The `pH` of `0.004M` hydrazine `(NH_(2).NH_(2))` solution is `9.7`. Calculate its ionisation 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 concentration and the ionic product ion concentration and the ionic product of water we can calculate the concentration of hydroxyl ions. Thus we have :
`[H^(+)] = "antilog"(-pH)`
`= "antilog" (-9.7) = 1.67 xx 10^(-10)`
`[OH^(-)] = K_(w)//[H^(+)] = 1 xx 10^(-14)//1.67 xx 10^(-10)`
`= 5.98 xx 10^(-5)`
The concentration of the correpending hydrazinium ion is also the same as that of hydroxyl ion.The concentration of both these ions is very small so the concentration of the undissociated base can be taken equal to `0.004 M`.
Thus,
`K_(b) = [NH_(2)NH_(3^(+))]//[NH_(2)NH_(2)]`
`= (5.98 xx 10^(-5))^(2)//0.004 = 8.96 xx 10^(-7)`
`pK_(b) = - "log"K_(b) = - "log"(8.96 xx 10^(-7)) = 6.04`.