The rate constant at `25^(@)C` for the reaction of `NH_(4)^(+)` and `OH^(-)` to form `NH_(4)OH` is `4xx10^(10) M^(-1) sec^(-1)` and ionisation constant of aq. `NH_(3)` is `1.8xx10^(-5)`. The rate constant of proton transfer to `NH_(3)` is

Options

A. `1.8 xx 10^(-5)`
B. `7.2 xx 10^(+5)`
C. `3.6 xx 10^(5)`
D. `4.2 xx 10^(-5)`

Answer

2

Explanation

`NH_(4)^(+) + OH^(-) overset(K_(1))to NH_(4)OH, k_(1) = 4 xx 10^(10)`
`NH_(3) + H_(2)O overset(k_(1))iffNH_(4)^(darr) + OH^(-)`
`K_(eq)=(K_(f))/(K_(b))`
`rArr 1.8 xx 10^(-5)=(K_(f))/(4 xx 10^(10))`
`rArr K_(f) = 7.2 xx 10^(5)`