The `pH` of pure water at `25^(@)C` and `35^(@)C` are `7` and `6`, respectively. Calculate the heat of formation of water from `H^(o+)` and `overset(Θ)OH`.

Explanation

The dissociation reaction is : `H_(2)O hArr H^(+)+OH^(-)`
At `25^(@)C`, pH = 7 means `[H^(+)]=10^(-7)M :. K_(w)=10^(-14)`
At `35^(@)C`, pH =6 means `[H^(+)]=10^(-6)M :. K_(w)=10^(-12)`
As equilibrium constant for the dissociation of `H_(2)O` are in the same ratio as ionic products of water, we can apply the relation
`log.(K_(w_(2)))/(K_(w_(1)))=(Delta H)/(2.303 R)((1)/(T_(1))-(1)/(T_(2))):. log. (10^(-12))/(10^(-14))=(Delta H)/(2.303xx8.314 JK^(-1)"mol"^(-1))((1)/(298K)-(1)/(3-8K))`
or `DeltaH=52898 J "mol"^(-1) = 52.898 kJ "mol"^(-1)`.