# For a given strong acid HA, is the concentration of A^- likely to be high or low at equilibrium?

Feb 23, 2017

This is a good question, and it is clear that $H Y$ is the STRONGER ACID..........

#### Explanation:

The strength of an acid, $H A$, in an aqueous medium, is dictated SOLELY by the extent the given equilibrium lies to the RIGHT (as we face the page):

$H A \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + {A}^{-}$

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 ${H}_{3} {O}^{+} \left(\text{hydronium ion}\right)$.

$\left(a\right)$ In the diagram, for $H Y$, there is a greater number (i.e. concentration) of such hydronium ions, and thus the equilibrium lies farther to the right. On this basis, $H Y$ is a stronger acid than $H X$ in that there are more hydronium ions..........

$\left(b\right)$ And if $H Y$ 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.

$\left(c\right)$ Given equal concentrations of $H X$ and ${Y}^{-}$, since ${X}^{-}$ is the stronger base, the GIVEN equilibrium,

$H X \left(a q\right) + {Y}^{-} r i g h t \le f t h a r p \infty n s {X}^{-} + H Y \left(a q\right)$

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.