Predicting Ka/Kb values using values for solubility in water Given that $K_\mathrm{a}$ values are determined by the position of equilibrium for a species such as $\ce{HCl}$ where $K_\mathrm{a} = \ce{\frac{{[H^{+}]}{[Cl^{-}]}}{[HCl]}}$, I was wondering whether it would be possible to work out $K_\mathrm{a}$ values using 'Solubility in Water' values.

For example: $\ce{HCl}$ has a solubility of $\pu{720g/L}$ in water at 20 degrees Celsius. Is this possible or is there no direct correlation between solubility in water and $K_\mathrm{a}$?
Your example, $\ce {HCl}$, is considered to be a strong acid. Thus, it it dissociate $100\%$ in water. Thus, the molarity of the given solution (if it exists):

$$\frac{\pu{720 g}}{\pu{1.0 L}}\times\frac{\pu{1.0 mol}}{\pu{36.5 g}} = \pu{19.7 mol L^-1}$$
That means, since $\ce {HCl (aq) -&gt; H+(aq) + Cl-(aq)}$, $[\ce {H+}]$ of the solution is $19.7$, and $[\ce {Cl-}]=19.7$ as well so you can find $\mathrm{pH}$ of the solution. But, since no $\ce {HCl}$ remains in the solution, $K_\mathrm{a}$ has no meaning.