Question

Problem Page Fill in the left side of this equilibrium constant equation for the reaction of hypobromous acid with water.

Answer 1

Answer: The equilibrium constant for the given reaction is
`K_a=([BrO^-][H_3O^+])/([HBrO])`

Step-by-step 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
`K_(eq)`

For a general chemical reaction:

`aA+bB\rightleftharpoons cC+dD`

The expression for
`K_(eq)` is written as:

`K_(eq)=([C]^c[D]^d)/([A]^a[B]^b)`

The concentration of pure liquids and pure solids are taken as 1.

The chemical equation for the reaction of hypobromous acid with water follows:

`HBrO(aq.)+H_2O(l)\rightarrow BrO^-(aq.)+H_3O^+(aq.)`

The expression of equilibrium constant for above equation follows:

`K_a=([BrO^-][H_3O^+])/([HBrO])`

Hence, the equilibrium constant for the given reaction is
`K_a=([BrO^-][H_3O^+])/([HBrO])`