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

Explanation

`CN^(-) + H_2O hArr OH^(-) + HCN`
`K_b=([OH^-][HCN])/([CN^-])=[OH^-]^2/([CN^-])`
`therefore pK_b=-log K_b=-2 log [OH^-]+log [CN^-]`
`[CN^-]` =[NaCN]
4.70 =-2 log `[OH^-]` +log (0.5)
2 log `[OH^-]` =-4.70 -0.30
or 2 log `[OH^-]` =-5.00
-log `[OH^-]` =2.50
pOH=2.50
`therefore` pH=14-2.5 =11.5