Question

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

Explanation:

Step1: Write the dissociation reaction of HCN

Hydrocyanic acid (\(HCN\)) reacts with water (\(H_2O\)) as follows: \(HCN(aq)+H_2O(l)
ightleftharpoons H_3O^{+}(aq)+CN^{-}(aq)\)

Step2: Recall the formula for acid - dissociation constant (\(K_a\))

For a general acid - dissociation reaction \(HA + H_2O
ightleftharpoons H_3O^{+}+A^{-}\), the acid - dissociation constant is given by \(K_a=\frac{[H_3O^{+}][A^{-}]}{[HA]}\)

Step3: Substitute the species from the \(HCN\) reaction into the \(K_a\) formula

Here, \(HA = HCN\), \(A^{-}=CN^{-}\). So, \(K_a=\frac{[H_3O^{+}][CN^{-}]}{[HCN]}\)

Answer:

\(\frac{[H_3O^{+}][CN^{-}]}{[HCN]}\)