what-is-the-correct-equilibrium-expression-for-the-dissociation-of-a-weak-acid-h

Question

What is the correct equilibrium expression for the dissociation of a weak acid HA in water?

Accepted Answer

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

Answer

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

Answer

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

Answer

K_{a} = [$$H_{2}O$$][$$A^{-}$$] / [HA]