Ionic product of water at `310 K` is `2.7xx10^(-14)`. What is the `pH` of netural water at this temperature?

Explanation

`H_(2)O hArr H^(+)+OH^(-)`
`.: [H^(+)][OH^(-)]= K_(w)` or `[H^(+)]^(2)= K_(w)`
`:. [H^(+)]=[OH^(-)]`
`:. [H^(+)]= sqrt((2.7xx10^(-14)))`
`= 1.64xx10^(-7))`
`:. pH = 6.785`