The ionisation constant of `NH_4^+` in water is `5.6xx10^(-10)` at `25^@C` the rate constant for the reaction of `NH_4^+` and `OH^-` to form `NH_3` and `H_2O` at `25^@C` IS `3.4xx10^10`L mol/sec Find the rate constant for proton transfer from water to `NH_3`?

Explanation

`NH_(3) + H_(2) O to NH_(4)^(+) + OH, K_(b) = 3.4 xx 10^(-10)"L mol"^(-1)" s"^(-1)`
`NH_(4)^(+) + H_(2) O to NH_(4) OH + H^(+) , K_(a) = 5.6 xx 10^(-10) `
`K_(NH_3) . (K_("base")) = (K_f)/( K_b)`
`(K_w)/( K_("acid" ) ) = (K_f)/( K_b)" "["As" K_("acid") xx K_("base") = K_(w) ]`
`(10^(-14) )/( 5.6 xx 10^(-10) ) = (K_f)/( 3.4 xx 10^(-10) ) therefore K_f = 6.07 xx 10^(-15)" s"^(-1)`