K a for HF is 3.5 × 10 –4 . Calculate K b for the fluoride ion.
Chemistry · JEE Main · NTA Exams — Equilibrium
Ka for HF is 3.5 × 10–4. Calculate Kb for the fluoride ion.
- 3.5 × 10–4
- 1.0 × 10–7
- 2.9 × 10–11
- 1.0 × 10–14
Answer
(C) 2.9 × 10 –11
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