What is the pH of a `0.50M` aqueous NaCN solution ? `(pK_(b)of CN^(-)=4.70)`

Explanation

`pK_(a)for HCN=14-4.70=9.30`
`{:(NaCN+H_(2)OhArr,NaOH+,HCN),(1,0,0),(1-h,h,h):}`
`:. [OH^(-)]=c.h=Csqrt([(K_(H))/(C )])`
`=sqrt(K_(H).C)=sqrt([(K_(w).C)/(K_(a))])`
or `pOH= (1)/(2)[pK_(w)=logC-pK_(a)]`
`=(1)/(2)[14+0.3010-9.30]=2.5`
`:. pH=14-2.5=11.5`