Why does the high charge density of a proton prevent it from existing alone in a solution? My chemistry lecture notes say that when an acid is in a solution, it can give off protons, but only when there is another molecule that can accept those protons. It is said that the reason why the proton cannot exist separately in the solution is the high charge density of the proton. What does this mean? Why high charge density, and not high charge itself?
 A: Your chemistry lectures look quite ambiguous.
Firstly, (arrhenious) acids do ionize and donate protons in solutions. This is because the water molecules are amphoteric and act as a base here. The water molecules accept a proton from the acid to form hydronium H30+ ions, also called hydrated H+ or H+(aq) ions.
Secondly, a bare proton/H+ cannot exist in a solution while an a sodium ion with the same charge can. Modelling the ions as a sphere of charge, by gauss' law, the electric field at the surface of the sphere is given by ke/r^2 where r is the radius of the sphere.

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*in a solution of Na+ ions, there is an electrostatic attraction between the Na+ ions and the negatively charged lone pairs of H2O molecules. Due to the relatively large cationic radius of sodium, the lone pair orbitals cannot get very close to the ion. As a result, only weak ion dipole interactions are formed  the two

*in a hypothetical solution of bare H+ ions, the charge is the same but the lone pairs can get a lot closer to the ion due to the smaller cationic radius and thus experience a much stronger electric force. This force is strong enough to result in a formal dative covalent bond between H2O and H+ to form H3O+

here, the deciding factor is the electric field strength at the surface, given by E=1/4πε (Q/R^2). Since the surface charge density is Q/4πR^2, the Electric field can be written as a function of the charge density and thus the deciding factor can be stated as the charge density
