K a for hydrofluoric acid is 6.9 × 10 –4 . What is the equilibrium constant K for the…
Chemistry · JEE Main · NTA Exams — Equilibrium
Ka for hydrofluoric acid is 6.9 × 10–4. What is the equilibrium constant K for the following reaction ?
\(\mathrm{F}^{-}(\mathrm{aq} .)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HF}(\mathrm{aq} .)+\mathrm{OH}^{-}(\mathrm{aq} .)\)
\(\mathrm{F}^{-}(\mathrm{aq} .)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HF}(\mathrm{aq} .)+\mathrm{OH}^{-}(\mathrm{aq} .)\)
- 6.9 × 10–11
- 1.4 × 10–11
- 2.6 × 10–9
- 8.3 × 10–6
Answer
(B) 1.4 × 10 –11
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