What is the pH of a 0.235 M 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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The correct answer is A because the pH of a 0.235 M solution of potassium fluoride can be calculated using the following equation:pH = 14 - pOH = 14 - (1/2)pKa - (1/2)log[HF]where pKa is the dissociation constant for HF and [HF] is the concentration of HF in the solution.Substituting the given values into the equation, we get:pH = 14 - (1/2)log(6.6 × 10^-4) - (1/2)log(0.235) = 8.28

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