[qwerty44]

Here is the question:
The acid dissociation constant for an acid dissolved in water is equal to the...

I said b) equilibrium constant

The correct answer was d) the equilibrium constant times the concentration of water

Can someone explain?

[Borek]

I don't like neither the question nor the answer. I suppose it is meant to take into account fact that there are no free H⁺ cations in the solution, they are all usually considered to be in the form of H₃O⁺. Thus the dissociation reaction is not

HA <-> H⁺ + A^- (K_dis)

but

HA + H₂O <-> H₃O⁺ + A^- (K_dis1)

There are several problems with this approach. First of all dissociation constant described this way should be not multiplied, but divided by the water concentration:

K_dis1 = K_dis/[H₂O]

which makes answer d) obviously incorrect.

But that's not all. H₃O⁺ is only an approximation. You may find information about H⁺ being present in the solution in the form of H₃O⁺, H₅O₂⁺, H₇O₃⁺ or even H₉O₄⁺; this is probably multi step equilibrium, in a way similar to many complexation reactions. If so, why not take one step more and go for - say

HA + 3H₂O <-> H₇O₃⁺ + A^- (K_dis3)

in which case

K_dis3 = K_dis/[H₂O]³

So I can write at least 5 different versions of this equation with 5 different powers (0 to 4). I wonder how your teacher will try to convince me which one - and why - is correct.

See additional bit of information at water dissociation constant discussion.

[enahs]

There are several problems with this approach. First of all dissociation constant described this way should be not multiplied, but divided by the water concentration:

Multiplication and division are really the same thing though, as multiplying by the inverse is the same as dividing  . But, that is a poor justification for this question/answer.