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:
- 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.
- 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.