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^(-14)`
`therefore [H^(+)] [OH^(-)]=10^(-14) (? K_(W)=[H^(+)] [OH^(-)])`
`therefore [H^(+)]=10^(-7)M therefore pH=7?`
Now, `K_(W)" for "H_(2)O at 60^(@)C=9.62 xx 10^(-14)`
`therefore [H^(+)] [OH^(-)]=9.62 xx 10^(-14)`
For pure water `[H^(+)=[OH^(+)]`
`therefore [H^(+)]^(2)=9.62 xx 10^(-14)`
`therefore [H^(+)]=sqrt((9.92 xx 10^(-14))=3.10 xx 10^(-7)M`
`therefore pH=-LOG H^(+)=-log 3.10 xx 10^(7)`
`therefore pH=6.51`
Thus, pH of water becomes 6.51 at `60^@` but the nature is neutral since calculation for pure water has been made, i.e., pH scale at `60^@C` becomes in be- tween 0 to 13.02.