a math question dealing with fractions - please help

[freeziepleasy]

If you have two things, say 4 apples and 2 oranges, and you wanted to know how many more apples you have than oranges, you'd divide 4 apples/2 oranges and get that there are twice as many apples as there are oranges.

In chemistry, an acid dissociates into a proton and its conjugate base:

HA <--> H+ + A-

If you start off with 0.04 M of HA, then at equilibrium, the concentrations of each specie are as follows,

[HA] = 0.04 - x
[H+] = x
[A-] = x

(I know a lot of people in this guild are math majors or beyond that, so I think most of you already make sense of what this is already without needing any chemistry explanation. However, if chemistry explanation is needed, I'll try my best... but I really doubt anyone has trouble.)

The dissociation constant (how easily an acid can give up a proton), K, is defined as

K = [H][A-] / [HA]

IF... the dissociation constant is comparing the concentrations of [H] and [A-] to [HA], then why would they want to multiply [H] and [A]? Wouldn't adding them be a better choice when trying to compare?

I know where to use the dissociation value, and the definition, but when it comes to the math, I have no idea why they have [H] multiplied with [A]. What does the fraction [H][A-] / [HA] even tell me??

Thanks to anyone who answers this.
Again, I don't think any chemistry explanation is needed because I'm just asking for an explanation to what the fraction means...

And as for the apple/orange example, it was supposed to be a bit analogous to the chemistry equation and the comparison between species in the reaction. I wrote it in there so that no one thinks that I'm asking a chem. question, but rather, a math question related to chemistry.  

[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-.
The probability that a solution full of protons will accept another proton depends on how many protons it already has, i.e. it depends on [H+]. Similarly for the conjugate base. So we can say that the value of [H+] at equilibrium corresponds to the probability that the solution will accept a proton. Similarly, we say the same of the conjugate base, i.e. that [A-] corresponds to the probability that the solution will accept a molecule of the conjugate base.
Since the dissociation of a proton from its conjugate base adds both a proton and a molecule of conjugate base, to get the probability of this dissociation occurring you need to calculate the probability of the solution accepting both a proton and a molecule of conjugate base. Since these two events are mostly independent, you get that you have to multiply the probabilities. Hence you multiply [H+] and [A-], instead of adding them.  

[freeziepleasy]

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-.
The probability that a solution full of protons will accept another proton depends on how many protons it already has, i.e. it depends on [H+]. Similarly for the conjugate base. So we can say that the value of [H+] at equilibrium corresponds to the probability that the solution will accept a proton. Similarly, we say the same of the conjugate base, i.e. that [A-] corresponds to the probability that the solution will accept a molecule of the conjugate base.
Since the dissociation of a proton from its conjugate base adds both a proton and a molecule of conjugate base, to get the probability of this dissociation occurring you need to calculate the probability of the solution accepting both a proton and a molecule of conjugate base. Since these two events are mostly independent, you get that you have to multiply the probabilities. Hence you multiply [H+] and [A-], instead of adding them.

Thank you very much!! smile  

[VorpalNeko]

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.)