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The pH of pure water at `25^(@)C and35^(@)C` are 7 and 6 respectively. Calculate the heat of dissociation of `H_(2)O` into `H^(+) and OH^(-)` ions. |
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Answer» 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)`. |
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