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

At `25^(@)C: [H^(o+)] = 10^(-7), :. K_(w) = 10^(-14)`
At `35^(@)C: [H^(o+)] = 10^(-6), :. K_(w) = 10^(-12)`
Now using
`2.303 log (K_(w_(2)))/(K_(w_(1))) = (DeltaH)/(R) = [(T_(2)-T_(1))/(T_(1)T_(2))]`
`2.303 log (10^(-12))/(10^(-14)) = (DeltaH)/(2cal) [(10)/(298xx308)]`
`DeltaH = 84551.4 cal mol^(-1) = 84.551 kcal mol^(-1)`
Thus, `H_(2) O hArr H^(oplus) + overset(Theta)(O)H , Delta H = 84.551` kcal `"mol"^(-1)`
`:. H^(oplus) + overset(Theta)(O)H hArr H_(2)O , Delta H = -84.551` kcal `"mol"^(-1)`.