The ionization constant of `overset(o+)(NH_(4))` ion in water is `5.6 xx 10^(-10)` at `25^(@)C`. The rate constant the reaction of `overset(o+)NH_(4)` and `overset(ɵ)(OH)` ion to form `NH_(3)` and `H_(2)O` at `25^(@)C` is `3.4 xx 10^(10)L mol^(-1) s^(-1)`. Calculate the rate constant for proton transfer from water to `NH_(3)`.

Explanation

`K_(a)(NH_(4)^(+)) = 5.6 xx 10^(-10)`
`K_(b)(NH_(3)) = K_(w)//K_(a) = 10^(-14)/5.6xx10^(-16) = 1.8 xx 10^(-5)`
`i.e., NH_(3) + H_(2)O overset("K_(1)")underset("K_(2)")to NH_(4)^(+) + OH^(-)`
`K = K_(1)/K_(2) = 1.8 xx 10^(-5) implies K_(1) = KK_(2) = 1.8 xx 10^(-5) xx 3.4 xx 10^(10) = 6.12 xx 10^(5)`