Derive ionic product of water. Also find its value at 25^(@)C.

1.	Derive ionic product of water. Also find its value at 25^(@)C.
Answer» Solution :When HIGH current is passed through water, it UNDERGOES partial dissocitaion. `H_(2)OhArrH^(+)+OH^(-)` Applying law of mass ACTION, `K=([H^(+)][OH^(-)])/([H_(2)O]),K[H_(2)O]=[H^(+)][OH^(-)]` But `K[H_(2)O]=Kw""thereforeKw=[H^(+)][OH^(-)]` Where, `K_(w)` is the ionic product of water. Value of `K_(w)" at "25^(@)C` : It is found at `25^(@)C[H^(+)]=[OH^(-)]=10^(-7)"mol"//dm^(3)` `therefore" "Kw=[H^(+)][OH^(-)]=10^(-7)xx10^(-7)=10^(-14)("mol"//dm^(3))^(2)` At `25^(@)C` the value of Kw is `10^(-4)("mol"//dm^(3))^(2)`

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