 '16 at 18:54
• Alex, water evaporates as whole molecules. It would take extremely high temperatures to start breaking down water molecules to their constituent atoms. As for the energy considerations involved here with water and dissolved NaCl, there's a nice, easy-to-understand explanation here: van.physics.illinois.edu/qa/listing.php?id=1490 – Samuel Weir Oct 15 '16 at 19:22
• Actually the sea is salty because the vapor is not. – Roman Reiner Oct 15 '16 at 22:05
• @RomanReiner not really; it is salty because minerals have dissolved in it. If the vapor would be salty (whatever that might mean...), then the rain would be salty, and the rivers would be salty, bringing back the salt to the sea, with slightly more dissolved in it, building up the salt level over time. It would possibly reach an equilibrium after a long time, but not at the current salt levels. – fishinear Oct 16 '16 at 19:17

Water molecules are polar; this basically means that they have a "positive side" and a "negative side"

Salt is composed by Na$^+$ and Cl$^-$ ions held together by electrostatic forces (is a ionic compound). When salt is put into water, it dissociates, i.e. the Na$^+$ ions are separated from the Cl$^-$ ions.

When in the water, such ions are surrounded by water molecules facing them with the side which has charge opposite to that of the ion; this is because this way they can reach a lower energy state, being their electrostatic field screened by that of the water molecules (picture below [source]).

Water evaporates when the thermal energy of the molecules is high enough to break about half the hydrogen bonds between them [source]. For the ions, it is much more difficult to evaporate, because their thermal energy would have to be enough to compensate the effect of the water molecules which surround them.

Basically, both the water molecules and the ions are in what is called a potential energy well: to "kick them out" of the well, we have to provide them with an energy as high as the depth of the energy well $\Delta E$ (picture below).

The depth of the well in which the water molecules are (due to hydrogen bonding) is much lower than the depth of the well where the ions are. So, a far higher thermal energy is needed to take an ion out of the water. Since thermal energy is proportional to $k_B T$, where $k_B$ is Boltzmann's constant, this means that a far higher temperature is needed.

Update: Some numbers

To get an idea of the order of magnitudes of the energies involved, we should consider the following:

• At room temperature ($T_r\simeq298$K), $k_B T_r = 0.026$ eV (however, we should keep in mind that this is just an order of magnitude...)
• The energy of an hydrogen bond (hydrogen bond enthalpy) in water is around $23.3$ kJ/mol = $0.24$ eV = $9.3 \ k_B T_r$, and in order a volume for water to evaporate, about half of all the hydrogen bonds in the volume must be broken:

There is no standard  definition for the hydrogen bond energy. In liquid water, the energy of attraction between water  molecules (hydrogen bond enthalpy) is optimally about $23.3$ kJ/mol (Suresh and Naik, 2000)  and  almost five times the average thermal collision fluctuation at $25$°C. This is the energy required for  breaking and completely separating the bond, and equals about half the enthalpy of vaporization ($44$ kJ/mol  at $25$°C), as an average of just under two hydrogen bonds per molecule are broken when water  evaporates. [source]

• The energy gained by putting an ion in water (technical term "hydrating" the ion) is known as the hydration energy or hydration enthalpy ($\Delta H_{hyd}$). Since we are interested in the opposite process (the remotion or de-hydration of the ion), we have to take $-\Delta H_{hyd}$. Here we can find some numbers. We can see that

$$\Delta H_{hyd}(\text{Na}^+) = -406 \ \text{kJ/mol} = -4.2 \ \text{eV} = -162\ k_B T_r$$

$$\Delta H_{hyd}(\text{Cl}^-) = -363 \ \text{kJ/mol} = -3.8 \ \text{eV} = -145 \ k_B T_r$$

• Comments are not for extended discussion; this conversation has been moved to chat. – rob Oct 16 '16 at 21:47
• I'd love if some ballpark figure could be offered in this answer (or as a comment) in eV, just so there is an order of magnitude to which we can compare the relevant energies. I take that kT is about 0.025 eV, while the hydrogen bond has an energy close to some 0.21 eV (from wikipedia). What would be the energy in eV for the ions? About their hydration energy (~38,5 eV, if my math and understanding is right)? I am curious if those numbers are close to what your answer suggests. To me, those seem to be important complements to the answer... – Vendetta Oct 18 '16 at 18:57
• @Vendetta You are right. Some numbers would be good! I will try to add them. And by the way, you are correct, it is exactly the hydration energy, only I'm not sure about the order of magnitude, I shall see what I can find online. – valerio Oct 18 '16 at 19:50
• Unless the very nice images here are your work would you cite their source? – pentane Oct 18 '16 at 20:59
• @pentane Added reference for the first one. The second one is my work. – valerio Oct 18 '16 at 21:16

At any given temperature and pressure, liquids (like water) and solids (like salt) are in equilibrium with their respective gaseous form. The vapour pressure describes this phenomenon. Often, solids have a lower vapour pressure as liquids. The higher the vapour pressure, the more volatile a compound is. And in the instance of pure sodium chloride / NaCl, this vapour pressure is much lower than the one of water. This alone contributes why heating sea water, and condensing the vapour, will first yield water.

