% ionisation

[dinosaur93]

Given a reaction of weak acid in equilibrium, what happens to the % ionisation when it has been diluted?

Which side does the equilibrium favour to oppose the compensate for the change?

[kenhung123]

I think it increases?

[sin0001]

Yeah the % ionisation increases, because equilibrium shift to the right upon dilution (to re-establish equilibrium), so more acid is ionised

[Niskii]

Question relating to that...

For example, if you have ethanoic acid or another weak acid. When it's diluted with water, doesn't it decrease the concentration of everything? How do you know it shifts to the right to oppose the change, thus increasing the % ionisation?

Thanks

[sin0001]

If the concentration of everything is decreased that means, initially, the concentration fraction (Q) would be less than K, so to re-establish equilibrium the concentration fraction must increase to = K, so it shifts to the right because more products = higher Q

[nooshnoosh95]

isnt it also that a dilution is an addition of water which is a reactant
so the system tries to oppose this change by increasing the amount of product ie. favouring the forward reaction

[sin0001]

No the system only favours particular directions when there's a change in concentration. In this case, adding water won't change the concentration of water, because it's always a constant (around 56 M), so dilution does not cause a net forward reaction because of the change in concentration of the water, but to increase the concentration fraction & re-establish equilibrium.

[Niskii]

Ahh that makes sense. Thank you!