Question

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

Answer 1

`pH= 9.95`, Thus codeine is a base.
`{:(C_(18)+H_(21)NO_(2)rarr,"Codenie"^(+), +OH^(-)), (1,0,0), ((1-alpha),alpha,alpha):}`
`:. [OH^(-)]=Calpha`
or `(10^(-4))/([H^(+)])=C alpha [:.[H^(+)][OH^(-)]=10^(-14)]`
Also `pH=9.95` or `-Iog H^(+)=9.95`
or `[H^(+)]=1.12xx10^(-10)`
`:. (10^(-4))/(1.12xx10^(-10))=0.005xxalpha)`
`:. alpha=0.0179` or `1.79%`
Now `K_(b)= (["Codeine"]^(+)[OH^(-)])/([Codeine"])`
`= (CalphaxxCalpha)/(C(1-alpha))=(Calpha^(2))/((1-alpha))`
`K_(b)= (0.05xx(0.0179)^(2))/((1-0.0179))`
`=1.63xx10^(-6)`
Also `pK_(b) = 5.78`