Question

fill in the left side of this equilibrium constant equation for the reaction of phosphoric acid (h₃po₄) with water.
= kₐ

Explanation:

Step1: Write the chemical equation for the reaction

Phosphoric acid (\(H_{3}PO_{4}\)) donates a proton (\(H^{+}\)) to water (\(H_{2}O\)):
\(H_{3}PO_{4}(aq)+H_{2}O(l)
ightleftharpoons H_{2}PO_{4}^{-}(aq)+H_{3}O^{+}(aq)\)

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

For a general acid - dissociation reaction \(HA + H_{2}O
ightleftharpoons A^{-}+H_{3}O^{+}\), \(K_{a}=\frac{[A^{-}][H_{3}O^{+}]}{[HA]}\)

Step3: Substitute the species from the \(H_{3}PO_{4}\) reaction into the \(K_{a}\) formula

Here, \(HA = H_{3}PO_{4}\), \(A^{-}=H_{2}PO_{4}^{-}\)
\(K_{a}=\frac{[H_{2}PO_{4}^{-}][H_{3}O^{+}]}{[H_{3}PO_{4}]}\)

Answer:

\(\frac{[H_{2}PO_{4}^{-}][H_{3}O^{+}]}{[H_{3}PO_{4}]}\)