2H_(2)O longrightarrow H_(3)O^(+)+OH,K_(w)=1times10^(-14)at25^(@)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

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

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 ?

The standard potential of the reaction H_(2)O + e^(-) rightarrow (1/2)H_(2) + OH^(-) at 298 K by using k_(w) (H_(2)O) = 10^(-14) , is: