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The K(w) for 2H(2)O hArr H(3)O^(+)OH^(-)...

The `K_(w)` for `2H_(2)O hArr H_(3)O^(+)_OH^(-)` changes from `10^(-14)` at `25^(@)C` to `9.62xx10^(-14)` at `60^(@)C`. What is pH of water at `60^(@)C` ? What happens to its neutrality ?

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2H_2 O hArr H_(3)O^(+) + OH^(-) The ionic product changes from 10^(-14) at 25^@ C to 9.62 xx 10^(-14) at 60^@ C. What is the pOH of water at 60^@ C ?

2H_(2)OhArrH_(3)O^(+)+OH^(-) K_(w)=1xx10^(-14) at 25^(@)C . Hence, K_(a) is:

2H_(2)O hArr H_(3)O^(o+) + overset(Theta)OH,K_(w) = 10^(-14) at 25^(@)C , hence K_(a) is

2H_(2)O hArr H_(3)O^(o+) + overset(Theta)OH,K_(w) = 10^(-14) at 25^(@)C , hence K_(a) is

H_(2)O+H_(2)OhArrH_(3)O^(+)+OH^(-) The ionic product of water is 1xx10^(-14) at 25^(@)C and 3.0xx10^(-14) at 40^(@)C is the above process endothermic or exothermic?

K_w for water at 25^(@)C is equal to 10^(-14) . What is its value at 90^@C -

The ionic product of warter at 60^@C is 9.61 xx 10^-14 . The pH of water at 60^@C is :