What is the pH of a 0.235 M aqueous solution of potassium fluoride? Ka (HF)= 6.6 × 10^-4 A)8.28 B)11.45 C)8.59 D)5.41 E)5.72

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Potassium fluoride (KF) dissociates completely in water to form K+ and F- ions. The F- ion then reacts with water to form HF and OH- according to the equation F- + H2O ⇌ HF + OH-. The concentration of F- is given as 0.235 M, and the Ka for HF is 6.6 × 10^-4. To find the pH, we first need to calculate the Kb for F- because it's the base in this reaction. Kb can be found using the relation Kw = Ka × Kb, where Kw (the ion-product constant for water) is 1.0 × 10^-14 at 25°C. Thus, Kb = Kw / Ka = (1.0 × 10^-14) / (6.6 × 10^-4) = 1.515 × 10^-11. Using the formula for Kb, [OH-] = sqrt(Kb × [F-]), we can calculate the concentration of OH- ions. Then, pOH = -log[OH-], and pH = 14 - pOH. Following these steps with the given values leads to a pH of approximately 8.28.

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