Why do ionic substances dissolve in polar solvents, when ionic bonding is so much stronger than H-bonding?

Question

I know there will be attraction between the dipoles and the ions, but isn't the ionic bond stronger this?

Answer 1

Lets say the ionic molecule is NaCl and the solvent is polar, e.g. water.

Now, if a substance is polar, it has both + and - poles, right?

NaCl is formed from Na+ and Cl-

H is slightly positive, because O is slightly negative. (because of electronegativity)

So O attracts Na+ and H attracts Cl-. Since a water molecule is bent (because of lone pairs of electrons, or those not used in bonding)

The molecule gets "torn" apart. (--- indicate attractive forces)

H  H----Cl-

\  /

O  ---Na+.

So you can see the ions are separated, and thus they can dissolve in water (they are no longer solids).

Whereas consider a non polar molecule, an organic molecule, in such a case, electronegativity of O and H would be of no use, moreover non-polar tends to stay away from polar, so there is no interaction, or more accurately, there is a hydrophobic interaction or an interaction that 'fears' water. The hydrogen bonds cannot be broken as there is no force acting on it.

Interaction of NaCl and H2O is a hydrophilic interaction, just FYI.

Answer 2

Soham gave a good description of *what* happens when you dissolve an ionic compound in a polar solvent, but didn't actually tell you *why*, which seems to be what you're after.  

That's actually a really good question, and of course it's more interesting than even your question suggests, because sometimes the answer is "it doesn't": phosphates, carbonates, sulfates, and hydroxide salts are generally insoluble in water, and don't dissolve.  Among those that do, most of them dissolve more readily at higher temperature, and precipitate back out at lower temperature.

Consider Soham's case.  What happens?

NaCl(s) --H2O--> [Na]+(aq) + [Cl]–(aq)

On the left you have strong ionic interactions, each Na+ surrounded by six Cl- and each Cl- surrounded by six Na+ (NaCl is a face centred cubic lattice of chlorides with sodium ions in each octahedral hole).  You also have pure water, with hydrogen bonding among all the water atoms.  When it dissolves, you give up the Na+ -- Cl– interactions, and you lose some of the H-bonding among some of the water molecules, because you're sticking ions in between them.  Both those things are enthalpically uphill, which works against dissolving a compound.  

But in return, you get the new interactions that Soham described: strong polar-ionic attractive forces between each ion and a bunch of water molecules.  In fact, each ion gets surrounded by a cloud of six waters, so that Na+ has six water O atoms bound to it, and each Cl- has six "H ends" of water molecules surrounding it.  (And there are secondary and tertiary solvent spheres beyond that, where the H atoms of the six waters around the Na+ in trun get surrounded by the O atoms of more waters oriented towards them, and so on.)  These new interactions are enthalpically *down*hill, which helps to dissolve the compound.

So you're trading ion-ion interactions for ion-dipole interactions.  If the dipoles in the solvent are particularly strong, or the ionic forces in the solid not especially so, then dissolving the compound is enthalpically favoured.  Usually, that's not the case.  How does it dissolve, then?

There's another driving force, that of entropy.  The dissolved solution has a higher entropy than the ordered solid and the pure solvent, so dissolving a solid is favoured entropically.  That counteracts the uphil enthalpy, and becomes more important at high temperatures, so most materials become more soluble as temperature increases.

So, basically, if a substance dissolves, you trade strong ion-ion forces in the ionic solid (and some of the intermolecular attractions in the solvent) for ion-dipole forces in the solution which are, usually, not quite as strong, but enough to get you close to breaking even.  Entropy, the favourability of greater disorder in the solution, gets you the rest of the way and makes dissolution overall favourable.  In many cases, it doesn't work, the ion-ion are too strong, the dipole-ion too weak, and it doesn't dissolve.

An aside:

"moreover non-polar tends to stay away from polar, so there is no interaction, or more accurately'', there is a hydrophobic interaction or an interaction that 'fears' water"

This is a common misconception about what really causes a so-called "hydrophobic" interaction (a very stupid name that contributes to the confusion).  Add hexane (or gasoline, or oil, or liquid silicone, or whatever) to water.  The hexane spreads out evenly, forming a uniform layer over the surface of the water, because every hexane molecule wants to be next to a water molecule.  Non-polar does NOT "stay away" from polar, it tries to get as close as possible, and the nonpolar molecules separate from each other in order to MAXimize the nonpolar-to-polar interactions.  Which makes sense.  A dipole-dispersion interaction is stronger and more favourable than a dispersion-dispersion, so the system makes as many of them as possible.

So why doesn't hexane dissolve?  It wants to.  It would love that, because then each hexane molecule would be surrounded by better polar-dispersion interactions.  But water doesn't let it, because it would also involve breaking apart the really strong water-water interactions, which is too uphill.  You'd have to separate hexane molecules from each other (dispersion-dispersion, weak), separate water molecules from each other (H-bonding, very strong), and surround hexanes with waters (polar-dispersion, stronger than disp-disp, but not H-bonding).  The new interactions you get aren't enough to overcome the ones you'd have to break, so hexane cannot penetrate the water.  Not because it "fears" or avoids the water -- it doesn't -- but because it's shunned by the water.