An aqueous solution of NaF is prepared by dissolving 0.350 mol of NaF in sufficient water to yield 1.0 L of solution.The pH of the solution was 8.93 at 25.0 °C.The K_b of F^-is ________. A)1.2 × 10^-5 B)2.1 × 10^-10 C)6.9 × 10^-9 D)2.8 × 10^-12 E)9.9 × 10^-2

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NaF is a salt of a weak base, F⁻, and a strong acid, Na⁺. Therefore, the solution will be basic due to the hydrolysis of F⁻ ions. The equilibrium for the hydrolysis of F⁻ is: F⁻ + H₂O ⇌ HF + OH⁻ The equilibrium constant expression is: Kb = [HF][OH⁻]/[F⁻] Since the pH is given, we can use it to find the [OH⁻] concentration: pH = 14 - pOH 8.93 = 14 - pOH pOH = 5.07 [OH⁻] = 10⁻⁵.⁰⁷ M Assuming that all the NaF dissolves and dissociates in water completely, the [F⁻] concentration is 0.350 M. Now we can substitute the concentrations into the Kb expression and solve for Kb: Kb = [HF][OH⁻]/[F⁻] Kb = x²/0.350 Kb = (10⁻⁸.⁴⁵)²/0.350 Kb = 2.1 × 10⁻¹⁰ Therefore, the K_b of F^- is 2.1 × 10^-10.

An Aqueous Solution of NaF Is Prepared by Dissolving 0