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

Answer

A

Explanation

For a conjugate acid-base pair,
`pK_(a) +pK_(b) = 14`
or `pK_(a) (HCN) + pK_(b) (CN^(Theta)) = 14`
or `pK_(a) (HCN) + 4.7 = 14`
or `pK_(a) (HCN_ = 9.3`
`NaCN` is a salt of a weal acid `HCN` and a strong base `NaOH`.
For a salt of a weak acid and storng base,
`pH = 7 +(1)/(2) pK_(a) +(1)/(2) log C = 7 +((1)/(2)xx9.7) +(1)/(2)log0.5`
` = 11.7`