Determine the molar solubility of BaF2 in pure water given that the Ksp for BaF2 is 2.45 × 10⁻⁵.

Let the molar solubility of BaF2 be 's'. Therefore, the concentration of Ba²⁺ ions will be 's' and the concentration of F⁻ ions will be '2s' because two fluoride ions are produced for each formula unit of BaF2 that dissolves.

Substitute these concentrations into the Ksp expression: $K_{\mathrm{sp}}=\; (s)(2s)²\; =\; 4s³$. Set this equal to the given Ksp value: $4s³\; =\; 2.45\; \times \; 10⁻⁵$.