Question

fill in the left side of this equilibrium constant equation for the reaction of carbonic acid (h₂co₃) with water.
= kₐ

Explanation:

Step1: Write the dissociation reaction of carbonic acid

Carbonic acid (\(H_{2}CO_{3}\)) donates a proton (\(H^{+}\)) to water (\(H_{2}O\)) to form \(H_{3}O^{+}\) and \(HCO_{3}^{-}\). The reaction is \(H_{2}CO_{3}(aq)+H_{2}O(l)
ightleftharpoons H_{3}O^{+}(aq)+HCO_{3}^{-}(aq)\).

Step2: Recall the formula for acid - dissociation constant (\(K_{a}\))

For a general acid - dissociation reaction \(HA + H_{2}O
ightleftharpoons H_{3}O^{+}+A^{-}\), the acid - dissociation constant is defined as \(K_{a}=\frac{[H_{3}O^{+}][A^{-}]}{[HA]}\).
In the case of \(H_{2}CO_{3}\), \(HA = H_{2}CO_{3}\), \(A^{-}=HCO_{3}^{-}\). So, \(K_{a}=\frac{[H_{3}O^{+}][HCO_{3}^{-}]}{[H_{2}CO_{3}]}\)

Answer:

\(\frac{[H_{3}O^{+}][HCO_{3}^{-}]}{[H_{2}CO_{3}]}\)