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 form water to `NH_(3)`.

Options

A. `6.07xx10^(5)s^(-1)`
B. `6.07xx10^(-10)s^(-1)`
C. `6.07xx10^(-5)s^(-1)`
D. `6.07xx10^(10)s^(-1)`

Answer

A

Explanation

`NH_(3)+H_(2)O overset(K_(f))underset(K_(b))hArr NH_(4)^(+)+OH^(-)`, , `(K_(b)= 3.4xx10^(10))`
`NH_(4)^(+)+H_(2)OhArr NH_(4)OH+H^(+)`, , `(K_(a)=5.6xx10^(-10))`
`K_((base))NH_(3)= (K_(f))/(k_(b))=(K_(w))/(K_(acid(NH_(4)^(+)))` , `( :' K_(acid)xxK_(base)= K_(w))`
or `(K_(f))/(3.4xx10^(10))=(10^(-14))/(5.6xx10^(-10))`
or `K_(f)=6.07xx10^(5)`