The enthalpy of neutralization of HCN by NaOH  is  $-12.13 \mathrm{kJ} {\mathrm{mol}}^{-1}$. The enthalpy of ionization of HCN will be  

$-57.4+x=-12.13$

    $x=45.2$

Reaction is:

$\mathrm{HCN}+\mathrm{NaOH}\to \mathrm{NaCN}+{\mathrm{H}}_2\mathrm{O}$

The enthalpy change for this reaction is -57.62 kJ/mol at 25°C.

$$\begin{gathered} \mathrm{enthalpy} \mathrm{of} \mathrm{neutralization}+\mathrm{enthalpy} \mathrm{of} \mathrm{ionization}=-12.13 \mathrm{kJ}/\mathrm{mol} \\ -57.4 \mathrm{kJ}/\mathrm{mol}+\mathrm{enthalpy} \mathrm{of} \mathrm{ionization}=-12.13 \mathrm{kJ}/\mathrm{mol} \\ \mathrm{enthalpy} \mathrm{of} \mathrm{ionization}=45.2 \mathrm{kJ}/\mathrm{mol} \end{gathered}$$