After 0.050 mole of hydrofluoric acid is dissolved in 200 mL of aqueous solution, it is found that 7.90% of the acid is dissociated at equilibrium. Calculate the equilibrium concentrations of H3O , undissociated hydrofluoric acid (HF), and conjugate weak base of hydrofluoric acid.

[HF] = 0.23M

0.02M = [F⁻]

[H₃O⁺] = 0.02M

Explanation:

The equilibrium of HF with water is:

HF + H₂O ⇄ H₃O⁺ + F⁻

Where F⁻ is the conjugate base of the HF

The initial concentration of HF is:

[HF] = 0.0500mol / 0.200L = 0.25M

Now, in equilibrium, if the 7.90% is dissociated, the undissociated HF is the 100 - 7.90 = 92.1%

0.25M*92.1% = 0.23M = [HF]

The conjugate base will be 0.25M - 0.23M = 0.02M = [F⁻]

And as the H₃O⁺ ion comes from the same equilibrium of F⁻ and is produced in the same proportion:

[H₃O⁺] = 0.02M