Predicting Ka/Kb values using values for solubility in water
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.