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Chapter 15: Problem 47

Hydrofluoric acid is a weak acid. Write the mass action expression for its acid ionization reaction.

Short Answer

Expert verified
Answer: The mass action expression for the ionization reaction of hydrofluoric acid (HF) is Ka = K' * [H3O+] * [F-] / [HF], where Ka is the equilibrium constant, K' is the modified equilibrium constant, [H3O+] is the equilibrium concentration of hydronium ions, [F-] is the equilibrium concentration of fluoride ions, and [HF] is the equilibrium concentration of hydrofluoric acid.

Step by step solution

01

Write the ionization reaction of hydrofluoric acid

The chemical equation for the ionization of hydrofluoric acid (HF) in water is: HF (aq) + H2O (l) <=> H3O+ (aq) + F- (aq)
02

Identify the equilibrium concentrations of all species

In an equilibrium state, the concentrations of all species (HF, H2O, H3O+, and F-) do not change. Let's represent the equilibrium concentrations as follows: [HF] - equilibrium concentration of hydrofluoric acid [H2O] - equilibrium concentration of water molecules (this will remain constant as water is the solvent) [H3O+] - equilibrium concentration of hydronium ions [F-] - equilibrium concentration of fluoride ions
03

Write the mass action expression for the ionization reaction

The mass action expression for the ionization reaction is given by the equilibrium constant, Ka. The expression for Ka can be written as: Ka = \frac{[H3O+][F-]}{[HF][H2O]} As the concentration of water remains constant in a dilute solution, we can rewrite the expression as: Ka = K' * [H3O+] * [F-] / [HF] Here K' is the modified equilibrium constant. This is the final mass action expression for the ionization reaction of hydrofluoric acid.

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