If $\mathrm{HCl}$ is a weaker acid than $\mathrm{HBr}$, why is $\mathrm{ClCH}_{2} \mathrm{COOH}$ a stronger acid than $\mathrm{BrCH}_{2} \mathrm{COOH}$ ?
Solution To compare the acidities of $\mathrm{HCl}$ and $\mathrm{HBr}$, we need to compare the stabilities of their conjugate bases, $\mathrm{Cl}^{-}$and $\mathrm{Br}^{-}$. (Notice that an $\mathrm{H}-\mathrm{Cl}$ bond breaks in one compound and an $\mathrm{H}$-Br bond breaks in the other.) Because we know that size is more important than electronegativity in determining stability, we know that $\mathrm{Br}^{-}$is more stable than $\mathrm{Cl}^{-}$. Therefore, $\mathrm{HBr}$ is a stronger acid than $\mathrm{HCl}$.
In comparing the acidities of the two carboxylic acids, we again need to compare the stabilities of their conjugate bases, $\mathrm{ClCH}_{2} \mathrm{COO}^{-}$and $\mathrm{BrCH}_{2} \mathrm{COO}^{-}$. (Notice that an $\mathrm{O}-\mathrm{H}$ bond breaks in both compounds.) The only way the conjugate bases differ is in the electronegativity of the atom that is drawing electrons away from the negatively charged oxygen. Because $\mathrm{Cl}$ is more electronegative than $\mathrm{Br}, \mathrm{Cl}$ exerts greater inductive electron withdrawal. Thus, it has a greater stabilizing effect on the base that is formed when the proton leaves, so the chloro-substituted compound is the stronger acid.