The strength of an acid, HA, in an aqueous medium, is dictated SOLELY by the extent the given equilibrium lies to the RIGHT (as we face the page):
HA(aq) + H_2O(l) rightleftharpoons H_3O^(+) + A^-
And thus for strong acids, the equilibrium lies to the right, and more of the solvent molecules are conceived to be protonated to give the H_3O^+("hydronium ion").
(a) In the diagram, for HY, there is a greater number (i.e. concentration) of such hydronium ions, and thus the equilibrium lies farther to the right. On this basis, HY is a stronger acid than HX in that there are more hydronium ions..........
(b) And if HY is the stronger acid, it is clear that Y^- is the WEAKER conjugate base (why? again because Y^- competes poorly for the proton). So X^- is the STRONGER base.
(c) Given equal concentrations of HX and Y^-, since X^- is the stronger base, the GIVEN equilibrium,
HX(aq) + Y^(-) rightleftharpoons X^(-) + HY(aq)
will lie to the LEFT as written.......
This question is certainly non-trivial, and would tax most 2nd year/3rd year inorganic chemistry students......anyway, please criticize my reasoning if there is an issue.
