1.

The ionization constant of `NH_(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» `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)`


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