`0.16`g of `N_(2)H_(4)` are dissolved in water and the total volume made upto 500 mL. Calculate the percentage of `N_(2)H_(4)` that has reacted with water in this solution. `(K_(b)for N_(2)H_(4)=4.0xx10^(-6)lt)`

Explanation

`N_(2)H_(4)+H_(2)O hArr N_(2)H_(5)^(+) + OH^(-)`
Conc. Of `N_(2)H_(4) = 0.16 g ` in 500 ml = 0.32 g `L^(-1) = (0.32)/(32) ` mol `L^(-1) = 0.01 M = 10^(-2) M`
Suppose x mol `L^(-1)` is the amount of hydrazine reacted. Then
`{:(,N_(2)H_(4) ,+ ,H_(2)O,hArr,N_(2)H_(5)^(+),+,OH^(-),,),("Initial conc.",10^(-2)M,,,,,,,,),("Conc. after reaction",10^(-2) - x ,,x,,x,,,,):}`
`K_(b) = ([N_(2)H_(5)^(+)][OH^(-)])/([N_(2)H_(4)])`
`4.0xx10^(-6) = (x^(2))/(10^(2)-x) ~= (x^(2))/(10^(2)) or x^(2)=4.0xx10^(-8) or x=2.0xx10^(-4)` mol `L^(-1)`
`:.` % of hydrazine reacted with water `=(2.0xx10^(-4))/(10^(-2))xx100=2%`