HX + H_{2}O <-> H_{3}O^{+} + X^{-}

K_{a} = [ H_{3}O^{+} ] [ X^{-} ] / [ HA ]

The K_{a} term excludes the [H_{2}O] in the denominator as [H_{2}O] is fairly a constant in an aqueous system.

Just include [H_{2}O] in the denominator to convert K_{a} to K_{eq}

K_{eq} = [ H_{3}O^{+} ] [ X^{-} ] / [ HA ] [H_{2}O]