Calculate the pH of 0.5 M aqueous solution of NaCN, the ${\mathrm{pK}}_{\mathrm{b}}\text{ of }{\mathrm{CN}}^{-}\text{is }4.70$

NaCN is a salt of strong base and weak acid ; pH 
$$\begin{gathered} =7+\frac{1}{2}{\mathrm{pK}}_{\mathrm{a}}+\frac{1}{2}\log \mathrm{C} \\ {\mathrm{pK}}_{\mathrm{a}}\text{ for }\mathrm{HCN}=14-4.70=9.30 \\ \therefore \mathrm{pH}=7+\frac{1}{2}\times 9.30+\frac{1}{2}\log 0.5;\mathrm{pH}=11.5 \end{gathered}$$