The equilibrium constant for this reaction is 3.6×10 –7 . OCl ^ - ( aq .)+ H _ 2 O ( l )…
Chemistry · JEE Main · NTA Exams — Equilibrium
The equilibrium constant for this reaction is 3.6×10–7.
\(\mathrm{OCl}^{-}(\mathrm{aq} .)+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{HOCl}(\mathrm{aq} .)+\mathrm{OH}^{-}(\mathrm{aq} .)\)
What is Ka for HOCl ?
\(\mathrm{OCl}^{-}(\mathrm{aq} .)+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{HOCl}(\mathrm{aq} .)+\mathrm{OH}^{-}(\mathrm{aq} .)\)
What is Ka for HOCl ?
- 2.8 × 10–8
- 3.6 × 10–7
- 6 × 10–4
- 2.8 × 10–6
Answer
(A) 2.8 × 10 –8
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