Answer: The equilibrium constant for the given reaction is [tex]K_a=\frac{[BrO^-][H_3O^+]}{[HBrO]}[/tex]
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_{eq}[/tex]
For a general chemical reaction:
[tex]aA+bB\rightleftharpoons cC+dD[/tex]
The expression for [tex]K_{eq}[/tex] is written as:
[tex]K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]
The concentration of pure liquids and pure solids are taken as 1.
The chemical equation for the reaction of hypobromous acid with water follows:
[tex]HBrO(aq.)+H_2O(l)\rightarrow BrO^-(aq.)+H_3O^+(aq.)[/tex]
The expression of equilibrium constant for above equation follows:
[tex]K_a=\frac{[BrO^-][H_3O^+]}{[HBrO]}[/tex]
Hence, the equilibrium constant for the given reaction is [tex]K_a=\frac{[BrO^-][H_3O^+]}{[HBrO]}[/tex]
