$\mathrm{HOCl}\left(\mathrm{aq}\right)⟶{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OCl}}^{-}\left(\mathrm{aq}\right)$

The ionization of hypochlorous acid represented above has $\mathrm{K}=3.0 \mathrm{x}{10}^{-8}$ at 25° C. What is K for this reaction?

${\mathrm{OCl}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons \mathrm{HOCl}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

Explanation :

$\begin{array}{l}\mathrm{HOCl}\rightleftarrows {\mathrm{H}}^{+}+{\mathrm{OCl}}^{-}{\mathrm{K}}_{\mathrm{a}}=3\times {10}^{-8}\\ {\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{OH}}^{-}{\mathrm{K}}_{\mathrm{w}}={10}^{-14}\end{array}$

Reversing the first equation 

$\begin{array}{r}\frac{\begin{array}{l}{\mathrm{H}}^{+}+{\mathrm{OCl}}^{-}\rightleftharpoons {\mathrm{HOCl}}^2 {{\mathrm{K}}_{\mathrm{a}}}^{\mathrm{\prime }}=\frac{1}{{\mathrm{K}}_a}\\ {\mathrm{H}}_2{\mathrm{H}}^{-4}+{\mathrm{OH}}^{-} {\mathrm{K}}_w=1{\mathrm{O}}^{-14}\end{array}}{{\mathrm{H}}_2\mathrm{O}+{\mathrm{OCl}}^{-}\rightleftharpoons \mathrm{HOCl}+{\mathrm{OH}}^{-}}\end{array}$

${\mathrm{K}}_{\mathrm{b}}$ for reaction,

${\mathrm{k}}_{\mathrm{eq}}=\frac{{\mathrm{k}}_{\mathrm{w}}}{{\mathrm{k}}_{\mathrm{a}}}=\frac{{10}^{-14}}{3\times {10}^{-8}}$

${\mathrm{K}}_{\mathrm{b}}=3.3\times {10}^{-7}$