# How does the strength of an acid relate to the magnitude of the equilibrium constant K_a?

Jan 7, 2017

${\text{HA"(aq) + "H"_2"O"(l) rightleftharpoons "H"_3"O"^+ + "A}}^{-}$

#### Explanation:

The forward reaction represents the ionization of the acid $\text{HA}$ in aqueous solution. This is an equilibrium reaction, and represented numerically by the expression:

${\text{K}}_{a}$ $=$ $\left(\left[\text{H"_3"O"^+]["A"^(-)(aq)])/(["HA} \left(a q\right)\right]\right)$,

where ${\text{K}}_{a}$ is the so-called $\text{acid dissociation constant}$.

Strong acids, ${\text{HX" ("X"!="F"), "H"_2"SO"_4, "HClO}}_{4}$, have high values of ${\text{K}}_{a}$ so that there tends to be minimal concentration of free acid $\text{HA}$ at equilibrium.

And a base in water competes for the proton:

${\text{A"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HA"(aq) + "HO}}^{-}$

${\text{K}}_{b}$ $=$ $\left(\left[{\text{HA"]["HO"^(-)(aq)])/(["A}}^{-} \left(a q\right)\right]\right)$

${\text{K}}_{b}$ is the so-called $\text{base association constant}$.

${\text{K}}_{a}$ is generally quoted, because it can be shown that under standard conditions, $p {\text{K"_a+p"K}}_{b} = 14$