Theionization constantofNH_(4)^(+) in water is 5.6 xx 10^(-10)at 25^(@)C. The rate constant for reaction of NH_(4)^(+) and OH^(-)to form NH_(3) and H_(2)O "at"25^(@)C "is " 3.4xx10^(10)litre"mol"^(-1) "sec^(-1)". Calculate the rate constant for proton transfer from proton transfer from water to NH_(3).

Theionization constantofNH_(4)^(+) in water is 5.6 xx 10^(-10)at 25^(@

1.	Theionization constantofNH_(4)^(+) in water is 5.6 xx 10^(-10)at 25^(@)C. The rate constant for reaction of NH_(4)^(+) and OH^(-)to form NH_(3) and H_(2)O "at"25^(@)C "is " 3.4xx10^(10)litre"mol"^(-1) "sec^(-1)". Calculate the rate constant for proton transfer from proton transfer from water to NH_(3).
Answer» Solution :`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)`