What is the pH of a 0.05 M sodium fluoride solution, given that F⁻ has a Kb of 1.59 x 10⁻¹¹?

Set up the equilibrium expression for the base dissociation of fluoride: F⁻ + H₂O ⇌ HF + OH⁻. The expression for Kb is: $K_b=\frac{\mathrm{[HF]}\mathrm{[OH]}}{\mathrm{[F]}}$.

Assume that the initial concentration of F⁻ is 0.05 M and that the change in concentration due to dissociation is small. Use the approximation method to solve for [OH⁻] by substituting into the Kb expression: $K_b=\frac{xx}{0.05-x}$, where x is [OH⁻].