The ionisation constant of `NH_(4)^+` in water is `5.6 xx 10^(-10)" at "25^@C`. The rate constant for the rection of `NH_(4)^+ and OH^-` to form `NH_(3) and H_(2)O" at "25^@C 3.4 xx 10^(10)" L mol"^(-1) s^(-1)`. Calculate the rate constant for proton from water to `NH_(3)`.

Explanation

`NH_(4)^(+) + H_(2)O hArr NH_(4)OH + H^(+) , K_(a) = 5.6xx10^(-10)`
`NH_(3) + H_(2)O underset(k_(b))overset(k_(f))hArr NH_(4)^(+) + OH^(-) , k_(b) = 3.4xx10^(10)`
Ai,=m, To find `k_(f)`
we know that for a conjugate acid-base pair
`K_("acid")xxK_("base") = K_(w), i.e., K_(a)xxK_(b)=K_(w)`
`:. K_(b) = (k_(w))/(K_(a))=(10^(-14))/(5.6xx10^(-10))`
But `K_(b) = (k_(f))/(k_(b))`
`k_(f) = K_(b) xx k_(b)=(10^(-14))/(5.6xx10^(-10))xx 3.4xx10^(10)=0.607xx10^(6)=6.07xx10^(5)`