`0.00135` mole of `NH_(3)` dissociates in `1.0` litre solution of `0.10M`. Calculate the dissociation constant of `NH_(3)`.

Explanation

`underset((0.1-0.00135))(NH_(3))+H_(2)OhArr underset(0.00135)(NH_(4)^(+))+underset(0.00135)(OH^(-))`
`:. K_(b)= ([NH_(4)^(+)][OH^(-)])/([NH_(3)])`
`= (0.00135xx0.00135)/((0.1-0.00135))`
`= ((0.00135)^(2))/(0.1)`
`= 1.823xx10^(-5)`