The `pH` of `0.004M` hydrazine solution is `9.7`. Calculate its ionization constant `K_(b)` and `pK_(b)`.

Explanation

`NH_(2)NH_(2)+H_(2)O hArr NH_(2)NH_(3)^(+)+OH^(-)`
From the pH we can calculate the hydrogen ion cencentration, Knowing hydrogen on concentration and the ionic product of water we can calculate the concentration of hydroxyi ions. Thus we have :
`[H^(+)]`= antilog (-pH)
=antilog `(-9.7)=1.67xx10^(-10)`
`[OH^(-)]=K_(w)//[H^(+)]=1xx10^(-14)//1.67xx10^(-10)`
`=5-98xx10^(-5)`
The cocentration of the correspondinghydrazinium ion is also the same as that of hydroxyl ion. the cocentrations of both these ions are very samall so the concentration of the undissociated base can be taken equal to 0.004M.
Thus,
`K_(b)=[NH_(2)NH_(3)^(+)][OH^(-)]//[NH_(2)NH_(2)]`
`=(5.98xx10^(-5))^(2)//0.004=8.96xx10^(-7)`
`pK_(b)=-logK_(b)=log(8.96xx10^(-7))=6.04`