`0.16 g N_(2)H_(4)` is dissoolved in `H_(2)O` and total volume is made upto `500 mL`. Calculate the percentage of `N_(2)H_(4)` that has reacted with `H_(2)O` in this solution. `K_(b)` for `N_(2)H_(4) = 4.0 xx 10^(-6)M`.

Explanation

`{:(,N_(2)H_(4)+,H_(2)OhArr,N_(2)H_(5)^(o+)+,OH^(Theta)),("Before dissociation",1,,0,0),("After dissociation",1-alpha,,alpha,alpha):}`
Also `K_(b) = (Calpha^(2))/((1-alpha))`
Assuming `1-alpha = 1`
`K_(b) = C alpha^(2)`
`[N_(2)H_(4)] = C = (0.16 xx 1000)/(32 xx500) = 0.01`
Given `K_(b) =4 xx 10^(-6)M :. alpha^(2) = (4 xx 10^(-6))/(0.01) = 4 xx 10^(-4)`
`alpha = 2 xx 10^(-2)` i.e., `alpha = 0.2 or 2%`