The ionization constant of HCN is 10^-5, then the equilibrium constant for its reaction with NaOH is

On reaction with NaOH,

NaOH + HCN → NaCN + H₂O_(aq)

_{Resultant solution contains a salt of strong base weak acid. Such a salt will undergo anionic hydrolysis in the following manner.}

${\mathrm{CN}}^{-}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons \mathrm{HCN}+{\mathrm{OH}}^{-}$

${\mathrm{K}}_{\mathrm{h}}={\left({\mathrm{K}}_{\mathrm{b}}\right)}_{{\mathrm{CN}}^{-}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}}={10}^{-9}$

_{So the reverse reaction will have an equilibrium constant of 10⁹.}