Why Is H2O Included in Keq Calculations?

[zmike]

So I see in my book that for an acid dissociation, the Ka = K eq [H 2 O]
and that the Keq includes H2O in the eqn. Shouldn't there be no liquids in Keq?

Keq = [H 3 O + ] [A - ]
[HA] [H 2 O]

Why is there H2O in Keq?
If I am asked to calculate Keq do I have I have to include [H2O]?

Thanks
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[DrDu]

That's a question of standard state. For diluted substances you use, simplified, the concentration c divided by the concentration $c^0$ =1 mole/l, e.g. $[HA]=c_{HA}/c^0_{HA}$. For the solvent you use as standard state the pure solvent that is for water $c^0=$ 55.5 mole/l.
So the concentration of H2O in a dilute solution of the acid is very nearly the concentration of H2O in pure water, hence the quotient which enters the equilibrium factor is nearly 1.