The ionization constant of `NH_(4)^(+)` in water is `5.6 xx 10^(-10)` at `25^(@)C`. The rate constant for the reaction of `NH_(4)^(+)` and `OH^(-)` 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 of `NH_(3)`.

Options

A. `6.07 xx 10^(5)s^(-1)`
B. `12.03 xx 10^(5)s^(-1)`
C. `9.82 xx 10^(5)s^(-1)`
D. `16.34 xx 10^(5)s^(-1)`

Answer

A

Explanation

`N_(4)^(+) + H_(2)O hArr NH_(3) + H_(3)O^(+) " "K_(a) = 5.6 xx 10^(-10)`
`NH_(4)^(+) + OH^(-) underset(k_(b))overset(k_(f))hAr NH_(3) + H_(2)O^(+) " "K_(f) = 3.4 xx 10^(10) L mol^(-1)s^(-1)`
The second equation can be generated as follows.
`NH_(4)^(+) + H_(2)O hArr NH_(3) + H_(3)O^(+) " "K_(a)`
Subtraction `{:(2H_(2)O hArr H_(3)O^(+) + OH^(-))/(NH_(4)^(+) + OH^(-) hArr NH_(3) + H_(2)O) {:(K_(w)),(K_(eq) = K_(a)//K_(w)):}`
Also `K_(eq) = k_(f)//k_(b)`. Hence `(k_(f))/(k_(b)) = (K_(a))/(K_(w))`
or or `k_(b) = k_(f)((K_(w))/(K_(a))) = (3.4 xx 10^(10) L mol^(-1)s^(-1))((1.0 xx 10^(-14) mol^(2) L^(-2))/(5.6 xx 10^(-10) mol L^(-1))) = 6.07 xx 10^(-5)s^(-1)`