The pH of 0.005(M) codeine `(C_(18)H_(21)NO_(3))` solution is 9.95. calculate its ionisation constant and `pK_(b)`.

Explanation

For a solution of a monoacidic base like codeine `(C_(18)H_(21)NO_(3))`,
`pOH=(1)/(2)pK_(b)-(1)/(2)logc`
pH of the solution=9.95,
So, `pOH=14-pH=14-9.95=4.05`
`therefore 4.05=(1)/(2)pK_(b)-(1)/(2)log0.005`
`therefore pK_(b)=5.8 and K_(b)=10^(-pK_(b))=10^(-5.8)=1.58xx10^(-6)`
`[because pK_(b)=-log_(10)K_(b)]`