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

Explanation

`NaCN + H_(2)O hArr NaOH + HCN`
or `CN^(-) + H_(2)O hArr OH^(-) HCN`
`K_(b)=([OH^(-)][HCN])/([CN^(_)])=([OH^(-)]^(2))/([CN^(-)])`
`:. pK_(b) = - log K_(b) = - 2 log [OH^(-)]+log [CN^(-)]`
or `4.70 = - 2 log [OH^(-)]+ log (0.5)` [`:' [CN^(-)]=[NaCN]]`
or` 2 log [OH^(-)]=-5.00 or log [OH^(-)]=-2.5`
`-log [OH^(-)]=2.5`
`p_(OH)=2.5`
`pH = 14-25 = 11.5`