[fletcoli]

Consider a solution of propanoic acid in water. Describe what happens to the Ka and degree of ionisation if the solution evaporates to half its volume at 25°C.

I'm struggling with this question. My confusion lies in that: halfing voume doubles concentration, yet it doesn't double 1 side instead both, therefore wouldn't you expect no change to the degree of ionisation as both sides concentrations are increasing by the same amount?

Here is my attempt in understanding it:

In propanoic acid's dissolution

C2H5COOH(aq) + H2O(l)  C2H5COO(aq) + H3O(aq)

I thought removing water from this system would actually have no effect as water exists as a liquid and so it essentially has a concentration of 1. Is that correct?

But is this a sound explanation: The degree of ionisation decreases as the increase in concentration is more significant for the product side that has 2mols vs 1 on reactant side. This then causes the reverse reaction rate to become faster then the forward reaction rate shifting the equilibrium left. Also is this true: Decreasing the amount of water slows the forward reaction rate as there are less water molecules for the propanoic acid to react with.

[Hunter2]

halfing volume doubles concentration, yet it doesn't double 1 side instead both, therefore wouldn't you expect no change to the degree of ionisation as both sides concentrations are increasing by the same amount

No the half volume of water is gone. Check the law of mass action. Here water is included in the equation.
Consider what you would have if all water is gone.

[Borek]

halfing voume doubles concentration

Risky statement.

Yes, if you remove half of the solvent, concentration of the solute goes up. Same can't be said about the solvent though.