The Relationship Between Dilution and Dissociation

H: Some Definitions

Be advised the equilibrium constant, $K$ ⊕The equilibrium constant relates reactant and product concentrations, representing the position of equilibrium for a reaction., of a weak acid:

$K_a = \frac{[\mathrm H^+][\mathrm A^-]}{[\mathrm H\mathrm A]}$

where $\mathrm H^+$ is the hydrogen ion, $\mathrm A^-$ is the conjugate base, and $\mathrm H\mathrm A$ is the weak acid. If $x$ is the concentration of the hydronium, then because $\mathrm H\mathrm A$ is monoprotic,

$x = [\mathrm H^+] = [\mathrm A^-] ,$

and because $\mathrm H\mathrm A$ is a weak acid,

⊕Where $[\mathrm H\mathrm A]_0$ is the initial acid concentration.

$[\mathrm H^+] = [\mathrm A^-] = x \ll [\mathrm H\mathrm A]_0\ .$

This leads to

$[\mathrm H\mathrm A] = [\mathrm H\mathrm A]_0 - x \approx [\mathrm H\mathrm A]_0\ \leadsto$

$K_a \approx \frac{x^2}{[\mathrm H\mathrm A]_0} = \frac{[\mathrm H^+]^2}{[\mathrm H\mathrm A]_0}$

Exciting stuff, I know! But this is where the fun starts. Firstly, note two things:

The equilibrium constant is roughly the square of the hydronium concentration, divided by initial acid concentration.
The reaction of the weak acid can be said to have some active describer ∂ representing the percent of the acid dissociated at equilibrium.

Okay. So consider the following:

He: What will happen to ∂ if we dilute the solution?

Hastily, most will say ,

Well, that's obvious, clearly the percent dissociation must decrease!

Why?

If it's diluted, the solution is less concentrated with hydronium ions, so a lesser percentage of the acid dissociates.

In a haste, this makes sense - a sort-of 'duh' derivation. Unfortunately, it's a fallacy.

Li: Percent dissociation is inversely proportional to the concentration.

With some thought, we can understand why this is - first intuitively, then mathematically. When we dilute an acid, we are essentially adding more area upon which the acid can react, thus leading to a higher rate of dissociation. To prove this:

Let's say we dilute our solution by a factor of ten. The reaction quotient $Q$ ⊕The reaction quotient describes the direction a reaction will proceed to establish equilibrium. can then be described as

$Q = \frac{\frac{[\mathrm H^+]}{10}\frac{[\mathrm A^-]}{10}}{\frac{[\mathrm H\mathrm A]}{10}} \approx \frac{\frac{[\mathrm H^+]}{10}\frac{[\mathrm H^+]}{10}}{\frac{[\mathrm H\mathrm A]_0}{10}} = \frac{1}{10}\frac{[\mathrm H^+]^2}{[\mathrm H\mathrm A]_0} = \frac{1}{10}K_a$

$Q \approx \frac{1}{10}K_a$

Because the reaction quotient is an order of magnitude less than the equilibrium constant, the reaction will shift to the right ⊕The acid will dissociate further, increasing the numerator of the reaction quotient. in order to reach equilibrium.

QED - diluting a weak acid shifts the concentration of a reaction's products to the left, leading the acid to dissociate more in order to reestablish equilibrium, thus increasing the percent dissociation.

Phew!