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 ?

Explanation

`K_(w)` for `H_(2)O` at `25^(@)C=10^(-4)`
`:. [H^(+)][OH^(-)]= 10^(-14)(.: K_(w)= [H^(+)][OH^(-)])`
`:. [H^(+)]= 10^(-7)M`
`:. pH= 7`
Now `K_(w)` for `H_(2)O` at `60^(@)C= 9.62xx10^(-14)`
`:. [H^(+)][OH^(-)]= 9.62xx10^(-14)`
For pure water `[H^(+)]=[OH^(-)]`
`:. [H^(+)]^(2)= 9.62xx10^(-14)`
`:. [H^(+)]= sqrt((9.62xx10^(-14)))`
`=3.10xx10^(-7)M`
`:. pH= - Iog H^(+)= - Iog 3.10xx10^(-7)`
`:. pH= 6.51`
Thus, pH of water becomes `6.51` at `60^(@)C` but the nature is neutral since calulation for pure water has been made, i.e., pH scale at `60^(@)C` becomes in between 0 to 13.02`.