How do we rationalize the aqueous solubilities of organic and inorganic materials?

1 Answer
Jan 9, 2017

Answer:

The important distinction is between #"molecular"# and #"non-molecular"# bonding.

Explanation:

Certain salts have some solubility in aqueous solvents: sodium chloride dissolves in water to some extent, even tho its structure is non-molecular, and the bonds that form the ionic lattice are strong. As a polar solvent, water is capable of solvating individual ions, and break up the ionic lattice to form so-called aquated ions, #"[Na(OH"_2)_6]^+#, which we would normally represent as #"Na"^+(aq)# and #"Cl"^(-)(aq)#. Organic solvents tend to offer poor solvation of ionic species, and as a result, such salts tend to have miniscule solubility in these media.

On the other hand, giant covalent structures, such as those of #C_"graphite"# or #C_"diamond"# have strong #C-C# bonds in an infinite array across the entire solid, for which solvation by water cannot energetically compensate. These materials ALSO have exceptionally poor to zero solubility in organic solvents for the same reasons.

Of course, some organic molecules, such as #"sucrose"#, have substantial water solubility due to the presence of polar, hydrogen-bonding groups on the sugar molecule. Other organic molecules, such as the alkane series, will have limited water solubility, as there are no polar groups on the hydrocarbyl chain that could interact with the water molecule.

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