Calculating K_a from ΔE° for HBrO Reaction

[amcavoy]

Using \Delta \textrm{E}^{\circ} values and the fact that \textrm{K}_{\textrm{w}}=10^{-14}, how would I find the \textrm{K}_{\textrm{a}} value for the following reaction?:

\textrm{HBrO}\longrightarrow\textrm{H}^{+}+\textrm{BrO}^{-}​

 

[Noober]

dEº = -RTlnK

Dunno if that helps :/.

 

[Weave]

Noobler sounds right.
Rearrange to give:
\frac{\Delta E}{-RT}=ln K_{a}
and solve for K_{a]