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. `8.23 xx 10^(5)`
B. `6.07 xx 10^(5)`
C. `12.14 xx 10^(4)`
D. `10.3 xx 10^(4)`

Answer

B

Explanation

`NH_(4) hArr NH_(3) + HI^(+) K_(1) = 56 xx 10^(10)`
`H_(2)O hArr H^(+) + OH^(-) " "K_(2) = 1 xx 10^(14)`
From the above reactions
`N_(4) + OH^(-) underset(K_(2))overset(K_(1)) NH_(3) + H_(2)O " "K = (K_(1))/(K_(2)) , K = (56 xx 10^(-10))/(1 xx 10^(14)) = 56 xx 10^(4)`
Further at equilibrium , `K = (K_(1))/(K_(2)), K_(2) = (K_(1))/(K) = (34 xx 10^(10))/(56 xx 10^(4)) = 6.07 xx 10^(5)`