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acid dissociation constant weak acid

so I've just noticed that there are 2 generalised acid dissociation constant weak acid with water and without water
HA(aq)+H2O(l) < --> A- (aq)+H3O+(aq)
and
HA(aq) <--> H+(aq) + A- (aq)
what's different between these 2 equations?

and further question is when we use HA(aq)+H2O(l) < --> A- (aq)+H3O+(aq) this to calculate Ka
water is excluded why? is it because concentration of water is relatively stable?
Original post by universalcj
so I've just noticed that there are 2 generalised acid dissociation constant weak acid with water and without water
HA(aq)+H2O(l) < --> A- (aq)+H3O+(aq)
and
HA(aq) <--> H+(aq) + A- (aq)
what's different between these 2 equations?

and further question is when we use HA(aq)+H2O(l) < --> A- (aq)+H3O+(aq) this to calculate Ka
water is excluded why? is it because concentration of water is relatively stable?


There are two reasons.
As you correctly surmise the water is in great excess and changes by tiny proportions.
Also the law of mass action excludes pure substances (the equilibrium law), as they are not integrated into the system.

Water is also the solvent in an aqueous system, so it remains a moot point whether or not it's "pure"

However, these are all semantics. The first reason is good enough.

For acid dissociation in solution we clearly need water. Several dissociation equations are acceptable in most specifications:

HA <==> H+ + A-

HA(aq) <==> H+(aq) + A-(aq)

HA + (aq) <==> H+(aq) + A-(aq)

HA + nH2O <==> H+(aq) + A-(aq)

HA + nH2O <==> H3O+(aq) + A-(aq)

They all get the basic (or acidic ) idea across. Personally, I would go for equation 2.

The equilibrium expression assumes water "concentration" to be constant and just factors the other components:

ka = [H+(aq)][A-(aq)]/[HA(aq)]

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