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)`.

Answer

Rate constant `= 6.07 xx 10^(5) s^(-1)`

Explanation

`NH_(3) + H_(2)O underset(k_(b))overset(k_(f))hArr NH_(4)^(o+)+overset(Θ )(OH) (k_(b) = 3.4 xx 10^(10))` ...(i)
`NH_(4)^(o+) + H_(2)O hArr NH_(4)OH +H^(o+) (k_(a) = 5.6 xx 10^(-10))`
`k_(base)(NH_(3))=(k_(f))/(k_(b))=(k_(w))/(k_(acid)(NH_(4)^(o+)))`
`( :. k_(acid)xxk_(base)=k_w)`
`(k_(f))/(3.4xx10^(10))=(10^(-14))/(5.6xx10^(-10))impliesk_(f)= 6.07xx10^(5)`
(Alternative methof)
We are given that
`NH_(4)^(o+) +H_(2)O hArr NH_(3) + H_(3)O^(o+) (k_(a) = 5.6 xx 10^(-10))`
`NH_(4)^(o+) + overset(Θ)(OH) underset(k_(b))overset(k_(f))hArr NH_(3) + H_(2)O (k_(f) = 3.4 xx 10^(10) L mol^(-1) s^(-1))`
The second equation can be generalized as followes:
`{:(NH_(4)^(o+) + H_(2)O hArr NH_(3) + H_(3)O^(o+) (k_(a))),(" "2H_(2)O hArr H_(3)O^(o+) + overset(Θ )(OH) (k_(w))),(ul(bar("Subtract" : NH_(4)^(o+) + overset(Θ )(OH) hArr NH_(3)+H_(2)O))):}` ...(ii)
`k_(eq)=(k_(a))/(k_(w))` and also `k_(eq)=(k_(f))/(k_(b))`
`:. (k_(f))/(k_(b))=(k_(a))/(k_(w))`
`k_(b)=k_(f)[["Note here that equation(ii)is reversed",],["of Eq.(i) So" k_(b)"is calculated",]]`
`=(3.4xx10^(10)Lmol^(-1)s^(-1))((1.0xx10^(-14)mol^(2)L^(-2))/(5.6xx10^(-10)molL^(-1)))`
`= 6.07xx10^(5)s^(-1)`