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Calculating Ka

  1. May 6, 2007 #1
    Using [tex]\Delta \textrm{E}^{\circ}[/tex] values and the fact that [tex]\textrm{K}_{\textrm{w}}=10^{-14}[/tex], how would I find the [tex]\textrm{K}_{\textrm{a}}[/tex] value for the following reaction?:

    [tex]\textrm{HBrO}\longrightarrow\textrm{H}^{+}+\textrm{BrO}^{-}[/tex]​
     
  2. jcsd
  3. May 6, 2007 #2
    dEº = -RTlnK

    Dunno if that helps :/.
     
  4. May 6, 2007 #3
    Noobler sounds right.
    Rearrange to give:
    [tex]\frac{\Delta E}{-RT}=ln K_{a}[/tex]
    and solve for [tex]K_{a][/tex]
     
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