# If HCl is a strong acid, what is its Ka?

Oct 8, 2016

Large...............

#### Explanation:

${K}_{a}$ is a measure of the extent of the reaction:

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

And ${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[C {l}^{-}\right]}{\left[H C l \left(a q\right)\right]}$ $=$ $\text{a large numerical number}$

Now all acid behaviour is moderated by the identity of the solvent. In water, the given equilibrium lies strongly to the right hand side.

Of course, water is an important solvent, and it is the only solvent of biological relevance, but it is not the only SOLVENT. If we wished to move to a more acidic medium we could use $\text{acetic acid}$ or $\text{hydrogen fluoride}$

HCl(g) + "solvent" rightleftharpoons ("solvent")H^+ +Cl^(-)("solvent")

In a more acidic regime, the equilibrium would lie more to the left. And in fact, if we examined the behaviour of the hydrogen halide gases in acetic acid, or sulfuric acid, we would indeed find that the lower hydrogen halides, say $H I$ or $H B r$ were stronger acids than $H C l$ inasmuch as their equilibria lay further to the right. Even in water, the acidity of $H F$ is markedly reduced. Why?