I have learnt that in order for dissolution to occur, the solute-solvent intermolecular forces must be strong enough to overcome the solute-solute and solvent-solvent forces. However if this is the case, why are non-polar substances such as methane able to partially dissolve in polar solvents such as water? Aren't the dispersion forces always too weak to break the pre-existing hydrogen bonds present in water?


1 Answer 1


In solids, reorientational and especially translational molecular motion are highly hindered. The motion of individual molecules is constrained about mean fixed positions within a regular lattice. Such regularity is absent in liquids, where significant thermal energy is associated with rearrangements in molecular position and orientation. This is of course one defining property of a liquid (and more generally of fluids). The corollary of this structural irregularity is that it leads to a greater number of potential "defects" in the network of interactions within the substance.

Consider water. The wikipedia explains that in crystalline ice, water forms hydrogen bonds with 4 other water molecules. Meanwhile,

[from] TIP4P liquid water simulations at 25 °C, it was estimated that each water molecule participates in an average of 3.59 hydrogen bonds. At 100 °C, this number decreases to 3.24 due to the increased molecular motion and decreased density, while at 0 °C, the average number of hydrogen bonds increases to 3.69.[35]

The net result of this structural irregularity is that even molecules that interact weakly through dispersion interactions (or, in principle, not at all) with water have a significant if small chance of persisting within the liquid. Therefore, it is less a matter of dispersion interactions "breaking" hydrogen bonds, and more of apolar solutes occupying "defects" in the liquid.

References in wikipedia article

[35] Jorgensen, W. L.; Madura, J. D. (1985). "Temperature and size dependence for Monte Carlo simulations of TIP4P water". Mol. Phys. 56 (6): 1381.


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