Calculate the acid dissociation constant (Ka) of hydrofluoric acid if the following concentrations were found at equilibrium: [H3O+] = [F-] = 7.8 x 10-3 M, [HF] = 0.0922.

Question

HF (aq) + H2O (l) F-(aq) + H3O+(aq)  

Answer 1

HF(aq) + H₂O(l) ===> H₃O⁺(aq) + F^-(aq) ... ionization of HF in water

Ka = [H₃O⁺][F^-] / [HF]

Ka = (7.8x10^-3)(7.8x10^-3) / 0.0922

Ka = 6.59x10^-4