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Derive ionic product of water. Also find...

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

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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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