Water is a weak electrolyte so its ionisation is not complete and undissociated water is in equilibrium with its ions, `H_3O^(+)` and `OH^(-)` . Since water gives `H_3O^(+)` and `OH^(-)` both, it is called amphiprotic . Let us take pure water of density 1 g/ cc. Its concentration would be 1000/18 = 55.56 molar. Let a be the degree of ionization of water, then
`tt((2H_2O, ⇋ ,H_3O^(+), + OH^(-)),(c(1-alpha), quad , c alpha , calpha))`
`[H_2O] = c =1000/18 = 55.56 M`
`K_(eq) = ([H_3O^(+)][OH^(-)])/([H_2O]^2)`
where `K_(eq)` is ionisation (dissociation) constant of water.
Since water is present in large amount, there is no appreciable change in the concentration of water. So concentration of water is taken as constant and it is merged with the `K_(eq)` to give `K_w`.
`K_(eq) [H_2O]^2 = [H_3O^(+)][OH^(-)], K_w = [H_3O^(+)][OH^(-)] = (calpha)^2`
`K_w` is called ionic product of water. The value of `K_w` at `25^oC` is `1 xx 10^(-14)` and it is temperature dependent.
`1 xx 10^(-14) = [H_3O^(+)][OH^(-)]`
`1 xx 10^(-14) = [H_3O^(+)]^2 = [OH]^2` (as water on dissociation give equal number of `H_3O^(+)` and `OH^(-)` ions)
`[H_3O^(+)] = [OH]^(-) = sqrt(1 xx 10^(-14)) = 1 xx 10^(-7) M`, so water at `25^oC` has `10^(-7) M` of `H3O^(+)` and `10^(-7) M` of `OH^(-)` ions. As the number of `H_3O^(+)` and `OH^(-)` are same, so water is neutral.
The product of `H_3O^(+)` and `OH^(-)` concentration for any solution (acidic, basic or neutral) at any temperature is a constant, called ionic product (auto protolysis constant) of water.
For any acidic solution, `[H_3O^(+)] > [OH^(-)]`
or `[H_3O^(+)] > 10^(-7) M`
For any alkaline solution, `[H_3O^(+)] < [OH^(-)]`
or `[OH^(-)] >10^(-7) M.`
`=>` `K_w` is not same as the dissociation constant of water.
`=>` `K_w` is a constant irrespective of the presence of other ions in the solution.
`=>` `K_w` increases with temperature.
Temperature `(text()^oC) 0 quad 10 quad 20 quad 25 quad 30 quad 40 quad 50`
`K_w ( 10^(-14 )) quad 0.1 quad 0.2 quad 0.6 quad 1.0 quad 1.4 quad 2.9 quad 5.4`
Water is a weak electrolyte so its ionisation is not complete and undissociated water is in equilibrium with its ions, `H_3O^(+)` and `OH^(-)` . Since water gives `H_3O^(+)` and `OH^(-)` both, it is called amphiprotic . Let us take pure water of density 1 g/ cc. Its concentration would be 1000/18 = 55.56 molar. Let a be the degree of ionization of water, then
`tt((2H_2O, ⇋ ,H_3O^(+), + OH^(-)),(c(1-alpha), quad , c alpha , calpha))`
`[H_2O] = c =1000/18 = 55.56 M`
`K_(eq) = ([H_3O^(+)][OH^(-)])/([H_2O]^2)`
where `K_(eq)` is ionisation (dissociation) constant of water.
Since water is present in large amount, there is no appreciable change in the concentration of water. So concentration of water is taken as constant and it is merged with the `K_(eq)` to give `K_w`.
`K_(eq) [H_2O]^2 = [H_3O^(+)][OH^(-)], K_w = [H_3O^(+)][OH^(-)] = (calpha)^2`
`K_w` is called ionic product of water. The value of `K_w` at `25^oC` is `1 xx 10^(-14)` and it is temperature dependent.
`1 xx 10^(-14) = [H_3O^(+)][OH^(-)]`
`1 xx 10^(-14) = [H_3O^(+)]^2 = [OH]^2` (as water on dissociation give equal number of `H_3O^(+)` and `OH^(-)` ions)
`[H_3O^(+)] = [OH]^(-) = sqrt(1 xx 10^(-14)) = 1 xx 10^(-7) M`, so water at `25^oC` has `10^(-7) M` of `H3O^(+)` and `10^(-7) M` of `OH^(-)` ions. As the number of `H_3O^(+)` and `OH^(-)` are same, so water is neutral.
The product of `H_3O^(+)` and `OH^(-)` concentration for any solution (acidic, basic or neutral) at any temperature is a constant, called ionic product (auto protolysis constant) of water.
For any acidic solution, `[H_3O^(+)] > [OH^(-)]`
or `[H_3O^(+)] > 10^(-7) M`
For any alkaline solution, `[H_3O^(+)] < [OH^(-)]`
or `[OH^(-)] >10^(-7) M.`
`=>` `K_w` is not same as the dissociation constant of water.
`=>` `K_w` is a constant irrespective of the presence of other ions in the solution.
`=>` `K_w` increases with temperature.
Temperature `(text()^oC) 0 quad 10 quad 20 quad 25 quad 30 quad 40 quad 50`
`K_w ( 10^(-14 )) quad 0.1 quad 0.2 quad 0.6 quad 1.0 quad 1.4 quad 2.9 quad 5.4`