ACS General Chemistry 2 Practice Exam

What is Kb of F-? Ka of HF is 6.8 × 10^-4

6.8e10 6.8e-4 1.5e3 1.5e-11

Correct Answer:

1.5e-11

Explanation:

This question uses the relationship between Ka and Kb for conjugate acid–base pairs: Ka × Kb = Kw, with Kw = 1.0 × 10^-14 at 25°C. Since F− is the conjugate base of HF, its base-dissociation constant is Kb(F−) = Kw / Ka(HF). Plugging in Ka(HF) = 6.8 × 10^-4 gives Kb ≈ (1.0 × 10^-14) / (6.8 × 10^-4) ≈ 1.47 × 10^-11, which rounds to 1.5 × 10^-11. So the fluoride ion is a very weak base in water. The other numbers don’t fit the Kw/Ka relation: taking 6.8 × 10^-4 would just reproduce Ka, a reciprocal or a unrelated magnitude doesn’t come from the Ka–Kb relationship, and 1.5 × 10^3 is not consistent with the expected small Kb at this Ka.

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