Ionic product of water `(K_(w)` is `10^(-14))` at `25^(@)C`. What is the dissociation constant of water and auto protonation constatn of water?

Answer

A, B, C, D

Explanation

`H_(2)O hArr H^(o+) + OH^(Theta)` (molarity of water `= 55.6M)`
`:. K_(d) = ([H^(o+)][overset(Theta)OH])/([H_(2)O]) = (K_(w))/([H_(2)O]) = (10^(-14))/(55.6) = 1.8 xx 10^(-16)`
Also `H_(2)O + H_(2)O hArr H_(3)O^(o+) + OH^(Theta)`
`K_(a.p) = ([H_(3)O^(o+)][OH^(o+)])/([H_(2)O]^(2)) = (K_(w))/([H_(2)O]^(2)) = (10^(-14))/((55.6)^(2)) = 3.24 xx 10^(-16)`