Why does the degree of dissociation change when we dilute a weak acid even though the equilibrium constant is constant?

Question

$K$ represents the ratio of concentrations of molecules in a solution at equilibrium, which means that $Q_\mathrm{r}$ (that ratio at any given point) looks to be identical to $K$. In other words, the molecules in that solution react accordingly so that they reach equilibrium and the ratio of their concentrations is equal to $K$.

If $K$ is large enough (bigger than $10^4$ in my curriculum), this means that the the concentration of the reactants is almost zero. In other words, the equilibrium position of that solution looks very much like a reaction that went to completion.

The more we dilute an acidic/basic solution, the higher the degree of dissociation, even though $K$ stays the same. So, does that mean that the more we dilute a solution the harder it is for it to reach the point of equilibrium for that specific molecule/solution or what?

For instance, say you found $K$ of solution to be $10^{-5}$. This means that when the reaction happens there are lots of reactants left, and not much products produced, which means that the degree of dissociation is low. But the more we dilute a solution, the closer it gets to a "complete reaction" (if you pour a small amount of weak-acid molecules into a large tank of water, it's certain that all of the weak-acid molecules are going to react with the water, i.e. the degree of dissociation approaches $100\%$).

So, how come $K$ can be independent of the initial reactants concentrations, and tell if a reaction was complete or not, when the "completion" of a reaction (the degree of dissociation) depends on the initial concentrations of reactants?

Answer 1

See my comment above.

As a numerical example, take acetic acid ($\ce{AcOH}$), which has $K_\mathrm{a} = 1.8 \cdot 10^{-5}$.

This means that:

$$K_\mathrm{a} = \frac{[\ce{AcO-}][\ce{H+}]}{[\ce{AcOH}]}$$

And the total nominal concentration of acid is:

$$C_\mathrm{a} = [\ce{AcOH}] + [\ce{AcO-}]$$

Combining these two equations, you can see that the % of dissociated acid is:

$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} = \frac{K_\mathrm{a}}{K_\mathrm{a} + [\ce{H+}]}$$

i.e. not a constant, but a value that depends on the $\mathrm{pH}$. So no, there is no need for the 'degree of completion' of the reaction to be a constant.

You can even calculate the explicit concentrations of all the species.

Given that:

$$C_\mathrm{a} = \left(\frac{1}{[\ce{H+}]} + \frac{1}{K_\mathrm{a}}\right) \cdot ( [\ce{H+}]^2 - K_\mathrm{w}^2)$$

you can see that, for $\mathrm{pH} = 5$ you need:

$$C_\mathrm{a} = \left(\frac{1}{10^{-5}} + \frac{1}{1.8 \cdot 10^{-5}}\right) \cdot \left((10^{-5})^2 - (10^{-14})^2\right) = 1.5554 \cdot 10^{-5}$$

Then:

$$[\ce{AcOH}] = \frac{C_\mathrm{a} \cdot [\ce{H+}]}{K_\mathrm{a} + [\ce{H+}]} = 5.555 \cdot 10^{-6}$$

$$[\ce{AcO-}] = C_\mathrm{a} - [\ce{AcOH}] = 9.999 \cdot 10^{-6}$$

So the % of dissociated $\ce{AcOH}$ is:

$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} \approx 64 \%$$

Now repeat for $\mathrm{pH} = 3$:

$$C_\mathrm{a} = \left(\frac{1}{10^{-3}} + \frac{1}{1.8 \cdot 10^{-5}}\right) \cdot \left((10^{-3})^2 - (10^{-14})^2\right) \approx 0.05656$$

$$[\ce{AcOH}] = \frac{C_\mathrm{a} \cdot [\ce{H+}]}{K_\mathrm{a} + [\ce{H+}]} \approx 0.05556$$

$$[\ce{AcO-}] = C_\mathrm{a} - [\ce{AcOH}] \approx 0.001$$

$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} \approx 1.8 \%$$

So you see, in a more dilute mixture the $\mathrm{pH}$ is higher, and the dissociation reaction of acetic acid is 'more complete', whereas at higher concentration the pH is lower, and there is proportionally less acetate anion.

All of these systems are at equilibrium.

Answer 2

$K$, the equilibrium constant, is independent of the composition of the system. It simply describes the preference of the system for reactants or products.

On the other hand, there is the reaction quotient, $Q$, which describes the position of the system relative to equilibrium. $Q$ is the value that is variable and that you adjust to be closer to $K$. Technically, $K$ can also be computed from the standard free energy change of the reaction, $\Delta G ^{\circ}$.

