`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?

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 equilibrium equation and K_(a) values for the three acids are given at 25^(@)C : HA(aq)+H_(2)OhArrH_(3)O^(+)(aq)+A^(-)(aq),K_(a)=2xx10^(-5) HB(aq)+H_(2)OhArrH_(3)O^(+)(aq)+B^(-)(aq),K_(a)=4xx10^(-6) HC(aq)+H_(2)OhArrH_(3)O^(+)(aq)+C^(-)(aq),K_(a)=1xx10^(-4) The pH of 0.1M aqueous NaC solution is : (NaC is sodium salt of acid HC)

From equation (i) and (ii), CohArrCO+(1)/(2)O_(2)[K_(c1)=9.0xx10^(-12) at 1000^(@)C] (i) H_(2)OhArrH_(2)+(1)/(2)O_(2)[K_(c2)=7.0xx10^(-12) at 1000^(@)C] (ii) the equilibrium for the reaction CO_(2)+H_(2)hArrCO+H_(2)O at the same temperature is

The equilibrium equation and K_(a) values for the three acids are given at 25^(@)C : HA(aq)+H_(2)OhArrH_(3)O^(+)(aq)+A^(-)(aq),K_(a)=2xx10^(-5) HB(aq)+H_(2)OhArrH_(3)O^(+)(aq)+B^(-)(aq),K_(a)=4xx10^(-6) HC(aq)+H_(2)OhArrH_(3)O^(+)(aq)+C^(-)(aq),K_(a)=1xx10^(-4) The pH of 0.2M aqueous HA solution is :