Question

A monoprotic weak acid, HA , dissociates in water according to the reaction HA(aq)+H2O(l)↽−−⇀H3O+(aq)+A−(aq) The equilibrium concentrations of the reactants and products are [HA]=0.260 M , [H3O+]=4.00×10−4 M , and [A−]=4.00×10−4 M . Calculate the Ka value for the acid HA.

Answer 1

Ka = 6.15x10⁻⁷

Step-by-step explanation:

Ka is defined as dissociation constant in the equilibrium of a weak acid with water. The general reaction is:

HA(aq) + H₂O(l) ⇆ H₃O⁺(aq) + A⁻(aq)

And Ka is defined as the ratio between molar concentrations in equilibrium of products over reactants as follows:

Ka = [H₃O⁺] [A⁻] / [HA]

You don't take water in the equilibrium beacuse is a pure liquid

Replacing with the concentrations of the problem:

Ka = [H₃O⁺] [A⁻] / [HA]

Ka = [4.00x10⁻⁴] [4.00x10⁻⁴] / [0.260]

Ka = 6.15x10⁻⁷