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Acetic acid is a weak acid. It is in equilibrium with acetate and hydronium ions in aqueous solution:

$$\ce{CH3COOH(aq) + H2O(l) <=> CH3COO-(aq) + H3O+(aq)}$$

Ostwald's law states that the degree of ionisation of a weak acid increases upon dilution (in quantitative terms: as the square root of the volume).

Consider two scenarios:

  1. We start out with 1.001 moles of water and 1.000 moles of acetic acid (so water is the solvent and acetic acid is the solute), and dilute it with water to twice the volume.
  2. We start out with 100 mM acetic acid, and dilute it with water to twice the volume.

How do the rates of the forward and reverse reaction change in these two scenarios, and how do these changes help to explain Ostwald's dilution law? (I know that once equilibrium has been reached after dilution, the forward and reverse rates will be equal again, so I am asking about the rates after dilution, but before changes to concentrations by the reaction going towards equilibrium. If mixing is slower than the acid-base reactions and you can't do the actual experiment, treat it as a thought experiment.)

This question was inspired by Why degree of dissociation/ionisation affected by dilution? . I am specifically interested in addressing the following ideas from the OP of that question:

as there are more water molecules so the chances that say a CH3COO- ion attracting an H+ from it's surrounding water should increase

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since we are adding more water to it and since water has high dielectric constant so it'll ionise CH3COOH more

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since water is more so it'll hydrate the broken ion and surround it,make it harder for them to recombine

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What motivates or pushes [the acid] to dissociate more?

If it helps, it would be fine to discuss what happens with a single acetic acid molecule and a single acetate anion upon dilution.

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    $\begingroup$ How about Le Chatelier's principle? To dilute, you need to add water to LHS of the equilibrium equation. $\endgroup$ – Mathew Mahindaratne Jan 31 at 16:18
  • $\begingroup$ Yes, but this is not the textbook Le Châtelier situation where you change just one concentration. By adding water, all of the concentrations change. In fact, in scenario 2 the concentration of water is the one that changes the least. $\endgroup$ – Karsten Theis Jan 31 at 16:38
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Let's consider case 2. To make the math easier, I'm going to assume that acetic acid and any mixture of acetic acid/water has the same density as water. Thus, the molarity of pure acetic acid is approximated as 16.7 mol/L. That isn't exactly right, but conceptually that doesn't change anything. Also, since the mole fraction of water is a bit less than 1, let's use its actual molarity of approx 55.6 M.

So, to start, we have 0.1 M of acetic acid in water, which means only ~6 mL/L water, and we can assume the water is still 55.6 M. And assume a total volume of 1 L. At equilibrium, we know that the forward and reverse rates of ionization are the same, so

$k_f[\ce{H2O}][\ce{AcOH}]=k_r[\ce{AcO-}][\ce{H3O+}]$

Based on the published $K_a$, we can calculate that [AcOH]~0.1 M and [AcO-]=[H3O+]=0.0013 M.

If we instantly add an additional 1 L water to double the volume, the AcOH, AcO- and H3O+ concentrations all drop by half, but the water concentration stays at about 55.6 M. Intuitively, it should be clear that the result is a much greater drop in the rate of reverse reaction than the rate of the forward reaction. More rigorously, we relative rate of the forward reaction is (55.6x0.05)/(55.6x0.1)=0.5, while that of the reverse reaction is (0.00065)^2/(0.0013)^2=0.25. Since the rates were equal before dilution, the forward rate is now approx twice that of the reverse, and in order to return to equilibrium, the AcOH concentration must get even lower and the ion concentrations must come back up.

Conceptually, this comes about because the chance of an acetate ion and a hydronium ion encountering each other diminishes as the solution gets more dilute. The dissociation of the acid in bulk water, on the other hand, is just a function of how many undissociated molecules are present.

With regard to the quotes you included, some brief responses are

as there are more water molecules so the chances that say a CH3COO- ion attracting an H+ from it's surrounding water should increase

The concentration of water is not the same as H3O+, which is decreased by dliution. Water self-ionization will partially compensate, but not very much.

since we are adding more water to it and since water has high dielectric constant so it'll ionise CH3COOH more

The dielectric constant of water does have an influence on the $K_a$, but it doesn't change when you add more water.

since water is more so it'll hydrate the broken ion and surround it,make it harder for them to recombine

If we start with dilute acetic acid in water, we assume that the acetate ions are far enough apart from each other that they do not interfere with each other's solvation, so additional dilution doesn't factor in.

What motivates or pushes [the acid] to dissociate more?

I think it's more appropriate to say that it reassociates less rather than dissociates more, but that's mostly semantics.

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  • $\begingroup$ I think you are considering case 2 (you said 1). I don't understand why you say the concentration of water goes from 50 M to 53 M when the (total) acetic acid concentration merely decreases from 100 mM to 50 mM. Otherwise, excellent points. $\endgroup$ – Karsten Theis Jan 31 at 19:30
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    $\begingroup$ Thanks for pointing that out. I was being dumb there and counting 100 mL of acetic acid instead of 100 mM. The 100 mM acetic acid is only ~ 6 mL, so the water is essentially 55 M in both. I'll fix that and the case 1 typo $\endgroup$ – Andrew Jan 31 at 20:02
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Let's consider case 1. First, we will calculate the concentrations before and after dilution. Then, we will estimate the activities of the solutions. Finally, we will discuss what will happen to the rates of dissociation and association.

  1. For simplicity, let's say we are starting with an equimolar mixture of water and acetic acid. This corresponds to about 77% of acetic acid by mass, a solution with a density of about 1.07 g/mL 1. The concentration of acetic acid is about 13.7 M, and the concentration of water is the same (equimolar). If we dilute with water to twice the volume, the concentration of acetic acid will be about half, i.e. 6.8 M (if volumes are roughly additive). The concentration of water will increase significantly to about 34 M. So the dilution cuts the acetic acid concentration in half while it almost triples the water concentration
  2. According to Hansen et al (1955), the activity coefficients of acetic acid and water are both about 1.3 for an equimolar mixture (with the pure liquid as the reference state where the activity coefficient is one). For the diluted solution (molar ratio of acetic acid to water of about 1:5), the activity coeffient for acetic acid is about 1.6 and the one for water about 1.05. Because the other two relevant species, hydronium and acetate, are present in low concentrations and there are no other ionic species at high concentration, their activity coefficients probably are close to what they are in a dilute aqueous solution, but I have not found experimental data supporting this assumption.
  3. Different from case 2, where the forward rate was predicted to decrease by a factor 2, in case 1 we can predict that the forward rate will increase. The concentration of acetic acid drops by a factor two, but the concentration of water increases by more than a factor two. Similar to case 2, the reverse rate is predicted to decrease by a factor of about four.

What motivates or pushes [the acid] to dissociate more?

In case 1, it is correct to say acetic acid dissociates faster because the concentration of water is significantly higher after dilution. This contributes to the higher degree of dissociation upon dilution. At the same time, the rate of association drops by a factor of about four, a much larger change.

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  • $\begingroup$ According to an NMR study, the pH of water acetic acid mixtures is minimal at about equimolar ratio. $\endgroup$ – Karsten Theis Feb 3 at 3:00

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