The ionization constant of `NH_(4) ^(+)` in water is `5.6 xx10^(-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 xx10^(10)L mol^(-1)s^(-1)` calculate the rate constant for proton transfer from water to `NH_(3).`

Options

A. `8.23 xx 10 ^(5)`
B. `6.07 xx 10 ^(5)`
C. `12.14 xx 1 ^(4)`
D. `10.3 xx 10 ^(4)`

Answer

B

Explanation

`NH _(4) ^(+) hArr NH _(3) + OH ^(-) K _(1) = 5.6 xx 10 ^(-10)`
`H _(2) O hArr H ^(+) OH ^(-) K _(2) = 1 xx 10 ^(-14)`
From the above reactions
`NH_(4) ^(-) + OH ^(-) overset( K _(1)) underset( K _(2)) hArr + NH _(3) + H _(2) O K = ( K _(1))/( K_(2)) , K = ( 5.6 xx 10 ^(-10))/( 1 xx 10 ^(-14)) = 5.6 xx 10 ^(4)`
Further, at equilibrium, `K = ( K_(1))/( K_(2)) , K _(2) = (K_(1))/(K ) = ( 2. 4 xx 10 ^(10))/(5.6 xx 10 ^(4)) = 6. 07 xx 10 ^(5)`