What's confusing is the $Q$ and $K$ are computed in the exact same way, with the sole difference that for $Q$, you plug in the values you have right now, and for $K$, you plug in the values at any equilibrium.

If $Q=K$, you are at equilibrium.

If $Q < K$, you need to push the reaction towards more products.

If $Q > K$, you need to push the reaction towards more reactants.

When you dilute a solution (or make other changes), you may change the value of $Q$, which may put you farther from equilibrium, but the opposite may also be true. It depends on the specific reaction and the form of the equilibrium expression.

EDIT:

Your confusion is because you're confusing $Q$ measured at different points in time. At the end of the reaction $t = \infty$, $Q = K$. At the beginning of the reaction, $Q_{0} < K$. Over time, as the reaction goes towards equilibrium, the value of $Q$ will get closer to $K$. If at the beginning of the reaction, you dilute the solution, you may get a different value of $Q$. If it's smaller, then you have farther to go to get to equilibrium. But again, this depends on the molecularity of the reaction.

$$\ce{A + B -> C}$$

$$\ce{A -> C}$$

$$\ce{A -> C + D}$$

Here are three different reactions.

In the first reaction, at the beginning of the reaction, with $Q < K$, dilution pushes you towards equilibrium because reactant concentrations enter into the equilibrium expression twice.

$$Q = \frac{\ce{[C]}}{\ce{[A][B]}}$$

If you halve all three concentrations, the value of $Q$ will get bigger.

In a similar vein, the second equation is unaffected by dilution. In fact, at equilibrium, dilution preserves equilibrium.

And for reaction 3, dilution pushes you farther away from equilibrium.

For all three reactions, at the end of the reaction, you will have $Q = K$. But for reactions 1 and 3, dilution at equilibrium changes the value of $Q$ and the reaction will need to adjust again to reach equilibrium.

Answer 3

If K is large enough (bigger than $10^4$ in my curriculum), this means that the the concentration of the reactants is almost zero.

This statement is not always true - it depends on the stoichiometry of the reaction.

For the reaction $$\ce{A(aq) <=> B(aq)}$$

the concentration of A is much smaller (ten thousand times) than that of B if $K = 10^4$, and this is independent of diluting the solution. It's fair to say that there is almost no A compared to B.

For the reaction $$\ce{4A(aq) <=> 4B(aq)}$$

the concentration of A is just ten times smaller than that of B if $K = 10^4$, and this is also independent of diluting the solution. However, there is quite a bit of A compared to B, and it is a bit misleading to say the the concentration of A is almost zero.

Both of these reactions had the same number of reactant species as product species (all in the same solution, so all equally affected by dilution).

The more we dilute an acidic/basic solution, the more the reaction is "complete", the more the reactants disappear, but K stays the same.

Here, "complete" would mean the absence of reactants, i.e. a reaction that goes to completion. Weak acids are defined by not completely dissociating, in contrast to strong acids. The general reaction for a weak acid is:

$$\ce{AH(aq) <=> H+(aq) + A-(aq)}$$

Notice that there are more product particles than reactant particles. For those type of reactions, dilution favors the products. In fact, if $K = 10^-5$ for this reaction and all concentrations are $\pu{10^-5 M}$, the reaction is at equilibrium. This means even though the equilibrium constant is much smaller than one, you still can have reactants and products at the same concentrations.

On the other hand, if there are more reactant particles than product particles, dilution has the opposite effect. For a complex formation reaction between a metal M and a ligand L

$$\ce{M(aq) + 4 L(aq) <=> ML4(aq)}$$

dilution will cause the complex to fall apart. If $K = 10^4$ and the free ligand concentration is one molar, there will be a lot of complex and very little free metal. On the other hand, if the free ligand concentration is one millimolar (0.001 M), there is hardly any complex and most of the metal will be in the form of the free metal.

Why does the degree of dissociation change when we dilute a weak acid even though the equilibrium constant K is constant?

As the examples above illustrated, this is because for these specific reactions, the sum of exponents for the products is higher than that of the reactants in the equilibrium constant expression. As you dilute, Q will decrease, and the reaction will go forward to reach K again. At infinite dilution, the product will be favored no matter what the value of the equilibrium constant is.

If you want to have a general statement about what the value of K means, it would be something like "as K increases, the equilibrium concentration of products will increase and those of reactants will decrease".

Answer 4

A really handwavy explanation would be that the system, attempting to maximize its entropy, will prefer dissociated states as the system dilutes since there will be an increasing number of microscopic configurations with atoms $A$ and $B$ far apart (i.e. molecule $AB$ dissociated) as opposed to close together (i.e. atoms $A$ and $B$ bound together into molecule $AB$).