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

Explanation

Since water is neutral , `[H_(3)O^(+)] =[OH^(-)]`
we know that , `[H_(3)O^(+)] xx [OH^(-)] =K_(w) =2.7 xx 10^(-14)`
`[H_(3)O^(+)]^(2) =2.7 xx 10^(-14)`
`[H_(3)O^(+)] =(2.7 xx 10^(-14))^(1//2) =1.643 xx 10^(-7)`
`pH " of water " =- log [H_(3)O^(+)] =- log (1.643 xx 10^(-7))`
`=(-) (-7) + log (1.643) =(7 - log 1.643) =7 -0.2156 =6.78`