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.
