K is a ratio of reaction probabilities at equilibrium. As a side interpretation, pK = -log K is the pH of the solution in which the associated and dissociated concentrations are equal.
Layra-chan
The constant K represents a probability, not a proportion. You're not measuring how much [H+] and [A-] you have relative to the amount of [HA] you have; rather, you're measuring how likely it is that a given HA molecule will dissociate into an H+ and an A-.
That's not wrong, but it is misleading. It is indeed a ratio (which I gather what you meant instead of proportion), but of probabilities:
K = P(HA dissociates into H,A)/P(H,A yields HA)
rather than just the probability term in the numerator alone. To make sense of this, we have to fix some volume, and talk of the probabilities of the described event happening in this volume.
Allow me a bit of hand-waving. For HA⇄H+A, define the probabilities {p = P(HA dissociates), q = P(H,A yields HA)}, in some fixed small volume. Suppose we want to look at the rates at which these events there--they proportional to p[HA] and q[H][A], with the same volume-dependent constant of proportionality, as long as independence is assumed. Hence their ratio is ([H][A]/[HA])(q/p) = K(q/p). But at equilibrium, this ratio must be 1 as the reaction rates are equal on average; hence K = p/q.
For a reaction of the type A+B⇄X+Y, we don't have to handwave quite as furiously, as
K = P(collision of A,B reacts)/P(collision of X,Y reacts),
where "collision" means meeting in some small volume, but without having to include the probability of being found in said volume in the first place. (Although that's kind of obvious, since we're approximating the situation by the assumption of independence.)
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