In addition, to perform such a distillation, the Raoult's law applies, too. Basically, there is a total pressure $p$ in the system observed that is the sum of the individual partial pressures of the two components (water and salt). The partial pressure takes into account the vapour pressure the pure component possess (at this temperature and pressure) times the concentration of this component in the mixture (expressed as molar fraction):

$p (\mbox{total}) = p(\mbox{water}) * x(\mbox{water}) + p(\mbox{NaCl}) * x(\mbox{NaCl})$

where the molar fractions will add up to one. So if we simplify "seawater" as a solution of NaCl in water:

$x(\mbox{water}) + x({\mbox{NaCl}}) = 1$

The contribution of the total pressure $p$ by NaCl is lowered further as at ambient conditions, water may dissolve only 359 g NaCl per litre of water (reference), hence excluding a solution of higher concentration of NaCl than about 36 mass%.

This is a simplified view, leaving out any idea of "molecules" or "ions". Raoult's law is an idealization, too (hence, for example, azeotropes like ethanol/water). Of course, sea water contains more than NaCl; and to remove the later from the former, reverse osmosis is an energetically more favourable technique. Eventually, to distill NaCl, you need conditions of very high temperature and very low pressure, far distant from the ones to distill water.

• To distill NaCl you just let the water boil off. Put a quart of water on your stove add a cup of salt and let the water boil off. What did you get for vapor pressure of NaCl at 100 °C? – paparazzo Oct 16 '16 at 18:21
• @Paparazzi : The scenario of the Question is not that of salt poured into clean water. It is that of seawater. Separating the NaCl from the other non-water components of seawater does not happen if all one does is drive off the water. Put seawater at its boiling point (which moves as the non-water concentration changes) and what you get is not pure salt. – Eric Towers Oct 16 '16 at 21:50
• I think this answer would benefit from a comparison of some vapor pursues of typical salts vs. water. The difference is, of course, tremendous even without the Raoult's law effect – repurposer Oct 17 '16 at 1:49

Alcohol will boil off first
The boiling point is 78.37 °C
There will be some water but the ratio of alcohol in the vapor phase will be higher than the liquid. Water and alcohol are azeotropes so things flop at about 70% alcohol.
The size is not that much of a factor - hexane is volatile. wiki

Table salt (NA CL) is not much more volatile than a rock

melting point 801 °C (1,474 °F)
boiling point 1,413 °C (2,575 °F)
wiki

The lowest vapor pressure I could find is at 759.88 °K = 486.73 °C
1.88322997019E-006 kPa
Vapor Pressure of Sodium chloride

The vapor pressure of water is 101.32 kPa at 100 °C

Water is 53,801,182 times more volatile at 100 °C than salt at 486.73 °C

Salt is dissolved in water but that does not make the salt more volatile.

Basically no salt evaporates. With very rapid boiling you might get some entrainment. The vapor phase is going to have essentially not salts or minerals as they are solids are at 100 °C.

Take a quart of water and add a cup of salt. Put it on the stove and let all the water boil off. You will have a cup of salt.

• Comparing the melting point/vapor pressure of solid crystalline NaCl to the one of dissolved ions seems a bad idea to me. – Sanya Oct 16 '16 at 17:51
• @Sanya OK seems like a bad to you. That is the vapor pressure of salt. With heat applied to pure salt it does not disassociate before melting and/or evaporating. Those ions don't see the don't see the light of day (touch air). The ions only exists surrounded by water molecules. – paparazzo Oct 16 '16 at 18:02
• Although I kind of agree with @Sanya, this answer does illustrate intuitively why alcohol dissolved in water does evaporate (which was also mentioned in the question). It's a liquid at the relevant temperatures. – oerkelens Oct 17 '16 at 10:20
• @oerkelens Alcohol is not dissolved. It is mixed. It is the same molecule. – paparazzo Oct 17 '16 at 10:40
• @Paparazzi Wikipedia disagrees with your definition of solution. As above: en.wikipedia.org/wiki/Solution#Liquid_solutions – HardScale Oct 17 '16 at 12:57

There need to be clear distinctions between forces holding atoms (charged or neutral ) together and molecules (e.g., water, H$_2$O; Salt, NaCl) together. It is ionic or covalent bond that holds atoms together.

Covalent bond: Two processes for the formation of water molecule. 1) Two hydrogen atoms form H2 molecule, i.e,

H + H = H$_2$;

2) 2 hydrogen molecule and 1 Oxygen molecule form 2 molecules of water,i.e.,

2H$_{2}$ + O2 = 2H$_{2}$O)

Ionic bond:

Salt, NaCl if formed from 1 sodium ,Na$^+$ ion and 1 chlorine ion, Cl$^-$, i.e,

Na$^+$ + Cl$^-$ = NaCl

An attractive or repulsive force binds the molecules together (e.g., more than one H$_{2}$O molecules). This force is different from ionic or covalent bonds because it binds neutral molecules or atoms together and is called van der Waals force (wiki). This force ceases to exist when the molecules are further apart. During neutral and equilibrium condition, the water molecules and the dissolved salt get bound by van der Waals force. Additional thermal energy hence would create vibration and weakens this force between the molecules. Once this force weakens, H$_{2}$O (molar mass, ~ 18g) is much lighter than NaCl (molar mass, ~ 59g). Compare it with the molar mass of dry air (Nitrogen, and Oxygen and other trace gases), ~29g. So, water is lighter than air hence buoyed up into the air leaving behind the heavier NaCl.

In conclusion, it is not covalent or ionic bonding that brings about phase change. It is rather van Der Waals force (the force binding molecules together) that weakens when the molecules are not close enough by applying additional thermal energy (e.g., water starts to vaporize at 100 $^o$C and hence the H2O molecules fly into the air so easily because of their lighter weight than air).

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