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`.

Answer

`52.90`

Explanation

The dissociation reaction is
`H_(2)OhArrH^(+)+OH^(-)`
At `25^(@)C`, pH=7means `[H^(+)]=10^(-7)M :. K_(W)=10^(-14)`
At `35^(@)C`, pH = 6means `[H^(+)]=10^(-6)M :. K_(W)=10^(-12)`
As equilibrium constants 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)))=(DeltaH)/(2.303R)((1)/(T_(1))-(1)/(T_(2)))`
`:.log.(10^(-12))/(10^(-14))=(DeltaH)/(2.303xx8.314JK^(-1)mol^(-1)) ((1)/(298K)-(1)/(308K))`
or `DeltaH=52898J mol^(-1)=52.898kJmol^(-1)`.