The reaction equation is:
2HF(aq) + Ba(OH)₂(aq) → BaF₂(s) + H₂O (aq)
Writing the ionic form of this equation:
2H⁺ + 2F⁻ + Ba⁺² + 2OH⁻ → BaF₂ + H⁺ + OH⁻
The solubility of barium fluoride is 0.16 grams per 100 ml of water, which means that it is an insoluble compound. Moreover, spectator ions are those that remain unchanged before and after a reaction, so they "spectate" the reaction. In this case, the ions unchanged before and after the reaction are H⁺ and OH⁻.
